Formelsammlung Tensoralgebra

Diese Formelsammlung fasst Formeln und Definitionen der Tensoralgebra für Tensoren zweiter Stufe in der Kontinuumsmechanik zusammen. Es wird der dreidimensionale Raum zugrunde gelegt.

• Operatoren wie ${\displaystyle \mathrm {I} _{1}}$ werden nicht kursiv geschrieben.
• Buchstaben die als Indizes benutzt werden:
  • ${\displaystyle i,j,k,l,m,n\in \{1,2,3\}}$.
    Ausnahme:
    Die imaginäre Einheit ${\displaystyle \mathrm {i} ^{2}=-1}$ und die #Vektorinvariante ${\displaystyle {\vec {\mathrm {i} }}}$ werden in Abgrenzung zu den Indizes nicht kursiv geschrieben.
  • ${\displaystyle p,q,r,s\in \{1,2,\ldots ,9\}}$
  • ${\displaystyle u,v\in \{1,2,\ldots ,6\}}$
• Alle anderen Buchstaben stehen für reelle Zahlen oder komplexe Zahlen.
• Vektoren:
  • Alle hier verwendeten Vektoren sind geometrische Vektoren im dreidimensionalen euklidischen Vektorraum ${\displaystyle \mathbb {V} }$.
  • Vektoren werden mit Kleinbuchstaben bezeichnet.
    Ausnahme #Dualer axialer Vektor ${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {A} }}}}$
  • Einheitsvektoren mit Länge eins werden wie in ê mit einem Hut versehen. Die Standardbasis von ${\displaystyle \mathbb {V} }$ ist ê_1,2,3.
  • Vektoren mit unbestimmter Länge werden wie in ${\displaystyle {\vec {a}}}$ mit einem Pfeil versehen.
  • Dreiergruppen von Vektoren wie in ${\displaystyle {\vec {h}}_{1},{\vec {h}}_{2},{\vec {h}}_{3}}$ oder ${\displaystyle {\vec {g}}^{1},{\vec {g}}^{2},{\vec {g}}^{3}}$ bezeichnen eine rechtshändige Basis von ${\displaystyle \mathbb {V} }$.
  • Gleichnamige Basisvektoren mit unterem und oberem Index sind dual zueinander, z. B. ${\displaystyle {\vec {g}}_{1},{\vec {g}}_{2},{\vec {g}}_{3}}$ ist dual zu ${\displaystyle {\vec {g}}^{1},{\vec {g}}^{2},{\vec {g}}^{3}}$.
• Tensoren zweiter Stufe werden wie in A mit fetten Großbuchstaben notiert. Die Menge aller Tensoren wird mit ${\displaystyle {\mathcal {L}}:=\mathrm {Lin} (\mathbb {V} ,\mathbb {V} )}$ bezeichnet. Tensoren höherer Stufe werden mit einer hochgestellten Zahl wie in ${\displaystyle {\stackrel {4}{\mathbf {C} }}}$ geschrieben. Tensoren vierter Stufe sind Elemente der Menge ${\displaystyle {\stackrel {4}{\mathcal {L}}}:=\mathrm {Lin} ({\mathcal {L}},{\mathcal {L}})}$.
• Es gilt die Einstein'sche Summenkonvention ohne Beachtung der Indexstellung.
  • Kommt in einer Formel in einem Produkt ein Index doppelt vor wie in ${\displaystyle c=a_{i}b^{i}}$ wird über diesen Index summiert:
    ${\displaystyle c=a_{i}b^{i}=\sum _{i=1}^{3}a_{i}b^{i}}$.
  • Kommen mehrere Indizes doppelt vor wie in ${\displaystyle c=A_{pq}B_{q}^{p}}$ wird über diese summiert:
    ${\displaystyle c=A_{pq}B_{q}^{p}=\sum _{p=1}^{9}\sum _{q=1}^{9}A_{pq}B_{q}^{p}}$.
  • Ein Index, der nur einfach vorkommt wie ${\displaystyle u}$ in ${\displaystyle a_{u}=A_{uv}b_{v}}$, ist ein freier Index. Die Formel gilt dann für alle Werte der freien Indizes:
    ${\displaystyle a_{u}=A_{uv}b_{v}\quad \leftrightarrow \quad a_{u}=\sum _{v=1}^{6}A_{uv}b_{v}\quad \forall \;u\in \{1,\ldots ,6\}}$.

Formelzeichen	Abschnitt in der Formelsammlung	Wikipedia-Artikel
${\displaystyle \mathbf {I,1} }$	#Einheitstensor	Einheitstensor
${\displaystyle \mathbf {Q,R} }$	#Orthogonale Tensoren	Orthogonaler Tensor
${\displaystyle \lambda }$	#Eigenwerte	Eigenwertproblem
${\displaystyle \delta _{ij}}$, δij	#Kronecker-Delta	Kronecker-Delta
${\displaystyle \epsilon _{ijk}}$	#Permutationssymbol	Permutationssymbol
${\displaystyle {\stackrel {3}{\mathbf {E} }}}$	#Fundamentaltensor 3. Stufe	Epsilon-Tensor
${\displaystyle [{\vec {a}}]_{\times }}$	#Axialer Tensor oder Kreuzproduktmatrix	Kreuzprodukt
${\displaystyle \left[\mathbf {T} \right],\left[{\stackrel {4}{\mathbf {T} }}\right]}$	#Voigt-Notation symmetrischer Tensoren zweiter Stufe, #Voigt’sche Notation von Tensoren vierter Stufe	Voigtsche Notation
${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {A} }}},\mathbf {A} _{\times }}$	#Dualer axialer Vektor	Kreuzprodukt
${\displaystyle {\vec {\mathrm {i} }}}$	#Vektorinvariante	Vektorinvariante
${\displaystyle \mathrm {i} }$		Imaginäre Einheit

Formelzeichen	Abschnitt in der Formelsammlung	Wikipedia-Artikel
${\displaystyle (\cdot )\cdot (\cdot )}$	Skalarprodukt von Vektoren, #Vektortransformation, #Tensorprodukt	Skalarprodukt
${\displaystyle (\cdot )\times (\cdot )}$	#Kreuzprodukt eines Vektors mit einem Tensor, #Kreuzprodukt von Tensoren	Kreuzprodukt
${\displaystyle (\cdot ):(\cdot )}$	#Skalarprodukt von Tensoren	Frobenius-Skalarprodukt
${\displaystyle (\cdot )\otimes (\cdot )}$	#Dyadisches Produkt	Dyadisches Produkt
${\displaystyle (\cdot )\cdot \!\!\times (\cdot )}$	#Skalarkreuzprodukt von Tensoren
${\displaystyle (\cdot )\times \!\!\times (\cdot )}$	#Doppeltes Kreuzprodukt von Tensoren
${\displaystyle (\cdot )\#(\cdot )}$	#Äußeres Tensorprodukt	Äußeres Tensorprodukt
${\displaystyle \parallel (\cdot )\parallel }$	#Betrag	Frobeniusnorm
${\displaystyle |x|,|{\vec {v}}|,|\mathbf {A} |}$	Betrag der Zahl x oder des Vektors ${\displaystyle {\vec {v}}}$, #Determinante des Tensors A	Determinante

Formelzeichen	Abschnitt in der Formelsammlung	Wikipedia-Artikel
${\displaystyle \mathrm {Sp,tr,I} _{1}}$	#Spur	Spur (Mathematik), Hauptinvariante
${\displaystyle \mathrm {I} _{2}}$	#Zweite Hauptinvariante	Hauptinvariante
${\displaystyle \mathrm {det,I} _{3},|\mathbf {A} |}$	#Determinante	Determinante, Hauptinvariante
sym	#Symmetrischer Anteil	Symmetrische Matrix
skw, skew	#Schiefsymmetrischer Anteil	Schiefsymmetrische Matrix
adj	#Adjunkte	Adjunkte
cof	#Kofaktor	Minor (Mathematik)#Kofaktormatrix
dev	#Deviator	Deviator, Spannungsdeviator
sph	#Kugelanteil	Kugeltensor

Formelzeichen	Abschnitt in der Formelsammlung	Wikipedia-Artikel
${\displaystyle (\cdot )_{ij},(\cdot )^{ij},(\cdot )_{j}^{i}}$	#Tensorkomponenten
${\displaystyle (\cdot )^{\top }}$	#Transposition	Transponierte Matrix
${\displaystyle (\cdot )^{\stackrel {mn}{\top }}}$	Transpositionen von Tensoren vierter Stufe
${\displaystyle (\cdot )^{-1}}$	#Inverse	Inverse Matrix
${\displaystyle (\cdot )^{-\top },(\cdot )^{\top -1}}$	#Transposition der #Inverse
${\displaystyle (\cdot )^{\mathrm {S} }}$	#Symmetrischer Anteil	Symmetrische Matrix
${\displaystyle (\cdot )^{\mathrm {A} }}$	#Schiefsymmetrischer Anteil	Schiefsymmetrische Matrix
${\displaystyle (\cdot )^{\mathrm {D} }}$	#Deviator	Deviator, Spannungsdeviator
${\displaystyle (\cdot )^{\mathrm {K} }}$	#Kugelanteil	Kugeltensor
${\displaystyle {\stackrel {n}{(\cdot )}}}$	Tensor n-ter Stufe
${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {A} }}},\mathbf {A} _{\times }}$	#Dualer axialer Vektor	Kreuzprodukt

Formelzeichen	Elemente
${\displaystyle \mathbb {R} }$, ℝ	Reelle Zahlen
${\displaystyle \mathbb {C} }$, ℂ	Komplexe Zahlen
${\displaystyle \mathbb {V} }$,𝕍	Vektoren
${\displaystyle {\mathcal {L}}}$, 𝓛 = Lin(𝕍,𝕍)	Tensoren zweiter Stufe
${\displaystyle {\stackrel {4}{\mathcal {L}}}=\mathrm {Lin} ({\mathcal {L,L}})}$	#Tensoren vierter Stufe
SO(3)	Drehgruppe, eigentlich #Orthogonale Tensoren

δ_ij = ${\displaystyle \delta _{ij}=\delta ^{ij}=\delta _{i}^{j}=\delta _{j}^{i}={\begin{cases}1&\mathrm {falls} \quad i=j\\0&\mathrm {sonst} \end{cases}}}$

Für Summen gilt dann z. B.

v_iδ_ij = v_j

A_ijδ_ij = A_ii

Dies gilt für die anderen Indexgruppen entsprechend.

${\displaystyle \epsilon _{ijk}={\hat {e}}_{i}\cdot ({\hat {e}}_{j}\times {\hat {e}}_{k})={\begin{cases}1&{\text{falls}}\;(i,j,k)\in \{(1,2,3),(2,3,1),(3,1,2)\}\\-1&{\text{falls}}\;(i,j,k)\in \{(1,3,2),(2,1,3),(3,2,1)\}\\0&{\text{sonst, d.h. bei doppeltem Index}}\end{cases}}}$

${\displaystyle \epsilon _{ijk}\epsilon _{lmn}={\begin{vmatrix}\delta _{il}&\delta _{jl}&\delta _{kl}\\\delta _{im}&\delta _{jm}&\delta _{km}\\\delta _{in}&\delta _{jn}&\delta _{kn}\end{vmatrix}}}$

${\displaystyle \epsilon _{ijk}\epsilon _{klm}=\delta _{il}\delta _{jm}-\delta _{im}\delta _{jl}}$

${\displaystyle \epsilon _{ijk}\epsilon _{jkl}=2\delta _{il}}$

${\displaystyle \epsilon _{ijk}\epsilon _{ijk}=6}$

Kreuzprodukt:

${\displaystyle a_{i}{\hat {e}}_{i}\times b_{j}{\hat {e}}_{j}=\epsilon _{ijk}a_{i}b_{j}{\hat {e}}_{k}=\epsilon _{ijk}a_{j}b_{k}{\hat {e}}_{i}=\epsilon _{ijk}a_{k}b_{i}{\hat {e}}_{j}}$

${\displaystyle \epsilon _{ijk}{\hat {e}}_{k}={\hat {e}}_{i}\times {\hat {e}}_{j}}$

Die hier verwendeten Vektoren sind Spaltenvektoren

${\displaystyle {\vec {a}}=a_{i}{\hat {e}}_{i}={\begin{pmatrix}a_{1}\\a_{2}\\a_{3}\end{pmatrix}}}$

Drei Vektoren ${\displaystyle {\vec {a}},{\vec {b}},{\vec {c}}}$ können spaltenweise in einer 3×3-Matrix ${\displaystyle M}$ arrangiert werden:

${\displaystyle M={\begin{pmatrix}{\vec {a}}&{\vec {b}}&{\vec {c}}\end{pmatrix}}={\begin{pmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{pmatrix}}}$

Die Determinante der Matrix

${\displaystyle |M|={\begin{vmatrix}{\vec {a}}&{\vec {b}}&{\vec {c}}\end{vmatrix}}}$

ist

• ungleich null, wenn die Spaltenvektoren linear unabhängig sind und
• größer null, wenn die Spaltenvektoren zusätzlich ein Rechtssystem bilden.

Also gewährleistet ${\displaystyle {\begin{vmatrix}{\vec {a}}&{\vec {b}}&{\vec {c}}\end{vmatrix}}>0}$, dass die Vektoren ${\displaystyle {\vec {a}},{\vec {b}},{\vec {c}}}$ eine rechtshändige Basis bilden.

Die Spaltenvektoren bilden eine Orthonormalbasis, wenn

${\displaystyle M^{\top }M={\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}}$

worin ${\displaystyle M^{\top }}$ die transponierte Matrix ist. Bei der hier vorausgesetzten Rechtshändigkeit gilt dann zusätzlich ${\displaystyle |M|=+1}$.

Basisvektoren ${\displaystyle {\vec {g}}_{1},{\vec {g}}_{2},{\vec {g}}_{3}}$

Duale Basisvektoren ${\displaystyle {\vec {g}}^{1},{\vec {g}}^{2},{\vec {g}}^{3}}$

Beziehungen zwischen den Basisvektoren

${\displaystyle {\vec {g}}_{i}\cdot {\vec {g}}^{j}=\delta _{i}^{j}}$

${\displaystyle {\vec {g}}^{1}={\frac {{\vec {g}}_{2}\times {\vec {g}}_{3}}{({\vec {g}}_{1},{\vec {g}}_{2},{\vec {g}}_{3})}},\quad g^{2}={\frac {{\vec {g}}_{3}\times {\vec {g}}_{1}}{({\vec {g}}_{1},{\vec {g}}_{2},{\vec {g}}_{3})}},\quad g^{3}={\frac {{\vec {g}}_{1}\times {\vec {g}}_{2}}{({\vec {g}}_{1},{\vec {g}}_{2},{\vec {g}}_{3})}}}$

${\displaystyle {\vec {g}}_{1}={\frac {{\vec {g}}^{2}\times {\vec {g}}^{3}}{({\vec {g}}^{1},{\vec {g}}^{2},{\vec {g}}^{3})}},\quad g_{2}={\frac {{\vec {g}}^{3}\times {\vec {g}}^{1}}{({\vec {g}}^{1},{\vec {g}}^{2},{\vec {g}}^{3})}},\quad g_{3}={\frac {{\vec {g}}^{1}\times {\vec {g}}^{2}}{({\vec {g}}^{1},{\vec {g}}^{2},{\vec {g}}^{3})}}}$

mit dem Spatprodukt

${\displaystyle ({\vec {a}},{\vec {b}},{\vec {c}}):={\vec {a}}\cdot ({\vec {b}}\times {\vec {c}})={\vec {c}}\cdot ({\vec {a}}\times {\vec {b}})={\vec {b}}\cdot ({\vec {c}}\times {\vec {a}})={\begin{vmatrix}{\vec {a}}&{\vec {b}}&{\vec {c}}\end{vmatrix}}}$

Trägt man die Basisvektoren spaltenweise in eine Matrix ein, dann finden sich die dualen Basisvektoren in den Zeilen der Inversen oder den Spalten der #transponiert #Inversen ${\displaystyle ()^{\top -1}}$:

${\displaystyle {\begin{pmatrix}{\vec {g}}^{1}&{\vec {g}}^{2}&{\vec {g}}^{3}\end{pmatrix}}={\begin{pmatrix}{\vec {g}}_{1}&{\vec {g}}_{2}&{\vec {g}}_{3}\end{pmatrix}}^{\top -1}}$

In der Standardbasis wie in jeder Orthonormalbasis sind die Basisvektoren ${\displaystyle {\hat {e}}_{1},{\hat {e}}_{2},{\hat {e}}_{3}}$ zu sich selbst dual:

${\displaystyle {\hat {e}}_{i}={\hat {e}}^{i}}$

${\displaystyle {\vec {v}}=v_{i}{\hat {e}}_{i}\quad \rightarrow \;v_{i}={\vec {v}}\cdot {\hat {e}}_{i}}$

${\displaystyle {\vec {v}}=v^{i}{\vec {g}}_{i}\quad \rightarrow \;v^{i}={\vec {v}}\cdot {\vec {g}}^{i}}$

${\displaystyle {\vec {v}}=v_{i}{\vec {g}}^{i}\quad \rightarrow \;v_{i}={\vec {v}}\cdot {\vec {g}}_{i}}$

${\displaystyle ({\vec {g}}_{i}\cdot {\vec {g}}_{k})({\vec {g}}^{k}\cdot {\vec {g}}^{j})={\vec {g}}_{i}\cdot ({\vec {g}}^{j}\cdot {\vec {g}}^{k}){\vec {g}}_{k}={\vec {g}}_{i}\cdot {\vec {g}}^{j}=\delta _{i}^{j}}$

Wechsel von

Basis ${\displaystyle {\vec {g}}^{1},{\vec {g}}^{2},{\vec {g}}^{3}}$ mit dualer Basis ${\displaystyle {\vec {g}}_{1},{\vec {g}}_{2},{\vec {g}}_{3}}$

nach

Basis ${\displaystyle {\vec {h}}^{1},{\vec {h}}^{2},{\vec {h}}^{3}}$ mit dualer Basis ${\displaystyle {\vec {h}}_{1},{\vec {h}}_{2},{\vec {h}}_{3}}$:

${\displaystyle {\vec {v}}=v_{i}\,{\vec {g}}^{i}=v_{i}^{\ast }\,{\vec {h}}^{i}\quad \rightarrow \;v_{i}^{\ast }=({\vec {h}}_{i}\cdot {\vec {g}}^{j})v_{j}}$

Matrizengleichung:

${\displaystyle {\begin{aligned}{\begin{pmatrix}v_{1}^{\ast }\\v_{2}^{\ast }\\v_{3}^{\ast }\end{pmatrix}}=&{\begin{pmatrix}{\vec {h}}_{1}\cdot {\vec {g}}^{1}&{\vec {h}}_{1}\cdot {\vec {g}}^{2}&{\vec {h}}_{1}\cdot {\vec {g}}^{3}\\{\vec {h}}_{2}\cdot {\vec {g}}^{1}&{\vec {h}}_{2}\cdot {\vec {g}}^{2}&{\vec {h}}_{2}\cdot {\vec {g}}^{3}\\{\vec {h}}_{3}\cdot {\vec {g}}^{1}&{\vec {h}}_{3}\cdot {\vec {g}}^{2}&{\vec {h}}_{3}\cdot {\vec {g}}^{3}\end{pmatrix}}{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}\\=&{\begin{pmatrix}{\vec {h}}_{1}&{\vec {h}}_{2}&{\vec {h}}_{3}\end{pmatrix}}^{\top }{\begin{pmatrix}{\vec {g}}^{1}&{\vec {g}}^{2}&{\vec {g}}^{3}\end{pmatrix}}{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}\end{aligned}}}$

Die grundlegenden Eigenschaften des dyadischen Produkts „⊗“ sind:

Abbildung 𝕍×𝕍→𝓛

${\displaystyle {\vec {a}}\otimes {\vec {g}}=\mathbf {T} \in {\mathcal {L}}}$

Multiplikation mit einem Skalar:

${\displaystyle x({\vec {a}}\otimes {\vec {g}})=(x{\vec {a}})\otimes {\vec {g}}={\vec {a}}\otimes (x{\vec {g}})=x{\vec {a}}\otimes {\vec {g}}}$

Distributivität:

${\displaystyle (x+y){\vec {a}}\otimes {\vec {g}}=x{\vec {a}}\otimes {\vec {g}}+y{\vec {a}}\otimes {\vec {g}}}$

${\displaystyle ({\vec {a}}+{\vec {b}})\otimes {\vec {g}}={\vec {a}}\otimes {\vec {g}}+{\vec {b}}\otimes {\vec {g}}}$

${\displaystyle {\vec {a}}\otimes ({\vec {g}}+{\vec {h}})={\vec {a}}\otimes {\vec {g}}+{\vec {a}}\otimes {\vec {h}}}$

Skalarprodukt:

${\displaystyle ({\vec {a}}\otimes {\vec {g}}):({\vec {b}}\otimes {\vec {h}})=({\vec {a}}\cdot {\vec {b}})({\vec {g}}\cdot {\vec {h}})}$

Weitere Eigenschaften von Dyaden siehe #Dyade und den folgenden Abschnitt.

Durch die Eigenschaften des dyadischen Produktes wird ${\displaystyle {\mathcal {L}}}$ zu einem euklidischen Vektorraum und entsprechend kann jeder Tensor komponentenweise bezüglich einer Basis von ${\displaystyle {\mathcal {L}}}$ dargestellt werden:

${\displaystyle \mathbf {A} \in {\mathcal {L}}\rightarrow \mathbf {A} =A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}=A^{ij}{\vec {a}}_{i}\otimes {\vec {g}}_{j}}$ mit Komponenten ${\displaystyle A_{ij},A^{ij}\in \mathbb {R} }$.

Die Dyaden ${\displaystyle \{{\hat {e}}_{i}\otimes {\hat {e}}_{j}|i,j=1,2,3\}}$ und ${\displaystyle \{{\vec {a}}_{i}\otimes {\vec {g}}_{j}|i,j=1,2,3\}}$ bilden Basissysteme von ${\displaystyle {\mathcal {L}}}$.

Abbildung ${\displaystyle {\mathcal {L}}\to {\mathcal {L}}}$

${\displaystyle ({\vec {a}}\otimes {\vec {g}})^{\top }:={\vec {g}}\otimes {\vec {a}}}$

${\displaystyle (A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})^{\top }=A_{ij}({\hat {e}}_{j}\otimes {\hat {e}}_{i})=A_{ji}({\hat {e}}_{i}\otimes {\hat {e}}_{j})}$

${\displaystyle (A^{ij}{\vec {a}}_{i}\otimes {\vec {g}}_{j})^{\top }=A^{ij}({\vec {g}}_{j}\otimes {\vec {a}}_{i})=A^{ji}({\vec {g}}_{i}\otimes {\vec {a}}_{j})}$

${\displaystyle \left(\mathbf {A} ^{\top }\right)^{\top }=\mathbf {A} }$

${\displaystyle (\mathbf {A+B} )^{\top }=\mathbf {A} ^{\top }+\mathbf {B} ^{\top }}$

${\displaystyle (\mathbf {A\cdot B} )^{\top }=\mathbf {B} ^{\top }\cdot \mathbf {A} ^{\top }}$

Abbildung ${\displaystyle {\mathcal {L}}\times \mathbb {V} \to \mathbb {V} }$ oder ${\displaystyle \mathbb {V} \times {\mathcal {L}}\to \mathbb {V} }$

Dyaden:

${\displaystyle ({\vec {a}}\otimes {\vec {g}})\cdot {\vec {h}}:=({\vec {g}}\cdot {\vec {h}}){\vec {a}}}$

${\displaystyle {\vec {b}}\cdot ({\vec {a}}\otimes {\vec {g}}):=({\vec {a}}\cdot {\vec {b}}){\vec {g}}}$

${\displaystyle ({\vec {a}}\otimes {\vec {g}})\cdot {\vec {h}}={\vec {h}}\cdot ({\vec {a}}\otimes {\vec {g}})^{\top }}$

${\displaystyle {\vec {b}}\cdot ({\vec {a}}\otimes {\vec {g}})=({\vec {a}}\otimes {\vec {g}})^{\top }\cdot {\vec {b}}}$

Allgemeine Tensoren:

${\displaystyle A_{ij}({\hat {e}}_{i}\otimes {\hat {e}}_{j})\cdot {\vec {v}}=A_{ij}({\vec {v}}\cdot {\hat {e}}_{j}){\hat {e}}_{i}}$

${\displaystyle A^{ij}({\vec {a}}_{i}\otimes {\vec {g}}_{j})\cdot {\vec {v}}=A^{ij}({\vec {v}}\cdot {\vec {g}}_{j}){\vec {a}}_{i}}$

${\displaystyle {\vec {v}}\cdot A_{ij}({\hat {e}}_{i}\otimes {\hat {e}}_{j})=A_{ij}({\vec {v}}\cdot {\hat {e}}_{i}){\hat {e}}_{j}}$

${\displaystyle {\vec {v}}\cdot A^{ij}({\vec {a}}_{i}\otimes {\vec {g}}_{j})=A^{ij}({\vec {v}}\cdot {\hat {a}}_{i}){\vec {g}}_{j}}$

Symbolisch:

${\displaystyle \mathbf {A} \cdot {\vec {v}}={\vec {v}}\cdot \mathbf {A} ^{\top }}$

${\displaystyle {\vec {v}}\cdot \mathbf {A} =\mathbf {A} ^{\top }\cdot {\vec {v}}}$

Abbildung ${\displaystyle {\mathcal {L}}\times {\mathcal {L}}\to {\mathcal {L}}}$

${\displaystyle ({\vec {a}}\otimes {\vec {g}})\cdot ({\vec {h}}\otimes {\vec {u}}):=({\vec {g}}\cdot {\vec {h}}){\vec {a}}\otimes {\vec {u}}}$

${\displaystyle ({\vec {a}}\otimes {\vec {g}})\cdot \mathbf {A} ={\vec {a}}\otimes ({\vec {g}}\cdot \mathbf {A} )={\vec {a}}\otimes {\vec {g}}\cdot \mathbf {A} ={\vec {a}}\otimes (\mathbf {A} ^{\top }\cdot {\vec {g}})}$

${\displaystyle \mathbf {A} \cdot ({\vec {a}}\otimes {\vec {g}})=(\mathbf {A} \cdot {\vec {a}})\otimes {\vec {g}}=\mathbf {A} \cdot {\vec {a}}\otimes {\vec {g}}}$

${\displaystyle (A_{ik}{\hat {e}}_{i}\otimes {\hat {e}}_{k})\cdot (B_{lj}{\hat {e}}_{l}\otimes {\hat {e}}_{j})=A_{ik}B_{kj}{\hat {e}}_{i}\otimes {\hat {e}}_{j}}$

${\displaystyle \left(A^{ij}{\vec {a}}_{i}\otimes {\vec {g}}_{j}\right)\cdot \left(B^{kl}{\vec {h}}_{k}\otimes {\vec {u}}_{l}\right)=A^{ij}({\vec {g}}_{j}\cdot {\vec {h}}_{k})B^{kl}{\vec {a}}_{i}\otimes {\vec {u}}_{l}}$

Abbildung ${\displaystyle {\mathcal {L}}\times {\mathcal {L}}\to \mathbb {R} }$

Definition über die #Spur:

${\displaystyle ({\vec {a}}\otimes {\vec {g}}):({\vec {b}}\otimes {\vec {h}}):=\mathrm {Sp} (({\vec {a}}\otimes {\vec {g}})^{\top }\cdot ({\vec {b}}\otimes {\vec {h}}))=({\vec {a}}\cdot {\vec {b}})({\vec {g}}\cdot {\vec {h}})}$

${\displaystyle \mathbf {A} :\mathbf {B} :=\mathrm {Sp} (\mathbf {A} ^{\top }\cdot \mathbf {B} )}$

Eigenschaften:

${\displaystyle \mathbf {A} :\mathbf {B} =\mathbf {B} :\mathbf {A} =\mathbf {A} ^{\top }:\mathbf {B} ^{\top }=\mathbf {B} ^{\top }:\mathbf {A} ^{\top }}$

${\displaystyle \mathbf {A} ^{\top }:\mathbf {B} =\mathbf {A} :\mathbf {B} ^{\top }}$

${\displaystyle \mathbf {A} :(\mathbf {B\cdot C} )=(\mathbf {B} ^{\top }\cdot \mathbf {A} ):\mathbf {C} =(\mathbf {A\cdot C} ^{\top }):\mathbf {B} }$

${\displaystyle (\mathbf {A\cdot B} ):\mathbf {C} =\mathbf {B} :(\mathbf {A} ^{\top }\cdot \mathbf {C} )=\mathbf {A} :(\mathbf {C\cdot B} ^{\top })}$

${\displaystyle ({\vec {u}}\otimes {\vec {v}}):\mathbf {A} ={\vec {u}}\cdot \mathbf {A} \cdot {\vec {v}}}$

Abbildung ${\displaystyle \mathbb {V} \times {\mathcal {L}}\to {\mathcal {L}}}$ oder ${\displaystyle {\mathcal {L}}\times \mathbb {V} \to {\mathcal {L}}}$

Dyaden:

${\displaystyle {\vec {a}}\times ({\vec {b}}\otimes {\vec {g}})=({\vec {a}}\times {\vec {b}})\otimes {\vec {g}}={\vec {a}}\times {\vec {b}}\otimes {\vec {g}}}$

${\displaystyle ({\vec {a}}\otimes {\vec {g}})\times {\vec {h}}={\vec {a}}\otimes ({\vec {g}}\times {\vec {h}})={\vec {a}}\otimes {\vec {g}}\times {\vec {h}}}$

${\displaystyle {\vec {a}}\times {\vec {b}}\otimes {\vec {g}}=-{\big (}({\vec {b}}\otimes {\vec {g}})^{\top }\times {\vec {a}}){\big )}^{\top }}$

${\displaystyle {\vec {a}}\otimes {\vec {g}}\times {\vec {h}}=-{\big (}{\vec {h}}\times ({\vec {a}}\otimes {\vec {g}})^{\top }{\big )}^{\top }}$

${\displaystyle a_{j}{\hat {e}}_{j}\times (A_{kl}{\hat {e}}_{k}\otimes {\hat {e}}_{l})=a_{j}A_{kl}({\hat {e}}_{j}\times {\hat {e}}_{k})\otimes {\hat {e}}_{l}=\epsilon _{ijk}a_{j}A_{kl}{\hat {e}}_{i}\otimes {\hat {e}}_{l}}$

${\displaystyle (A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})\times a_{k}{\hat {e}}_{k}=A_{ij}a_{k}{\hat {e}}_{i}\otimes ({\hat {e}}_{j}\times {\hat {e}}_{k})=\epsilon _{jkl}A_{ij}a_{k}{\hat {e}}_{i}\otimes {\hat {e}}_{l}}$

Allgemeine Tensoren:

${\displaystyle ({\vec {a}}\times \mathbf {A} )\cdot {\vec {g}}:={\vec {a}}\times (\mathbf {A} \cdot {\vec {g}})={\vec {a}}\times ({\vec {g}}\cdot \mathbf {A} ^{\top })}$

${\displaystyle {\vec {b}}\cdot ({\vec {a}}\times \mathbf {A} ):=({\vec {b}}\times {\vec {a}})\cdot \mathbf {A} }$

${\displaystyle {\vec {g}}\cdot (\mathbf {A} \times {\vec {a}}):=({\vec {g}}\cdot \mathbf {A} )\times {\vec {a}}=(\mathbf {A} ^{\top }\cdot {\vec {g}})\times {\vec {a}}}$

${\displaystyle (\mathbf {A} \times {\vec {a}})\cdot {\vec {b}}=\mathbf {A} \cdot ({\vec {a}}\times {\vec {b}})}$

${\displaystyle {\vec {a}}\times \mathbf {A} =-\left(\mathbf {A} ^{\top }\times {\vec {a}}\right)^{\top }}$

${\displaystyle \mathbf {A} \times {\vec {a}}=-\left({\vec {a}}\times \mathbf {A} ^{\top }\right)^{\top }}$

Symmetrische Tensoren: ${\displaystyle {\vec {a}}\times \mathbf {A} ^{\mathrm {S} }=-\left(\mathbf {A} ^{\mathrm {S} }\times {\vec {a}}\right)^{\top }}$

Insbesondere Kugeltensoren: ${\displaystyle {\vec {a}}\times \mathbf {A} ^{\mathrm {K} }=\mathbf {A} ^{\mathrm {K} }\times {\vec {a}}=-({\vec {a}}\times \mathbf {A} ^{\mathrm {K} })^{\top }}$

Schiefsymmetrische Tensoren: ${\displaystyle {\vec {a}}\times \mathbf {A} ^{\mathrm {A} }=\left(\mathbf {A} ^{\mathrm {A} }\times {\vec {a}}\right)^{\top }}$

#Axialer Tensor oder Kreuzproduktmatrix mit dem #Einheitstensor:

${\displaystyle ({\vec {a}}\times \mathbf {1} )\cdot {\vec {g}}={\vec {a}}\cdot ({\vec {g}}\times \mathbf {1} )={\vec {a}}\cdot (\mathbf {1} \times {\vec {g}})={\vec {a}}\times {\vec {g}}}$

Mehrfach:

${\displaystyle ({\vec {a}}\times ({\vec {b}}\times \mathbf {A} ))\cdot {\vec {g}}={\vec {a}}\times ({\vec {b}}\times (\mathbf {A} \cdot {\vec {g}}))=({\vec {a}}\cdot \mathbf {A} \cdot {\vec {g}}){\vec {b}}-({\vec {a}}\cdot {\vec {b}})\mathbf {A} \cdot {\vec {g}}}$

${\displaystyle {\vec {a}}\times ({\vec {b}}\times \mathbf {A} )={\vec {b}}\otimes {\vec {a}}\cdot \mathbf {A} -({\vec {a}}\cdot {\vec {b}})\mathbf {A} }$

Meistens ist aber:

${\displaystyle (\mathbf {A} \cdot {\vec {a}})\times {\vec {g}}\neq \mathbf {A} \cdot ({\vec {a}}\times {\vec {g}})=(\mathbf {A} \times {\vec {a}})\cdot {\vec {g}}}$

${\displaystyle {\vec {a}}\times ({\vec {g}}\cdot \mathbf {A} )\neq ({\vec {a}}\times {\vec {g}})\cdot \mathbf {A} ={\vec {a}}\cdot ({\vec {g}}\times \mathbf {A} )}$

Abbildung ${\displaystyle {\mathcal {L}}\times {\mathcal {L}}\to \mathbb {V} }$

${\displaystyle \mathbf {A\times B} ={\stackrel {3}{\mathbf {E} }}:(\mathbf {A\cdot B} ^{\top })=-{\stackrel {3}{\mathbf {E} }}:(\mathbf {B\cdot A} ^{\top })=-\mathbf {B\times A} \in \mathbb {V} }$

mit #Fundamentaltensor 3. Stufe ${\displaystyle {\stackrel {3}{\mathbf {E} }}}$.

${\displaystyle ({\vec {a}}\otimes {\vec {g}})\times ({\vec {b}}\otimes {\vec {h}})=({\vec {g}}\cdot {\vec {h}}){\vec {a}}\times {\vec {b}}}$

${\displaystyle {\begin{aligned}&A_{ik}({\hat {e}}_{i}\otimes {\hat {e}}_{k})\times {\big (}B_{jl}({\hat {e}}_{j}\otimes {\hat {e}}_{l}){\big )}=A_{ik}B_{jk}({\hat {e}}_{i}\times {\hat {e}}_{j})=\ldots \\&\ldots ={\begin{pmatrix}A_{21}B_{31}-A_{31}B_{21}+A_{22}B_{32}-A_{32}B_{22}+A_{23}B_{33}-A_{33}B_{23}\\A_{31}B_{11}-A_{11}B_{31}+A_{32}B_{12}-A_{12}B_{32}+A_{33}B_{13}-A_{13}B_{33}\\A_{11}B_{21}-A_{21}B_{11}+A_{12}B_{22}-A_{22}B_{12}+A_{13}B_{23}-A_{23}B_{13}\end{pmatrix}}\end{aligned}}}$

Zusammenhang mit #Dualer axialer Vektor und #Vektorinvariante:

${\displaystyle \mathbf {A\times B} =-2{\stackrel {A}{\overrightarrow {\mathbf {A\cdot B} ^{\top }}}}={\vec {\mathrm {i} }}(\mathbf {A\cdot B} ^{\top })}$

Mit #Einheitstensor:

${\displaystyle \mathbf {1\times A} =2{\stackrel {A}{\overrightarrow {\mathbf {A} }}}=-{\vec {\mathrm {i} }}(\mathbf {A} )}$

Mehrfachprodukte:

${\displaystyle (\mathbf {A\cdot B} )\times \mathbf {C} =\mathbf {A} \times (\mathbf {C\cdot B} ^{\top })}$

${\displaystyle \mathbf {A} \times (\mathbf {B\cdot C} )=(\mathbf {A\cdot C} ^{\top })\times \mathbf {B} }$

Zusammenhang mit dem #Skalarkreuzprodukt von Tensoren:

${\displaystyle \mathbf {A\times B} =\mathbf {A} \cdot \!\!\times (\mathbf {B} ^{\top })}$

Abbildung ${\displaystyle {\mathcal {L}}\times {\mathcal {L}}\to \mathbb {V} }$

${\displaystyle ({\vec {a}}\otimes {\vec {g}})\cdot \!\!\times ({\vec {h}}\otimes {\vec {u}})=-({\vec {u}}\otimes {\vec {h}})\cdot \!\!\times ({\vec {g}}\otimes {\vec {a}}):=({\vec {g}}\cdot {\vec {h}}){\vec {a}}\times {\vec {u}}}$

${\displaystyle {\begin{aligned}&A_{ik}({\hat {e}}_{i}\otimes {\hat {e}}_{k})\cdot \!\!\times {\big (}B_{lj}({\hat {e}}_{l}\otimes {\hat {e}}_{j}){\big )}=A_{ik}B_{kj}({\hat {e}}_{i}\times {\hat {e}}_{j})=\ldots \\&\ldots ={\begin{pmatrix}A_{21}B_{13}-A_{31}B_{12}+A_{22}B_{23}-A_{32}B_{22}+A_{23}B_{33}-A_{33}B_{32}\\A_{31}B_{11}-A_{11}B_{13}+A_{32}B_{21}-A_{12}B_{23}+A_{33}B_{31}-A_{13}B_{33}\\A_{11}B_{12}-A_{21}B_{11}+A_{12}B_{22}-A_{22}B_{21}+A_{13}B_{32}-A_{23}B_{31}\end{pmatrix}}\end{aligned}}}$

Das Skalarkreuzprodukt mit dem #Einheitstensor vertauscht das dyadische Produkt durch das Kreuzprodukt:

${\displaystyle \mathbf {1} \cdot \!\!\times ({\vec {a}}\otimes {\vec {b}})={\vec {a}}\times {\vec {b}}}$

Allgemein:

${\displaystyle \mathbf {A} \cdot \!\!\times \mathbf {B} =-(\mathbf {B} ^{\top })\cdot \!\!\times (\mathbf {A} ^{\top })}$

${\displaystyle \mathbf {A} \cdot \!\!\times (\mathbf {B\cdot C} )=(\mathbf {A\cdot B} )\cdot \!\!\times \mathbf {C} }$

${\displaystyle (\mathbf {A\cdot B} )\cdot \!\!\times \mathbf {C} =\mathbf {A} \cdot \!\!\times (\mathbf {B\cdot C} )}$

Zusammenhang mit dem #Kreuzprodukt von Tensoren:

${\displaystyle \mathbf {S} \cdot \!\!\times \mathbf {T} =\mathbf {S\times (T^{\top })} }$

Zusammenhang mit #Vektorinvariante und #Dualer axialer Vektor:

${\displaystyle \mathbf {A} \cdot \!\!\times \mathbf {B} ={\vec {\mathrm {i} }}(\mathbf {A} \cdot \mathbf {B} )=-2{\stackrel {A}{\overrightarrow {\mathbf {A} \cdot \mathbf {B} }}}}$

Siehe auch #Äußeres Tensorprodukt #

Abbildung ${\displaystyle {\mathcal {L}}\times {\mathcal {L}}\to {\mathcal {L}}}$

${\displaystyle ({\vec {a}}\otimes {\vec {g}})\times \!\!\times ({\vec {h}}\otimes {\vec {b}}):=({\vec {g}}\times {\vec {h}})\otimes ({\vec {a}}\times {\vec {b}})=({\vec {g}}\otimes {\vec {a}})\#({\vec {h}}\otimes {\vec {b}})}$

${\displaystyle A_{ij}({\hat {e}}_{i}\otimes {\hat {e}}_{j})\times \!\!\times {\big (}B_{kl}({\hat {e}}_{k}\otimes {\hat {e}}_{l}){\big )}:=A_{ij}B_{kl}({\hat {e}}_{j}\times {\hat {e}}_{k})\otimes ({\hat {e}}_{i}\times {\hat {e}}_{l})}$

${\displaystyle \mathbf {A} \times \!\!\times \mathbf {B} =\mathbf {A} ^{\top }\#\mathbf {B} }$

Abbildung ${\displaystyle {\mathcal {L}}\times {\mathcal {L}}\to {\mathcal {L}}}$

${\displaystyle ({\vec {a}}\otimes {\vec {g}})\#({\vec {b}}\otimes {\vec {h}}):=({\vec {a}}\times {\vec {b}})\otimes ({\vec {g}}\times {\vec {h}})=({\vec {g}}\otimes {\vec {a}})\times \!\!\times ({\vec {b}}\otimes {\vec {h}})}$

${\displaystyle {\begin{aligned}&(A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})\#(B_{kl}{\hat {e}}_{k}\otimes {\hat {e}}_{l})=A_{ij}B_{kl}({\hat {e}}_{i}\times {\hat {e}}_{k})\otimes ({\hat {e}}_{j}\times {\hat {e}}_{l})\\&\qquad \qquad \qquad \qquad \qquad \quad \;\;\;=\epsilon _{ikm}\epsilon _{jln}A_{ij}B_{kl}{\hat {e}}_{m}\otimes {\hat {e}}_{n}\end{aligned}}}$

Mit der Formel für das Produkt zweier #Permutationssymbole:

${\displaystyle {\begin{aligned}\mathbf {A} \#\mathbf {B} =&{\big (}\mathrm {Sp} (\mathbf {A} )\mathrm {Sp} (\mathbf {B} )-\mathrm {Sp} (\mathbf {A\cdot B} ){\big )}\mathbf {1} \\&+{\big (}\mathbf {A\cdot B} +\mathbf {B\cdot A} -\mathrm {Sp} (\mathbf {A} )\mathbf {B} -\mathrm {Sp} (\mathbf {B} )\mathbf {A} {\big )}^{\top }\end{aligned}}}$

Grundlegende Eigenschaften:

${\displaystyle \mathbf {A} \#\mathbf {B} =\mathbf {B} \#\mathbf {A} =(\mathbf {A} ^{\top }\#\mathbf {B} ^{\top })^{\top }}$

${\displaystyle (\mathbf {A+B} )\#\mathbf {C} =\mathbf {A} \#\mathbf {C} +\mathbf {B} \#\mathbf {C} }$

${\displaystyle \mathbf {A} \#(\mathbf {B+C} )=\mathbf {A} \#\mathbf {B} +\mathbf {A} \#\mathbf {C} }$

Kreuzprodukt und #Kofaktor:

${\displaystyle (\mathbf {A} \#\mathbf {B} )\cdot ({\vec {u}}\times {\vec {v}})=(\mathbf {A} \cdot {\vec {u}})\times (\mathbf {B} \cdot {\vec {v}})-(\mathbf {A} \cdot {\vec {v}})\times (\mathbf {B} \cdot {\vec {u}})}$

${\displaystyle {\frac {1}{2}}(\mathbf {A} \#\mathbf {A} )\cdot ({\vec {u}}\times {\vec {v}})=\mathrm {cof} (\mathbf {A} )\cdot ({\vec {u}}\times {\vec {v}})=(\mathbf {A} \cdot {\vec {u}})\times (\mathbf {A} \cdot {\vec {v}})}$

#Hauptinvarianten:

${\displaystyle {\frac {1}{2}}(\mathbf {A\#1} ):\mathbf {1} =\mathrm {Sp} (\mathbf {A} )}$

${\displaystyle {\frac {1}{2}}(\mathbf {A\#A} ):\mathbf {1} =\mathrm {I} _{2}(\mathbf {A} )}$

${\displaystyle {\frac {1}{6}}(\mathbf {A\#A} ):\mathbf {A} =\det(\mathbf {A} )}$

Weitere Eigenschaften:

${\displaystyle \mathbf {1} \#\mathbf {1} =2\,\mathbf {1} }$

${\displaystyle \mathbf {A} \#\mathbf {1} =\mathrm {Sp} (\mathbf {A} )\mathbf {1} -\mathbf {A} ^{\top }}$

${\displaystyle (\mathbf {A} \#\mathbf {B} ):\mathbf {C} =(\mathbf {B} \#\mathbf {C} ):\mathbf {A} =(\mathbf {C} \#\mathbf {A} ):\mathbf {B} }$

${\displaystyle \mathrm {Sp} (\mathbf {A} \#\mathbf {B} )=\mathrm {Sp} (\mathbf {A} )\mathrm {Sp} (\mathbf {B} )-\mathrm {Sp} (\mathbf {A\cdot B} )}$

${\displaystyle (\mathbf {A} \#\mathbf {B} )\cdot (\mathbf {C} \#\mathbf {D} )=(\mathbf {A\cdot C} )\#(\mathbf {B\cdot D} )+(\mathbf {A\cdot D} )\#(\mathbf {B\cdot C} )}$

Aber meistens:

${\displaystyle (\mathbf {A} \#\mathbf {B} )\#\mathbf {C} \neq \mathbf {A} \#(\mathbf {B} \#\mathbf {C} )}$

.

${\displaystyle \mathbf {A} \cdot ({\vec {a}}\otimes {\vec {g}})=(\mathbf {A} \cdot {\vec {a}})\otimes {\vec {g}}}$

${\displaystyle {\vec {a}}\otimes (\mathbf {A} \cdot {\vec {g}})=({\vec {a}}\otimes {\vec {g}})\cdot \mathbf {A} ^{\top }}$

${\displaystyle {\vec {a}}\cdot \mathbf {A} \cdot {\vec {g}}=\mathbf {A} :({\vec {a}}\otimes {\vec {g}})}$

Spatprodukt und #Determinante eines Tensors:

${\displaystyle (\mathbf {A} \cdot {\vec {a}})\cdot {\big (}(\mathbf {A} \cdot {\vec {b}})\times (\mathbf {A} \cdot {\vec {c}}){\big )}=\mathrm {det} (\mathbf {A} )\;{\vec {a}}\cdot ({\vec {b}}\times {\vec {c}})}$

Kreuzprodukt und #Kofaktor:

${\displaystyle (\mathbf {A} \cdot {\vec {a}})\times (\mathbf {A} \cdot {\vec {b}})=\mathrm {cof} (\mathbf {A} )\cdot ({\vec {a}}\times {\vec {b}})}$

${\displaystyle \mathbf {A} ^{\top }\cdot {\big (}(\mathbf {A} \cdot {\vec {a}})\times (\mathbf {A} \cdot {\vec {b}}){\big )}=\mathrm {det} (\mathbf {A} )\;{\vec {a}}\times {\vec {b}}}$

#Axialer Tensor oder Kreuzproduktmatrix, #Kreuzprodukt von Tensoren, #Skalarkreuzprodukt von Tensoren, #Dualer axialer Vektor und #Vektorinvariante:

${\displaystyle ({\vec {u}}\times \mathbf {1} )\cdot {\vec {v}}=({\vec {u}}\otimes {\vec {v}})\times \mathbf {1} =({\vec {u}}\otimes {\vec {v}})\cdot \!\!\times \mathbf {1} ={\stackrel {A}{\overrightarrow {({\vec {u}}\times {\vec {v}})\times \mathbf {1} }}}={\vec {\mathrm {i} }}({\vec {u}}\otimes {\vec {v}})={\vec {u}}\times {\vec {v}}}$

${\displaystyle \mathbf {A} =A_{ij}\,{\hat {e}}_{i}\otimes {\hat {e}}_{j}={\begin{pmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{pmatrix}}\quad \rightarrow \;A_{ij}={\hat {e}}_{i}\cdot \mathbf {A} \cdot {\hat {e}}_{j}}$

${\displaystyle \mathbf {A} =A^{ij}\,{\vec {a}}_{i}\otimes {\vec {g}}_{j}\quad \rightarrow \;A^{ij}={\vec {a}}^{i}\cdot \mathbf {A} \cdot {\vec {g}}^{j}=({\vec {a}}^{i}\otimes {\vec {g}}^{j}):\mathbf {A} }$

${\displaystyle \mathbf {A} =A_{ij}\,{\vec {a}}^{i}\otimes {\vec {g}}^{j}\quad \rightarrow \;A_{ij}={\vec {a}}_{i}\cdot \mathbf {A} \cdot {\vec {g}}_{j}}$

${\displaystyle \mathbf {A} =A_{j}^{i}\,{\vec {a}}_{i}\otimes {\vec {g}}^{j}\quad \rightarrow \;A_{j}^{i}={\vec {a}}^{i}\cdot \mathbf {A} \cdot {\vec {g}}_{j}}$

${\displaystyle \mathbf {A} =A_{i}^{j}\,{\vec {a}}^{i}\otimes {\vec {g}}_{j}\quad \rightarrow \;A_{i}^{j}={\vec {a}}_{i}\cdot \mathbf {A} \cdot {\vec {g}}^{j}}$

${\displaystyle \mathbf {A} =A_{ij}{\vec {a}}^{i}\otimes {\vec {a}}^{j}=A_{ij}^{\ast }{\vec {b}}^{i}\otimes {\vec {b}}^{j}}$

Die Komponenten ${\displaystyle A_{ij}^{\ast }}$ ergeben sich durch Vor- und Nachmultiplikation mit dem #Einheitstensor ${\displaystyle \mathbf {1} ={\vec {b}}^{i}\otimes {\vec {b}}_{i}}$:

${\displaystyle {\begin{aligned}\mathbf {A} =\mathbf {1\cdot A\cdot 1} ^{\top }=&({\vec {b}}^{i}\otimes {\vec {b}}_{i})\cdot (A_{kl}{\vec {a}}^{k}\otimes {\vec {a}}^{l})\cdot ({\vec {b}}_{j}\otimes {\vec {b}}^{j})\\=&({\vec {b}}_{i}\cdot {\vec {a}}^{k})A_{kl}({\vec {a}}^{l}\cdot {\vec {b}}_{j}){\vec {b}}^{i}\otimes {\vec {b}}^{j}=:A_{ij}^{\ast }{\vec {b}}^{i}\otimes {\vec {b}}^{j}\\\rightarrow A_{ij}^{\ast }=&({\vec {b}}_{i}\cdot {\vec {a}}^{k})A_{kl}({\vec {a}}^{l}\cdot {\vec {b}}_{j})\end{aligned}}}$

Allgemein:

${\displaystyle \mathbf {A} =A_{ij}{\vec {a}}^{i}\otimes {\vec {g}}^{j}=A_{ij}^{\ast }{\vec {b}}^{i}\otimes {\vec {h}}^{j}}$

Basiswechsel mit ${\displaystyle \mathbf {1} =({\vec {b}}_{i}\cdot {\vec {a}}^{k}){\vec {b}}^{i}\otimes {\vec {a}}_{k}=({\vec {h}}_{j}\cdot {\vec {g}}^{l}){\vec {h}}^{j}\otimes {\vec {g}}_{l}}$:

${\displaystyle {\begin{aligned}\mathbf {A} =\mathbf {1\cdot A\cdot 1} ^{\top }=&({\vec {b}}_{i}\cdot {\vec {a}}^{k})({\vec {b}}^{i}\otimes {\vec {a}}_{k})\cdot A_{mn}({\vec {a}}^{m}\otimes {\vec {g}}^{n})\cdot ({\vec {h}}_{j}\cdot {\vec {g}}^{l})({\vec {g}}_{l}\otimes {\vec {h}}^{j})\\=&({\vec {b}}_{i}\cdot {\vec {a}}^{k})A_{kl}({\vec {h}}_{j}\cdot {\vec {g}}^{l})({\vec {b}}^{i}\otimes {\vec {h}}^{j})=A_{ij}^{\ast }{\vec {b}}^{i}\otimes {\vec {h}}^{j}\\\rightarrow A_{ij}^{\ast }=&({\vec {b}}_{i}\cdot {\vec {a}}^{k})A_{kl}({\vec {g}}^{l}\cdot {\vec {h}}_{j})\end{aligned}}}$

Definition für einen Tensor A:

${\displaystyle \langle {\vec {u}},{\vec {v}}\rangle :={\vec {u}}\cdot \mathbf {A} \cdot {\vec {v}}=\mathbf {A} :({\vec {u}}\otimes {\vec {v}})}$

Zwei Tensoren A und B sind identisch, wenn

${\displaystyle {\vec {u}}\cdot \mathbf {A} \cdot {\vec {v}}={\vec {u}}\cdot \mathbf {B} \cdot {\vec {v}}\quad \forall \;{\vec {u}},{\vec {v}}\in \mathbb {V} }$

Definition

${\displaystyle \mathrm {cof} (\mathbf {A} ):=\mathbf {A^{\top }\cdot A^{\top }} -\mathrm {I} _{1}(\mathbf {A} )\mathbf {A} ^{\top }+\mathrm {I} _{2}(\mathbf {A} )\mathbf {1} }$

#Invarianten:

Wenn λ_1,2,3 die #Eigenwerte des Tensors A sind, dann hat cof(A) die Eigenwerte λ₁λ₂, λ₂λ₃, λ₃λ₁.

#Hauptinvarianten:

${\displaystyle \mathrm {I} _{1}(\mathrm {cof} (\mathbf {A} ))=\mathrm {I} _{2}(\mathbf {A} )}$

${\displaystyle \mathrm {I} _{2}(\mathrm {cof} (\mathbf {A} ))=\mathrm {I} _{1}(\mathbf {A} )\mathrm {det} (\mathbf {A} )}$

${\displaystyle \mathrm {det} (\mathrm {cof} (\mathbf {A} ))=\mathrm {det} ^{2}(\mathbf {A} )}$

#Betrag:

${\displaystyle \|\mathrm {cof} (\mathbf {A} )\|={\sqrt {\mathrm {I} _{2}(\mathbf {A^{\top }\cdot A} )}}={\frac {\sqrt {2}}{2}}{\sqrt {\|\mathbf {A} \|^{4}-\|\mathbf {A^{\top }\cdot A} \|^{2}}}}$

Weitere Eigenschaften:

${\displaystyle \mathrm {cof} (x\mathbf {A} )=x^{2}\mathrm {cof} (\mathbf {A} )}$

${\displaystyle \mathrm {det} (\mathbf {A} )\neq 0\quad \rightarrow \quad \mathrm {cof} (\mathbf {A} )=\det(\mathbf {A} )\mathbf {A} ^{\top -1}}$

${\displaystyle \mathbf {A} ^{\top }\cdot \mathrm {cof} (\mathbf {A} )=\mathrm {cof} (\mathbf {A} )\cdot \mathbf {A} ^{\top }=\mathrm {det} (\mathbf {A} )\mathbf {1} }$

${\displaystyle \mathrm {cof} (\mathbf {A\cdot B} )=\mathrm {cof} (\mathbf {A} )\cdot \mathrm {cof} (\mathbf {B} )}$

${\displaystyle \mathrm {cof} (\mathbf {A} ^{\top })=\mathrm {cof} (\mathbf {A} )^{\top }}$

${\displaystyle \mathrm {cof} \left(\mathrm {cof} (\mathbf {A} )\right)=\mathrm {det} (\mathbf {A} )\mathbf {A} }$

${\displaystyle {\begin{aligned}&\mathrm {cof} (A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})={\frac {1}{2}}(A_{kl}A_{mn}\epsilon _{kmi}\epsilon _{lnj})({\hat {e}}_{i}\otimes {\hat {e}}_{j})=\ldots \\&\ldots ={\begin{pmatrix}A_{22}A_{33}-A_{23}A_{32}&A_{23}A_{31}-A_{21}A_{33}&A_{21}A_{32}-A_{22}A_{31}\\A_{32}A_{13}-A_{33}A_{12}&A_{33}A_{11}-A_{31}A_{13}&A_{31}A_{12}-A_{32}A_{11}\\A_{12}A_{23}-A_{13}A_{22}&A_{13}A_{21}-A_{11}A_{23}&A_{11}A_{22}-A_{12}A_{21}\end{pmatrix}}\end{aligned}}}$

Kofaktor und #Äußeres Tensorprodukt:

${\displaystyle \mathrm {cof} (\mathbf {A} )={\frac {1}{2}}\mathbf {A} \#\mathbf {A} }$

${\displaystyle {\begin{aligned}\mathrm {cof} (\mathbf {A+B} )=&{\frac {1}{2}}(\mathbf {A} \#\mathbf {A} +2\mathbf {A} \#\mathbf {B} +\mathbf {B} \#\mathbf {B} )\\=&\mathrm {cof} (\mathbf {A} )+\mathrm {cof} (\mathbf {B} )+\mathbf {A} \#\mathbf {B} \end{aligned}}}$

Kreuzprodukt und Kofaktor:

${\displaystyle (\mathbf {A} \cdot {\vec {a}})\times (\mathbf {A} \cdot {\vec {b}})=\mathrm {cof} (\mathbf {A} )\cdot ({\vec {a}}\times {\vec {b}})}$

Definition:

${\displaystyle \mathrm {adj} (\mathbf {A} ):=\mathbf {A} ^{2}-\mathrm {I} _{1}(\mathbf {A} )\mathbf {A} +\mathrm {I} _{2}(\mathbf {A} )\mathbf {1} =\mathrm {cof} (\mathbf {A} )^{\top }}$

#Hauptinvarianten:

${\displaystyle \mathrm {I} _{1}(\mathrm {adj} (\mathbf {A} ))=\mathrm {I} _{2}(\mathbf {A} )}$

${\displaystyle \mathrm {I} _{2}(\mathrm {adj} (\mathbf {A} ))=\mathrm {I} _{1}(\mathbf {A} )\mathrm {det} (\mathbf {A} )}$

${\displaystyle \mathrm {det} (\mathrm {adj} (\mathbf {A} ))=\mathrm {det} ^{2}(\mathbf {A} )}$

#Betrag:

${\displaystyle \|\mathrm {adj} (\mathbf {A} )\|={\sqrt {\mathrm {I} _{2}(\mathbf {A^{\top }\cdot A} )}}={\frac {\sqrt {2}}{2}}{\sqrt {\|\mathbf {A} \|^{4}-\|\mathbf {A^{\top }\cdot A} \|^{2}}}}$

Weitere Eigenschaften:

${\displaystyle \mathrm {adj} (x\mathbf {A} )=x^{2}\mathrm {adj} (\mathbf {A} )}$

${\displaystyle \mathrm {det} (\mathbf {A} )\neq 0\quad \rightarrow \quad \mathrm {adj} (\mathbf {A} )=\det(\mathbf {A} )\mathbf {A} ^{-1}}$

${\displaystyle \mathbf {A} \cdot \mathrm {adj} (\mathbf {A} )=\mathrm {adj} (\mathbf {A} )\cdot \mathbf {A} =\mathrm {det} (\mathbf {A} )\mathbf {1} }$

${\displaystyle \mathrm {adj} (\mathbf {A\cdot B} )=\mathrm {adj} (\mathbf {B} )\cdot \mathrm {adj} (\mathbf {A} )}$

${\displaystyle \mathrm {adj} (\mathbf {A} ^{\top })=\mathrm {adj} (\mathbf {A} )^{\top }}$

${\displaystyle {\begin{aligned}\mathrm {adj} (\mathbf {A+B} )=&{\frac {1}{2}}(\mathbf {A} \#\mathbf {A} +2\mathbf {A} \#\mathbf {B} +\mathbf {B} \#\mathbf {B} )^{\top }\\=&\mathrm {adj} (\mathbf {A} )+\mathrm {adj} (\mathbf {B} )+\mathbf {A} ^{\top }\#\mathbf {B} ^{\top }\end{aligned}}}$

${\displaystyle \mathrm {adj} \left(\mathrm {adj} (\mathbf {A} )\right)=\mathrm {det} (\mathbf {A} )\mathbf {A} }$

${\displaystyle {\begin{aligned}&\mathrm {adj} (A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})={\frac {1}{2}}(A_{kl}A_{mn}\epsilon _{kmj}\epsilon _{lni})({\hat {e}}_{i}\otimes {\hat {e}}_{j})=\ldots \\&\ldots ={\begin{pmatrix}A_{22}A_{33}-A_{23}A_{32}&A_{32}A_{13}-A_{33}A_{12}&A_{12}A_{23}-A_{13}A_{22}\\A_{23}A_{31}-A_{21}A_{33}&A_{33}A_{11}-A_{31}A_{13}&A_{13}A_{21}-A_{11}A_{23}\\A_{21}A_{32}-A_{22}A_{31}&A_{31}A_{12}-A_{32}A_{11}&A_{11}A_{22}-A_{12}A_{21}\end{pmatrix}}\end{aligned}}}$

Definition

${\displaystyle \mathbf {A} ^{-1}:\quad \mathbf {A} ^{-1}\cdot \mathbf {A} =\mathbf {A\cdot A} ^{-1}=\mathbf {1} }$

Die Inverse ist nur definiert, wenn ${\displaystyle |\mathbf {A} |=\mathrm {det} (\mathbf {A} )=\mathrm {I} _{3}(\mathbf {A} )\neq 0}$

Zusammenhang mit dem adjungierten Tensor ${\displaystyle \mathrm {adj} (\mathbf {A} )}$:

${\displaystyle \mathbf {A} ^{-1}={\frac {1}{\det(\mathbf {A} )}}\mathrm {adj} (\mathbf {A} )}$

${\displaystyle {\begin{aligned}\mathbf {A} =&A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})\\\rightarrow \mathbf {A} ^{-1}=&{\frac {1}{|\mathbf {A} |}}{\begin{pmatrix}A_{22}A_{33}-A_{23}A_{32}&A_{32}A_{13}-A_{33}A_{12}&A_{12}A_{23}-A_{13}A_{22}\\A_{23}A_{31}-A_{21}A_{33}&A_{33}A_{11}-A_{31}A_{13}&A_{13}A_{21}-A_{11}A_{23}\\A_{21}A_{32}-A_{22}A_{31}&A_{31}A_{12}-A_{32}A_{11}&A_{11}A_{22}-A_{12}A_{21}\end{pmatrix}}\end{aligned}}}$

Werden die Spalten von A mit Vektoren bezeichnet, also ${\displaystyle \mathbf {A} ={\begin{pmatrix}{\vec {a}}_{1}&{\vec {a}}_{2}&{\vec {a}}_{3}\end{pmatrix}}}$, dann gilt:

${\displaystyle \mathbf {A} ^{-1}={\begin{pmatrix}{\vec {a}}^{1}&{\vec {a}}^{2}&{\vec {a}}^{3}\end{pmatrix}}^{\top }={\frac {1}{\mathrm {det} (\mathbf {A} )}}{\begin{pmatrix}{\vec {a}}_{2}\times {\vec {a}}_{3}&{\vec {a}}_{3}\times {\vec {a}}_{1}&{\vec {a}}_{1}\times {\vec {a}}_{2}\end{pmatrix}}^{\top }}$

Satz von Cayley-Hamilton:

${\displaystyle \mathbf {A} ^{-1}={\frac {1}{\mathrm {I} _{3}(\mathbf {A} )}}(\mathbf {A} ^{2}-\mathrm {I} _{1}(\mathbf {A} )\mathbf {A} +\mathrm {I} _{2}(\mathbf {A} )\mathbf {1} )}$

worin ${\displaystyle \mathrm {I} _{1,2,3}}$ die drei #Hauptinvarianten sind.

Inverse des transponierten Tensors:

${\displaystyle (\mathbf {A} ^{\top })^{-1}=(\mathbf {A} ^{-1})^{\top }=\mathbf {A} ^{\top -1}=\mathbf {A} ^{-\top }}$

Inverse eines Tensorprodukts:

${\displaystyle (\mathbf {A\cdot B} )^{-1}=\mathbf {B} ^{-1}\cdot \mathbf {A} ^{-1}}$

${\displaystyle (x\mathbf {A} )^{-1}={\frac {1}{x}}\mathbf {A} ^{-1}}$

#Äußeres Tensorprodukt und Inverse einer Summe:

${\displaystyle (\mathbf {A+B} )^{-1}={\frac {1}{\det(\mathbf {A+B} )}}\left(\mathrm {adj} (\mathbf {A} )+\mathrm {adj} (\mathbf {B} )+(\mathbf {A} \#\mathbf {B} )^{\top }\right)}$

Invertierungsformeln:

${\displaystyle (a\mathbf {1} +{\vec {b}}\otimes {\vec {c}})^{-1}={\frac {1}{a}}\left(\mathbf {1} -{\frac {1}{a+{\vec {b}}\cdot {\vec {c}}}}{\vec {b}}\otimes {\vec {c}}\right)}$

${\displaystyle {\begin{aligned}&(a\mathbf {1} +{\vec {b}}\otimes {\vec {c}}+{\vec {d}}\otimes {\vec {e}})^{-1}={\frac {1}{aD}}\left(D\mathbf {1} +{\vec {b}}\otimes (q{\vec {c}}+r{\vec {e}})+{\vec {d}}\otimes (s{\vec {c}}+t{\vec {e}})\right)\\&\qquad q=a+{\vec {d}}\cdot {\vec {e}},\quad r=-{\vec {c}}\cdot {\vec {d}},\quad s=-{\vec {b}}\cdot {\vec {e}},\quad t=a+{\vec {b}}\cdot {\vec {c}}\\&\qquad D=rs-qt\end{aligned}}}$

${\displaystyle ({\vec {a}}_{i}\otimes {\vec {g}}_{i})^{-1}={\vec {g}}^{i}\otimes {\vec {a}}^{i}}$

${\displaystyle \mathbf {A} \cdot {\hat {v}}=\lambda {\hat {v}}}$

mit Eigenwert ${\displaystyle \lambda }$ und Eigenvektor ${\displaystyle {\hat {v}}}$. Die Eigenvektoren werden auf die Länge eins normiert.

Jeder Tensor hat drei Eigenwerte und drei dazugehörige Eigenvektoren. Mindestens ein Eigenwert und Eigenvektor sind reell. Die beiden anderen Eigenwerte und -vektoren können reell oder komplex sein.

Charakteristische Gleichung

${\displaystyle \mathrm {det} (\mathbf {A} -\lambda _{i}\mathbf {1} )=-\lambda _{i}^{3}+\mathrm {I} _{1}(\mathbf {A} )\lambda _{i}^{2}-\mathrm {I} _{2}(\mathbf {A} )\lambda _{i}+\mathrm {I} _{3}(\mathbf {A} )=0}$

Lösung siehe Cardanische Formeln. Die Koeffizienten sind die #Hauptinvarianten :

${\displaystyle \mathrm {I} _{1}(\mathbf {A} ):=\mathrm {Sp} (\mathbf {A} )=\lambda _{1}+\lambda _{2}+\lambda _{3}}$

${\displaystyle \mathrm {I} _{2}(\mathbf {A} ):={\frac {1}{2}}{\big (}\mathrm {I} _{1}(\mathbf {A} )^{2}-\mathrm {I} _{1}(\mathbf {A} ^{2}){\big )}=\lambda _{1}\lambda _{2}+\lambda _{2}\lambda _{3}+\lambda _{3}\lambda _{1}}$

${\displaystyle \mathrm {I} _{3}(\mathbf {A} ):=\mathrm {det} (\mathbf {A} )=\lambda _{1}\lambda _{2}\lambda _{3}}$

Eigenvektoren ${\displaystyle {\vec {v}}}$ sind nur bis auf einen Faktor ≠ 0 bestimmt. Der Nullvektor ist kein Eigenvektor.

Bestimmungsgleichung: ${\displaystyle (\mathbf {A} -\lambda \mathbf {1} )\cdot {\vec {v}}={\vec {0}}}$

Tensor ${\displaystyle \mathbf {A} =A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}}$:

${\displaystyle {\begin{pmatrix}A_{11}-\lambda &A_{12}&A_{13}\\A_{21}&A_{22}-\lambda &A_{23}\\A_{31}&A_{32}&A_{33}-\lambda \end{pmatrix}}\cdot {\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\end{pmatrix}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}}$

Bestimmung mit gegebenem/angenommenem ${\displaystyle v_{1}}$:

${\displaystyle {\begin{pmatrix}A_{12}&A_{13}\\A_{22}-\lambda &A_{23}\\A_{32}&A_{33}-\lambda \end{pmatrix}}\cdot {\begin{pmatrix}v_{2}\\v_{3}\end{pmatrix}}=v_{1}{\begin{pmatrix}\lambda -A_{11}\\-A_{21}\\-A_{31}\end{pmatrix}}}$

Geometrische Vielfachheit 1:

${\displaystyle v_{2}=v_{1}{\frac {(\lambda -A_{33})A_{21}+A_{23}A_{31}}{(A_{22}-\lambda )(A_{33}-\lambda )-A_{23}A_{32}}}}$

${\displaystyle v_{3}=v_{1}{\frac {(\lambda -A_{22})A_{31}+A_{32}A_{21}}{(A_{22}-\lambda )(A_{33}-\lambda )-A_{23}A_{32}}}}$

Geometrische Vielfachheit 2:

${\displaystyle {\begin{pmatrix}A_{13}\\A_{23}\\A_{33}-\lambda \end{pmatrix}}v_{3}=-v_{1}{\begin{pmatrix}A_{11}-\lambda \\A_{21}\\A_{31}\end{pmatrix}}-v_{2}{\begin{pmatrix}A_{12}\\A_{22}-\lambda \\A_{32}\end{pmatrix}}}$

Die Formeln bleiben richtig, wenn die Indizes {1,2,3} zyklisch vertauscht werden.

Symmetrischen Tensoren: Für das Betragsquadrat der Komponenten ${\displaystyle v_{ij}}$ der auf Betrag 1 normierten Eigenvektoren ${\displaystyle {\vec {v}}_{i}}$ des (komplexen) Tensors ${\displaystyle A\in \mathbb {C} ^{n\times n}}$ gilt mit dessen Eigenwerten ${\displaystyle \lambda _{i}}$ und den Eigenwerten ${\displaystyle \mu _{jk}}$ der Hauptuntermatrizen von ${\displaystyle A}$:^[1]

${\displaystyle |v_{ij}|^{2}\prod _{k=1;k\neq i}^{n}{\big (}\lambda _{i}-\lambda _{k}{\big )}=\prod _{k=1}^{n-1}{\big (}\lambda _{i}-\mu _{jk}{\big )}}$

Sei ${\displaystyle \mathbf {A} =\mathbf {A} ^{\top }}$ symmetrisch.

Symmetrische Tensoren haben reelle Eigenwerte und paarweise zueinander senkrechte oder orthogonalisierbare Eigenvektoren, die also eine Orthonormalbasis aufbauen. Die Eigenvektoren werden so nummeriert, dass sie ein Rechtssystem bilden.

Hauptachsentransformation mit Eigenwerten ${\displaystyle \lambda _{i}}$ und Eigenvektoren ${\displaystyle {\hat {a}}_{i}}$ des symmetrischen Tensors A:

${\displaystyle {\begin{aligned}\mathbf {A} =&\sum _{i=1}^{3}\lambda _{i}{\hat {a}}_{i}\otimes {\hat {a}}_{i}=\left({\hat {a}}_{i}\otimes {\hat {e}}_{i}\right)\cdot \left(\sum _{j=1}^{3}\lambda _{j}{\hat {e}}_{j}\otimes {\hat {e}}_{j}\right)\cdot \left({\hat {e}}_{k}\otimes {\hat {a}}_{k}\right)\\=&{\begin{pmatrix}{\hat {a}}_{1}&{\hat {a}}_{2}&{\hat {a}}_{3}\end{pmatrix}}\cdot {\begin{pmatrix}\lambda _{1}&0&0\\0&\lambda _{2}&0\\0&0&\lambda _{3}\end{pmatrix}}\cdot {\begin{pmatrix}{\hat {a}}_{1}&{\hat {a}}_{2}&{\hat {a}}_{3}\end{pmatrix}}^{\top }\end{aligned}}}$

bzw.

${\displaystyle {\begin{pmatrix}{\hat {a}}_{1}&{\hat {a}}_{2}&{\hat {a}}_{3}\end{pmatrix}}^{\top }\cdot \mathbf {A} \cdot {\begin{pmatrix}{\hat {a}}_{1}&{\hat {a}}_{2}&{\hat {a}}_{3}\end{pmatrix}}={\begin{pmatrix}\lambda _{1}&0&0\\0&\lambda _{2}&0\\0&0&\lambda _{3}\end{pmatrix}}}$

Sei ${\displaystyle \mathbf {A} =-\mathbf {A} ^{\top }}$ schiefsymmetrisch.

Schiefsymmetrische Tensoren haben einen reellen und zwei konjugiert komplexe, rein imaginäre Eigenwerte. Der reelle Eigenwert von A ist null zu dem ein Eigenvektor gehört, der proportional zur reellen #Vektorinvariante ${\displaystyle {\vec {\mathrm {i} }}(\mathbf {A} )}$ ist. Siehe auch #Axialer Tensor oder Kreuzproduktmatrix.

Sei ${\displaystyle a,b,c\in \mathbb {R} }$ und ${\displaystyle {\vec {a}}_{1},{\vec {a}}_{2},{\vec {a}}_{3}\in \mathbb {R} ^{3}}$ eine Basis und ${\displaystyle {\vec {a}}^{1},{\vec {a}}^{2},{\vec {a}}^{3}}$ die dazu duale Basis.

Der Tensor

${\displaystyle \mathbf {T} =a\,{\vec {a}}_{1}\otimes {\vec {a}}^{1}+b\,{\vec {a}}_{2}\otimes {\vec {a}}^{2}+c\,{\vec {a}}_{3}\otimes {\vec {a}}^{3}}$

hat die Eigenwerte

${\displaystyle \lambda _{1}=a,\;\lambda _{2}=b,\;\lambda _{3}=c}$

und Eigenvektoren

${\displaystyle {\vec {v}}_{1}={\vec {a}}_{1},\;{\vec {v}}_{2}={\vec {a}}_{2},\;{\vec {v}}_{3}={\vec {a}}_{3}}$

Der #transponierte Tensor hat dieselben Eigenwerte zu den dualen Eigenvektoren

${\displaystyle {\vec {v}}_{1}={\vec {a}}^{1},\;{\vec {v}}_{2}={\vec {a}}^{2},\;{\vec {v}}_{3}={\vec {a}}^{3}}$

Der Tensor

${\displaystyle \mathbf {T} =c\,{\vec {a}}_{1}\otimes {\vec {a}}^{1}+a({\vec {a}}_{2}\otimes {\vec {a}}^{2}+{\vec {a}}_{3}\otimes {\vec {a}}^{3})+b({\vec {a}}_{2}\otimes {\vec {a}}^{3}-{\vec {a}}_{3}\otimes {\vec {a}}^{2})}$

hat die Eigenwerte

${\displaystyle \lambda _{1}=c,\;\lambda _{2}=a+\mathrm {i} \,b,\;\lambda _{3}=a-\mathrm {i} \,b}$

und Eigenvektoren

${\displaystyle {\vec {v}}_{1}={\vec {a}}_{1},\;{\vec {v}}_{2}={\vec {a}}_{2}+\mathrm {i} \,{\vec {a}}_{3},\;{\vec {v}}_{3}={\vec {a}}_{2}-\mathrm {i} \,{\vec {a}}_{3}}$

Der #transponierte Tensor hat dieselben Eigenwerte zu den Eigenvektoren

${\displaystyle {\vec {v}}_{1}={\vec {a}}^{1},\;{\vec {v}}_{2}={\vec {a}}^{2}-\mathrm {i} \,{\vec {a}}^{3},\;{\vec {v}}_{3}={\vec {a}}^{2}+\mathrm {i} \,{\vec {a}}^{3}}$

Die #Eigenwerte ${\displaystyle \lambda _{1},\lambda _{2},\lambda _{3}}$ sind Invarianten.

#Spur:	I1(A), Sp(A)
#Zweite Hauptinvariante:	I2(A)
#Determinante:	I3(A), det(A), │A│

Die Hauptinvarianten des Tensors A sind die Koeffizienten seines charakteristischen Polynoms:

${\displaystyle \mathrm {det} (\mathbf {A} -x\mathbf {1} )=-x^{3}+\mathrm {Sp} (\mathbf {A} )x^{2}-\mathrm {I} _{2}(\mathbf {A} )x+\mathrm {det} (\mathbf {A} )}$

Spezialfall:

${\displaystyle \mathrm {det} ({\vec {b}}\otimes {\vec {c}}+a\mathbf {1} )=a^{2}(a+{\vec {b}}\cdot {\vec {c}})}$

Satz von Cayley-Hamilton:

${\displaystyle -\mathbf {A} ^{3}+\mathrm {Sp} (\mathbf {A} )\mathbf {A} ^{2}-\mathrm {I} _{2}(\mathbf {A} )\mathbf {A} +\mathrm {det} (\mathbf {A} )\mathbf {1} =\mathbf {0} }$

Abbildung ${\displaystyle {\mathcal {L}}\to \mathbb {R} }$

${\displaystyle \mathrm {Sp} (\mathbf {A} )=\mathrm {I} _{1}(\mathbf {A} )={\frac {1}{2}}(\mathbf {A} \#\mathbf {1} ):\mathbf {1} =\lambda _{1}+\lambda _{2}+\lambda _{3}}$

mit #Eigenwerten λ_1,2,3 von A.

${\displaystyle \mathrm {Sp} ({\vec {a}}\otimes {\vec {g}})=\mathrm {Sp} ({\vec {g}}\otimes {\vec {a}}):={\vec {a}}\cdot {\vec {g}}}$

Linearität: ${\displaystyle x,y\in \mathbb {R} \rightarrow \quad \mathrm {Sp} (x\mathbf {A} +y\mathbf {B} )=x\mathrm {Sp} (\mathbf {A} )+y\mathrm {Sp} (\mathbf {B} )}$

${\displaystyle \mathrm {Sp} (\mathbf {A} )=\mathrm {Sp} (\mathbf {A} ^{\top })}$

${\displaystyle \mathrm {Sp} (\mathbf {A\cdot B} )=\mathrm {Sp} (\mathbf {B\cdot A} )}$

${\displaystyle \mathrm {Sp} (\mathbf {A^{\top }\cdot B} )=\mathrm {Sp} (\mathbf {A\cdot B^{\top }} )}$

${\displaystyle \mathrm {Sp} (\mathbf {A\cdot B\cdot C} )=\mathrm {Sp} (\mathbf {B\cdot C\cdot A} )=\mathrm {Sp} (\mathbf {C\cdot A\cdot B} )}$

In Komponenten:

${\displaystyle \mathrm {Sp} \left(A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}\right)=A_{ii}=A_{11}+A_{22}+A_{33}}$

${\displaystyle \mathrm {Sp} \left(A^{ij}{\vec {a}}_{i}\otimes {\vec {b}}_{j}\right)=A^{ij}{\vec {a}}_{i}\cdot {\vec {b}}_{j}}$

${\displaystyle \mathrm {Sp} \left(A_{j}^{i}{\vec {a}}_{i}\otimes {\vec {a}}^{j}\right)=\mathrm {Sp} \left(A_{i}^{j}{\vec {a}}^{i}\otimes {\vec {a}}_{j}\right)=A_{i}^{i}}$

Abbildung ${\displaystyle {\mathcal {L}}\to \mathbb {R} }$

${\displaystyle \mathrm {I} _{2}(\mathbf {A} ):={\frac {1}{2}}{\big (}\mathrm {Sp} (\mathbf {A} )^{2}-\mathrm {Sp} (\mathbf {A} ^{2}){\big )}={\frac {1}{2}}(\mathbf {A} \#\mathbf {A} ):\mathbf {1} =\lambda _{1}\lambda _{2}+\lambda _{2}\lambda _{3}+\lambda _{3}\lambda _{1}}$

mit #Eigenwerten λ_1,2,3 von A.

${\displaystyle \mathrm {I} _{2}(\mathbf {A} )=\mathrm {Sp(cof} (\mathbf {A} ))=\mathrm {Sp(adj} (\mathbf {A} ))}$

${\displaystyle \mathrm {I} _{2}(x\mathbf {A} )=x^{2}\mathrm {I} _{2}(\mathbf {A} )}$

${\displaystyle \mathrm {I} _{2}(\mathbf {A} ^{\top })=\mathrm {I} _{2}(\mathbf {A} )}$

${\displaystyle \mathrm {I} _{2}(\mathbf {A} \cdot \mathbf {B} )=\mathrm {I} _{2}(\mathbf {B} \cdot \mathbf {A} )}$

${\displaystyle \mathrm {I} _{2}(\mathbf {A} \cdot \mathbf {B} \cdot \mathbf {C} )=\mathrm {I} _{2}(\mathbf {B} \cdot \mathbf {C} \cdot \mathbf {A} )=\mathrm {I} _{2}(\mathbf {C} \cdot \mathbf {A} \cdot \mathbf {B} )}$

${\displaystyle \mathrm {I} _{2}(\mathbf {A} +\mathbf {B} )=\mathrm {I} _{2}(\mathbf {A} )+\mathrm {I} _{2}(\mathbf {B} )+\mathrm {Sp} (\mathbf {A} )\mathrm {Sp} (\mathbf {B} )-\mathrm {Sp} (\mathbf {A} \cdot \mathbf {B} )}$

In Komponenten:

${\displaystyle \operatorname {I} _{2}(A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})=A_{11}A_{22}+A_{11}A_{33}+A_{22}A_{33}-A_{12}A_{21}-A_{13}A_{31}-A_{23}A_{32}}$

${\displaystyle \operatorname {I} _{2}(A^{ij}{\vec {a}}_{i}\otimes {\vec {b}}_{j})={\frac {1}{2}}A^{ij}A^{kl}{\big (}({\vec {a}}_{i}\cdot {\vec {b}}_{j})({\vec {a}}_{k}\cdot {\vec {b}}_{l})-({\vec {a}}_{i}\cdot {\vec {b}}_{l})({\vec {a}}_{k}\cdot {\vec {b}}_{j}){\big )}}$

${\displaystyle \operatorname {I} _{2}\left(A_{j}^{i}{\vec {a}}_{i}\otimes {\vec {a}}^{j}\right)={\frac {1}{2}}(A_{i}^{i}A_{j}^{j}-A_{j}^{i}A_{i}^{j})}$

Abbildung ${\displaystyle {\mathcal {L}}\to \mathbb {R} }$

${\displaystyle \mathrm {I} _{3}(\mathbf {A} ):=\mathrm {det} (\mathbf {A} )={\frac {1}{6}}(\mathbf {A} \#\mathbf {A} ):\mathbf {A} =\lambda _{1}\lambda _{2}\lambda _{3}}$

mit #Eigenwerten λ_1,2,3 von A.

${\displaystyle \mathrm {det} (\mathbf {A} ^{\top })=\mathrm {det} (\mathbf {A} )}$

Determinantenproduktsatz:

${\displaystyle \mathrm {det} (\mathbf {A\cdot B} )=\mathrm {det} (\mathbf {B\cdot A} )=\mathrm {det} (\mathbf {A} )\mathrm {det} (\mathbf {B} )}$

${\displaystyle \mathrm {det} (\mathbf {A\cdot B\cdot C} )=\mathrm {det} (\mathbf {B\cdot C\cdot A} )=\mathrm {det} (\mathbf {C\cdot A\cdot B} )=\mathrm {det} (\mathbf {A} )\mathrm {det} (\mathbf {B} )\mathrm {det} (\mathbf {C} )}$

${\displaystyle \mathrm {det} (\mathbf {A} ^{-1})={\frac {1}{\mathrm {det} (\mathbf {A} )}}}$

Multiplikation mit Skalaren ${\displaystyle x\in \mathbb {R} }$:

${\displaystyle {\begin{vmatrix}x{\vec {a}}&{\vec {b}}&{\vec {c}}\end{vmatrix}}={\begin{vmatrix}{\vec {a}}&x{\vec {b}}&{\vec {c}}\end{vmatrix}}={\begin{vmatrix}{\vec {a}}&{\vec {b}}&x{\vec {c}}\end{vmatrix}}=x{\begin{vmatrix}{\vec {a}}&{\vec {b}}&{\vec {c}}\end{vmatrix}}}$

${\displaystyle \mathrm {det} (x\mathbf {A} )=x^{3}\mathrm {det} (\mathbf {A} )}$

In Komponenten:

${\displaystyle {\begin{aligned}\mathrm {det} \left(A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}\right)=&{\begin{vmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{vmatrix}}\\=&A_{11}(A_{22}A_{33}-A_{23}A_{32})+A_{12}(A_{23}A_{31}-A_{21}A_{33})\\&+A_{13}(A_{21}A_{32}-A_{22}A_{31})\end{aligned}}}$

${\displaystyle {\begin{aligned}\mathrm {det} (A^{ij}{\vec {a}}_{i}\otimes {\vec {g}}_{j})=&{\begin{vmatrix}A^{11}&A^{12}&A^{13}\\A^{21}&A^{22}&A^{23}\\A^{31}&A^{32}&A^{33}\end{vmatrix}}{\begin{vmatrix}{\vec {a}}_{1}&{\vec {a}}_{2}&{\vec {a}}_{3}\end{vmatrix}}{\begin{vmatrix}{\vec {g}}_{1}&{\vec {g}}_{2}&{\vec {g}}_{3}\end{vmatrix}}\end{aligned}}}$

${\displaystyle \operatorname {det} \left(A_{j}^{i}{\vec {a}}_{i}\otimes {\vec {a}}^{j}\right)={\begin{vmatrix}A_{1}^{1}&A_{2}^{1}&A_{3}^{1}\\A_{1}^{2}&A_{2}^{2}&A_{3}^{2}\\A_{1}^{3}&A_{2}^{3}&A_{3}^{3}\end{vmatrix}}}$

Zusammenhang mit den anderen Hauptinvarianten:

${\displaystyle {\begin{aligned}\mathrm {det} (\mathbf {A} )=&{\frac {1}{6}}{\big (}\mathrm {Sp} (\mathbf {A} )^{3}-3\mathrm {Sp} (\mathbf {A} )\mathrm {Sp} (\mathbf {A} ^{2})+2\mathrm {Sp} (\mathbf {A} ^{3}){\big )}\\[1ex]=&{\frac {1}{3}}{\big (}\mathrm {Sp} (\mathbf {A} ^{3})+3\mathrm {Sp} (\mathbf {A} )\mathrm {I} _{2}(\mathbf {A} )-\mathrm {Sp} (\mathbf {A} )^{3}{\big )}\end{aligned}}}$

Zusammenhang mit dem Spatprodukt:

${\displaystyle (\mathbf {A} \cdot {\vec {a}})\cdot {\big (}(\mathbf {A} \cdot {\vec {b}})\times (\mathbf {A} \cdot {\vec {c}}){\big )}=\mathrm {det} (\mathbf {A} ){\vec {a}}\cdot ({\vec {b}}\times {\vec {c}})}$

Zusammenhang mit #Äußeres Tensorprodukt:

${\displaystyle \det(\mathbf {A} )={\frac {1}{6}}(\mathbf {A} \#\mathbf {A} ):\mathbf {A} }$

${\displaystyle {\begin{aligned}\rightarrow \det(\mathbf {A+B} )=&\det(\mathbf {A} )+\det(\mathbf {B} )+\mathrm {Sp} (\mathbf {A} )\mathrm {I} _{2}(\mathbf {B} )+\mathrm {I} _{2}(\mathbf {A} )\mathrm {Sp} (\mathbf {B} )\\&+\mathrm {Sp} {\big (}\mathbf {A\cdot B\cdot (A+B)} {\big )}-\mathrm {Sp} (\mathbf {A\cdot B} )\mathrm {Sp} (\mathbf {A+B} )\end{aligned}}}$

Zusammenhang mit dem #Kofaktor:

${\displaystyle \det(\mathbf {A} +\mathbf {B} )=\det(\mathbf {A} )+\mathrm {cof} (\mathbf {A} ):\mathbf {B} +\mathbf {A} :\mathrm {cof} (\mathbf {B} )+\det(\mathbf {B} )}$

Abbildung ${\displaystyle {\mathcal {L}}\to \mathbb {R} }$

${\displaystyle \parallel \mathbf {A} \parallel :={\sqrt {\mathbf {A} :\mathbf {A} }}={\sqrt {\mathrm {Sp} (\mathbf {A} ^{\top }\cdot \mathbf {A} )}}}$

${\displaystyle \parallel {\vec {a}}\otimes {\vec {g}}\parallel =|{\vec {a}}|\,|{\vec {g}}|}$

${\displaystyle \parallel A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}\parallel ={\sqrt {A_{ij}A_{ij}}}}$

${\displaystyle \parallel A^{ij}{\vec {a}}_{i}\otimes {\vec {g}}_{j}\parallel ={\sqrt {A^{ij}A^{kl}({\vec {a}}_{i}\cdot {\vec {a}}_{k})({\vec {g}}_{j}\cdot {\vec {g}}_{l})}}}$

${\displaystyle \parallel A_{j}^{i}{\vec {a}}_{i}\otimes {\vec {a}}^{j}\parallel ={\sqrt {A_{j}^{i}A_{l}^{k}({\vec {a}}_{i}\cdot {\vec {a}}_{k})({\vec {a}}^{j}\cdot {\vec {a}}^{l})}}}$

Falls ${\displaystyle \mathbf {A} =\mathbf {A} ^{\top }}$:

${\displaystyle \quad \parallel \mathbf {A} \parallel ={\sqrt {\mathrm {Sp} ^{2}(\mathbf {A} )-2\mathrm {I} _{2}(\mathbf {A} )}}={\sqrt {\mathrm {Sp} (\mathbf {A} ^{2})}}={\sqrt {\lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}}}}$

Falls ${\displaystyle \mathbf {A} =-\mathbf {A} ^{\top }}$:

${\displaystyle \quad \parallel \mathbf {A} \parallel ={\sqrt {2\mathrm {I} _{2}(\mathbf {A} )}}={\sqrt {-\mathrm {Sp} (\mathbf {A} ^{2})}}}$

Für #Schiefsymmetrische Tensoren ${\displaystyle \mathbf {A} =-\mathbf {A} ^{\top }=\mathbf {A} ^{\mathrm {A} }}$ gibt es einen dualen axialen Vektor ${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {A} }}}}$ für den gilt:

${\displaystyle \mathbf {A} \cdot {\vec {v}}={\stackrel {A}{\overrightarrow {\mathbf {A} }}}\times {\vec {v}}}$ für alle ${\displaystyle {\vec {v}}\in \mathbb {V} }$

Der duale axiale Vektor ist proportional zur #Vektorinvariante:

${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {A} }}}:=-{\frac {1}{2}}{\vec {\mathrm {i} }}(\mathbf {A} )}$

Berechnung mit #Fundamentaltensor 3. Stufe ${\displaystyle {\stackrel {3}{\mathbf {E} }}}$, #Kreuzprodukt von Tensoren oder #Skalarkreuzprodukt von Tensoren:

${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {A} }}}=-{\frac {1}{2}}{\stackrel {3}{\mathbf {E} }}:\mathbf {A} =-{\frac {1}{2}}\mathbf {A} \times \mathbf {1} =-{\frac {1}{2}}\mathbf {A} \cdot \!\!\times \mathbf {1} }$

${\displaystyle {\stackrel {A}{\overrightarrow {A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}}}}=-{\frac {1}{2}}A_{ij}{\hat {e}}_{i}\times {\hat {e}}_{j}={\frac {1}{2}}{\begin{pmatrix}A_{32}-A_{23}\\A_{13}-A_{31}\\A_{21}-A_{12}\end{pmatrix}}}$

${\displaystyle {\stackrel {A}{\overrightarrow {A^{ij}({\vec {a}}_{i}\otimes {\vec {b}}_{j})}}}=-{\frac {1}{2}}A^{ij}{\vec {a}}_{i}\times {\vec {b}}_{j}}$

#Symmetrische Tensoren und #Kugeltensoren haben keinen dualen axialen Vektor: ${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {A} ^{\mathrm {S} }}}}={\stackrel {A}{\overrightarrow {\mathbf {A} ^{\mathrm {K} }}}}={\vec {0}}}$

Ein #Symmetrischer Anteil oder #Kugelanteil trägt nichts zum dualen axialen Vektor bei: ${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {A} }}}={\stackrel {A}{\overrightarrow {\mathbf {A} ^{\mathrm {D} }}}}={\stackrel {A}{\overrightarrow {\mathbf {A} ^{\mathrm {A} }}}}}$

Seien x eine beliebige Zahl, ${\displaystyle {\vec {u}},\,{\vec {v}}}$ beliebige Vektoren und A, B beliebige Tensoren zweiter Stufe. Dann gilt:

${\displaystyle {\stackrel {A}{\overrightarrow {{\vec {u}}\otimes {\vec {v}}}}}\;={\frac {1}{2}}{\vec {v}}\times {\vec {u}}}$

${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {A} }}}\times {\vec {v}}=\mathbf {A} ^{\mathrm {A} }\cdot {\vec {v}}}$

${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {A} ^{\top }}}}\quad \;=-{\stackrel {A}{\overrightarrow {\mathbf {A} }}}}$

${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {A+B} }}}={\stackrel {A}{\overrightarrow {\mathbf {A} }}}+{\stackrel {A}{\overrightarrow {\mathbf {B} }}}}$

${\displaystyle {\stackrel {A}{\overrightarrow {x\mathbf {A} }}}\quad \;=x\,{\stackrel {A}{\overrightarrow {\mathbf {A} }}}}$

${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {A} \#\mathbf {B} }}}\;=\mathbf {A} \cdot {\stackrel {A}{\overrightarrow {\mathbf {B} }}}+\mathbf {B} \cdot {\stackrel {A}{\overrightarrow {\mathbf {A} }}}}$

${\displaystyle \mathbf {A} \cdot {\stackrel {A}{\overrightarrow {\mathbf {A} }}}\;=\mathbf {A} ^{\top }\cdot {\stackrel {A}{\overrightarrow {\mathbf {A} }}}={\stackrel {A}{\overrightarrow {\mathbf {A} }}}\cdot \mathbf {A} }$

${\displaystyle {\stackrel {A}{\overrightarrow {\mathrm {cof} (\mathbf {A} )}}}=\mathbf {A} \cdot {\stackrel {A}{\overrightarrow {\mathbf {A} }}}}$

${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {A} ^{-1}}}}\quad =-{\frac {1}{\mathrm {det} (\mathbf {A} )}}\mathbf {A} \cdot {\stackrel {A}{\overrightarrow {\mathbf {A} }}}\quad {\text{falls}}\quad \mathrm {det} (\mathbf {A} )\neq 0}$

${\displaystyle {\stackrel {A}{\overrightarrow {{\vec {v}}\times \mathbf {A} }}}={\frac {1}{2}}(\mathrm {Sp} (\mathbf {A} )\mathbf {1} -\mathbf {A} )\cdot {\vec {v}}={\frac {1}{2}}(\mathbf {A} ^{\top }\#\mathbf {1} )\cdot {\vec {v}}}$

${\displaystyle {\stackrel {A}{\overrightarrow {{\vec {v}}\times \mathbf {1} }}}\;\;={\vec {v}}}$

${\displaystyle {\stackrel {A}{\overrightarrow {({\vec {u}}\times {\vec {v}})\times \mathbf {A} }}}={\frac {1}{2}}({\vec {u}}\cdot \mathbf {A} \times {\vec {v}}-{\vec {v}}\cdot \mathbf {A} \times {\vec {u}})}$

${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {B\cdot A\cdot B} ^{\top }}}}\;\;=\mathrm {cof} (\mathbf {B} )\cdot {\stackrel {A}{\overrightarrow {\mathbf {A} }}}}$

Darin ist „#“ ein #Äußeres Tensorprodukt, cof(·) ist der #Kofaktor.

${\displaystyle {\vec {\mathrm {i} }}(\mathbf {A} ):=\mathbf {A} \cdot \!\!\times \mathbf {1} =\mathbf {A} \times \mathbf {1} ={\stackrel {3}{\mathbf {E} }}:\mathbf {A} =-2{\stackrel {A}{\overrightarrow {\mathbf {A} }}}}$

${\displaystyle {\vec {\mathrm {i} }}(A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})=A_{ij}{\hat {e}}_{i}\times {\hat {e}}_{j}={\begin{pmatrix}A_{23}-A_{32}\\A_{31}-A_{13}\\A_{12}-A_{21}\end{pmatrix}}}$

${\displaystyle {\vec {\mathrm {i} }}(A^{ij}({\vec {a}}_{i}\otimes {\vec {b}}_{j}))=A^{ij}{\vec {a}}_{i}\times {\vec {b}}_{j}}$

Zusammenhang mit dem #Skalarkreuzprodukt von Tensoren:

${\displaystyle \mathbf {A} \cdot \!\!\times \mathbf {B} ={\vec {\mathrm {i} }}(\mathbf {A} \cdot \mathbf {B} )}$

#Symmetrische Tensoren haben keine Vektorinvariante: ${\displaystyle {\vec {\mathrm {i} }}(\mathbf {A} ^{\mathrm {S} })={\vec {0}}}$

Die Eigenschaften des dualen axialen Vektors sind hierher übertragbar. Seien x eine beliebige Zahl, ${\displaystyle {\vec {u}},\,{\vec {v}}}$ beliebige Vektoren und A, B beliebige Tensoren zweiter Stufe. Dann gilt:

${\displaystyle {\vec {\mathrm {i} }}({\vec {u}}\otimes {\vec {v}})\;\;={\vec {u}}\times {\vec {v}}}$

${\displaystyle {\vec {\mathrm {i} }}(\mathbf {A} )\times {\vec {v}}\;=-2\mathbf {A} ^{\mathrm {A} }\cdot {\vec {v}}=(\mathbf {A^{\top }-A} )\cdot {\vec {v}}}$

${\displaystyle {\vec {\mathrm {i} }}(\mathbf {A} ^{\top })\quad \;=-{\vec {\mathrm {i} }}(\mathbf {A} )}$

${\displaystyle {\vec {\mathrm {i} }}(\mathbf {A+B} )={\vec {\mathrm {i} }}(\mathbf {A} )+{\vec {\mathrm {i} }}(\mathbf {B} )}$

${\displaystyle {\vec {\mathrm {i} }}(x\mathbf {A} )\quad \;=x\,{\vec {\mathrm {i} }}(\mathbf {A} )}$

${\displaystyle {\vec {\mathrm {i} }}(\mathbf {A} \#\mathbf {B} )\;=\mathbf {A} \cdot {\vec {\mathrm {i} }}(\mathbf {B} )+\mathbf {B} \cdot {\vec {\mathrm {i} }}(\mathbf {A} )}$

${\displaystyle \mathbf {A} \cdot {\vec {\mathrm {i} }}(\mathbf {A} )\;=\mathbf {A} ^{\top }\cdot {\vec {\mathrm {i} }}(\mathbf {A} )={\vec {\mathrm {i} }}(\mathbf {A} )\cdot \mathbf {A} }$

${\displaystyle {\vec {\mathrm {i} }}(\mathrm {cof} (\mathbf {A} ))=\mathbf {A} \cdot {\vec {\mathrm {i} }}(\mathbf {A} )}$

${\displaystyle {\vec {\mathrm {i} }}(\mathbf {A} ^{-1})\quad =-{\frac {1}{\mathrm {det} (\mathbf {A} )}}\mathbf {A} \cdot {\vec {\mathrm {i} }}(\mathbf {A} )\quad {\text{falls}}\quad \mathrm {det} (\mathbf {A} )\neq 0}$

${\displaystyle {\vec {\mathrm {i} }}({\vec {v}}\times \mathbf {A} )\;=(\mathbf {A} -\mathrm {Sp} (\mathbf {A} )\mathbf {1} )\cdot {\vec {v}}=-(\mathbf {A} ^{\top }\#\mathbf {1} )\cdot {\vec {v}}}$

${\displaystyle {\vec {\mathrm {i} }}({\vec {v}}\times \mathbf {1} )\;\;=-2{\vec {v}}}$

${\displaystyle {\vec {\mathrm {i} }}(({\vec {u}}\times {\vec {v}})\times \mathbf {A} )={\vec {v}}\cdot \mathbf {A} \times {\vec {u}}-{\vec {u}}\cdot \mathbf {A} \times {\vec {v}}}$

${\displaystyle {\vec {\mathrm {i} }}(\mathbf {B\cdot A\cdot B} ^{\top })=\mathrm {cof} (\mathbf {B} )\cdot {\vec {\mathrm {i} }}(\mathbf {A} )}$

Darin ist „#“ ein #Äußeres Tensorprodukt, cof(·) ist der #Kofaktor.

Definition

${\displaystyle \mathbf {A} :={\vec {a}}\otimes {\vec {b}}}$

Kofaktor: ${\displaystyle \mathrm {cof} (\mathbf {A} )=\mathbf {0} }$

#Invarianten:

${\displaystyle \mathrm {Sp} (\mathbf {A} )={\vec {a}}\cdot {\vec {b}}}$

${\displaystyle \mathrm {I} _{2}(\mathbf {A} )=0}$

${\displaystyle \mathrm {det} (\mathbf {A} )=0}$

${\displaystyle \parallel \mathbf {A} \parallel =|{\vec {a}}|\,|{\vec {b}}|}$

#Eigensystem:

${\displaystyle {\begin{aligned}\lambda _{1}=&{\vec {a}}\cdot {\vec {b}},&{\vec {v}}_{1}=&{\frac {\vec {a}}{|{\vec {a}}|}}\\\lambda _{2}=&0,&{\vec {v}}_{2}=&{\frac {{\vec {a}}\times {\vec {b}}}{|{\vec {a}}\times {\vec {b}}|}}\\\lambda _{3}=&0,&{\vec {v}}_{3}=&{\frac {({\vec {a}}\times {\vec {b}})\times {\vec {b}}}{|({\vec {a}}\times {\vec {b}})\times {\vec {b}}|}}\end{aligned}}}$

Gegeben ein beliebiger Tensor 2. Stufe A. Dieser kann immer als Summe dreier Dyaden dargestellt werden:

${\displaystyle {\begin{aligned}\mathbf {A} =A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}=&{\vec {s}}_{j}\otimes {\hat {e}}_{j}={\begin{pmatrix}{\vec {s}}_{1}&{\vec {s}}_{2}&{\vec {s}}_{3}\end{pmatrix}}\\=&{\hat {e}}_{i}\otimes {\vec {z}}_{i}={\begin{pmatrix}{\vec {z}}_{1}&{\vec {z}}_{2}&{\vec {z}}_{3}\end{pmatrix}}^{\top }\\=&{\vec {a}}_{k}\otimes {\vec {g}}_{k}\end{aligned}}}$

mit Spaltenvektoren ${\displaystyle {\vec {s}}_{j}=A_{ij}{\hat {e}}_{i}=\mathbf {A} \cdot {\hat {e}}_{j}}$, Zeilenvektoren ${\displaystyle {\vec {z}}_{i}=A_{ij}{\hat {e}}_{j}={\hat {e}}_{i}\cdot \mathbf {A} }$ und ${\displaystyle {\vec {g}}_{k}=({\vec {a}}^{k}\cdot {\hat {e}}_{i})A_{ij}{\hat {e}}_{j}={\vec {a}}^{k}\cdot \mathbf {A} }$.

#Hauptinvarianten (${\displaystyle x_{m,n}:={\vec {x}}_{m}\cdot {\hat {e}}_{n}}$):

${\displaystyle \mathrm {I} _{1}(\mathbf {A} )=s_{i,i}=z_{i,i}={\vec {a}}_{i}\cdot {\vec {g}}_{i}}$

${\displaystyle {\begin{aligned}\mathrm {I} _{2}(\mathbf {A} )=&{\frac {1}{2}}(s_{i,i}s_{j,j}-s_{i,j}s_{j,i})={\frac {1}{2}}(z_{i,i}z_{j,j}-z_{i,j}z_{j,i})\\=&{\frac {1}{2}}{\big (}({\vec {a}}_{i}\cdot {\vec {g}}_{i})({\vec {a}}_{j}\cdot {\vec {g}}_{j})-({\vec {a}}_{i}\cdot {\vec {g}}_{j})({\vec {a}}_{j}\cdot {\vec {g}}_{i}){\big )}\end{aligned}}}$

${\displaystyle \mathrm {I} _{3}(\mathbf {A} )={\begin{vmatrix}{\vec {s}}_{1}&{\vec {s}}_{2}&{\vec {s}}_{3}\end{vmatrix}}={\begin{vmatrix}{\vec {z}}_{1}&{\vec {z}}_{2}&{\vec {z}}_{3}\end{vmatrix}}={\begin{vmatrix}{\vec {a}}_{1}&{\vec {a}}_{2}&{\vec {a}}_{3}\end{vmatrix}}{\begin{vmatrix}{\vec {g}}_{1}&{\vec {g}}_{2}&{\vec {g}}_{3}\end{vmatrix}}}$

#Betrag:

${\displaystyle \parallel \mathbf {A} \parallel ={\sqrt {{\vec {s}}_{i}\cdot {\vec {s}}_{i}}}={\sqrt {{\vec {z}}_{i}\cdot {\vec {z}}_{i}}}={\sqrt {({\vec {a}}_{i}\cdot {\vec {a}}_{j})({\vec {g}}_{i}\cdot {\vec {g}}_{j})}}}$

#Dualer axialer Vektor:

${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {A} }}}={\frac {1}{2}}{\hat {e}}_{i}\times {\vec {s}}_{i}={\frac {1}{2}}{\vec {z}}_{i}\times {\hat {e}}_{i}={\frac {1}{2}}{\vec {g}}_{i}\times {\vec {a}}_{i}}$

#Vektorinvariante:

${\displaystyle {\vec {\mathrm {i} }}(\mathbf {A} )={\vec {s}}_{i}\times {\hat {e}}_{i}={\hat {e}}_{i}\times {\vec {z}}_{i}={\vec {a}}_{i}\times {\vec {g}}_{i}}$

#Kofaktor:

${\displaystyle {\begin{aligned}\mathrm {cof} (\mathbf {A} )=&{\vec {z}}_{i}\otimes {\vec {s}}_{i}-\mathrm {I} _{1}(\mathbf {A} )\mathbf {A} ^{\top }+\mathrm {I} _{2}(\mathbf {A} )\mathbf {1} \\=&({\vec {a}}_{i}\cdot {\vec {g}}_{j}){\vec {g}}_{i}\otimes {\vec {a}}_{j}-({\vec {a}}_{i}\cdot {\vec {g}}_{i}){\vec {g}}_{j}\otimes {\vec {a}}_{j}+\mathrm {I} _{2}(\mathbf {A} )\mathbf {1} \end{aligned}}}$

#Inverse:

${\displaystyle \mathbf {A} ^{-1}={\hat {e}}_{i}\otimes {\vec {s}}^{i}={\vec {z}}^{i}\otimes {\hat {e}}_{i}={\vec {g}}^{i}\otimes {\vec {a}}^{i}}$

${\displaystyle \mathbf {1} ={\hat {e}}_{i}\otimes {\hat {e}}_{i}=\delta _{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}={\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}}$

${\displaystyle \mathbf {1} ={\vec {g}}_{i}\otimes {\vec {g}}^{i}={\vec {g}}^{i}\otimes {\vec {g}}_{i}=g^{ij}{\vec {g}}_{i}\otimes {\vec {g}}_{j}=g_{ij}{\vec {g}}^{i}\otimes {\vec {g}}^{j}}$

mit ${\displaystyle g_{ij}={\vec {g}}_{i}\cdot {\vec {g}}_{j}\,,\;g^{ij}={\vec {g}}^{i}\cdot {\vec {g}}^{j}}$

Allgemein:

${\displaystyle \mathbf {1} =({\vec {a}}^{i}\cdot {\vec {g}}^{j}){\vec {a}}_{i}\otimes {\vec {g}}_{j}}$

#Transposition und #Inverse:

${\displaystyle \mathbf {1} =\mathbf {1} ^{\top }=\mathbf {1} ^{-1}=\mathbf {1} ^{\top -1}}$

Kofaktor: ${\displaystyle \mathrm {cof} (\mathbf {1} )=\mathbf {1} }$

Vektortransformation

${\displaystyle \mathbf {1} \cdot {\vec {v}}={\vec {v}}\cdot \mathbf {1} ={\vec {v}}}$

Tensorprodukt

${\displaystyle \mathbf {A\cdot 1} =\mathbf {1\cdot A} =\mathbf {A} }$

Skalarprodukt

${\displaystyle \mathbf {A} :\mathbf {1} =\mathrm {Sp} (\mathbf {A} )}$

#Invarianten:

${\displaystyle \mathrm {Sp} (\mathbf {1} )=\mathbf {1} :\mathbf {1} =3}$

${\displaystyle \mathrm {I} _{2}(\mathbf {1} )=3}$

${\displaystyle \mathrm {det} (\mathbf {1} )=1}$

${\displaystyle \parallel \mathbf {1} \parallel ={\sqrt {3}}}$

#Eigenwerte:

${\displaystyle \lambda _{1,2,3}=1}$

Alle Vektoren sind #Eigenvektoren.

Definition

${\displaystyle \mathbf {H} :\quad \mathrm {det} (\mathbf {H} )=1}$

Kofaktor: ${\displaystyle \mathrm {cof} (\mathbf {H} )=\mathbf {H} ^{\top -1}}$

Determinantenproduktsatz:

${\displaystyle \mathrm {det} (\mathbf {A\cdot H} )=\mathrm {det} (\mathbf {H\cdot A} )=\mathrm {det} (\mathbf {A} )}$

Definition

${\displaystyle \mathbf {Q} :\quad \mathbf {Q} ^{-1}=\mathbf {Q} ^{\top }\quad {\textsf {oder}}\quad \mathbf {Q\cdot Q} ^{\top }=\mathbf {Q} ^{\top }\cdot \mathbf {Q} =\mathbf {1} }$

Kofaktor: ${\displaystyle \mathrm {cof} (\mathbf {Q} )=\mathrm {det} (\mathbf {Q} )\mathbf {Q} =\pm \mathbf {Q} }$

#Invarianten (${\displaystyle \alpha }$ ist der Drehwinkel):

${\displaystyle \mathrm {Sp} (\mathbf {Q} )=\mathrm {det} (\mathbf {Q} )+2\cos(\alpha )}$

${\displaystyle \mathrm {I} _{2}(\mathbf {Q} )=\mathrm {det} (\mathbf {Q} )\cdot \mathrm {Sp} (\mathbf {Q} )=1+2\,\mathrm {det} (\mathbf {Q} )\cos(\alpha )}$

${\displaystyle \mathrm {det} (\mathbf {Q} )=\pm 1}$

${\displaystyle \parallel \mathbf {Q} \parallel ={\sqrt {3}}}$

Eigentlich orthogonaler Tensor ${\displaystyle \mathrm {det} (\mathbf {Q} )=+1}$, entspricht einer Drehung und ist Element der Drehgruppe SO(3).

Uneigentlich orthogonaler Tensor ${\displaystyle \mathrm {det} (\mathbf {Q} )=-1}$, entspricht einer Drehspiegelung.

Spatprodukt:

${\displaystyle (\mathbf {Q} \cdot {\vec {a}})\cdot {\big (}(\mathbf {Q} \cdot {\vec {b}})\times (\mathbf {Q} \cdot {\vec {c}}){\big )}=\mathrm {det} (\mathbf {Q} ){\vec {a}}\cdot ({\vec {b}}\times {\vec {c}})}$

Kreuzprodukt und #Kofaktor:

${\displaystyle (\mathbf {Q} \cdot {\vec {a}})\times (\mathbf {Q} \cdot {\vec {b}})=\mathrm {det} (\mathbf {Q} )\mathbf {Q} \cdot ({\vec {a}}\times {\vec {b}})}$

${\displaystyle \mathrm {cof} (\mathbf {Q} )=\mathrm {det} (\mathbf {Q} )\mathbf {Q} }$

Gegeben ein Einheitsvektor ${\displaystyle {\hat {n}}={\begin{pmatrix}n_{1}&n_{2}&n_{3}\end{pmatrix}}^{\top }}$ und Drehwinkel α. Dann sind die folgenden Tensoren R zueinander gleich, orthogonal und drehen um die Achse ${\displaystyle {\hat {n}}}$ mit Winkel α:

Rodrigues-Formel:

${\displaystyle {\begin{aligned}\mathbf {R} =&\mathbf {1} +s_{\alpha }{\hat {n}}\times \mathbf {1} +d_{\alpha }({\hat {n}}\times \mathbf {1} )^{2}\\=&\mathbf {1} +s_{\alpha }{\hat {n}}\times \mathbf {1} +d_{\alpha }({\hat {n}}\otimes {\hat {n}}-\mathbf {1} )\end{aligned}}}$

${\displaystyle \mathbf {R} ={\begin{pmatrix}c_{\alpha }+d_{\alpha }n_{1}^{2}&-s_{\alpha }n_{3}+d_{\alpha }n_{1}n_{2}&s_{\alpha }n_{2}+d_{\alpha }n_{1}n_{3}\\s_{\alpha }n_{3}+d_{\alpha }n_{1}n_{2}&c_{\alpha }+d_{\alpha }n_{2}^{2}&-s_{\alpha }n_{1}+d_{\alpha }n_{2}n_{3}\\-s_{\alpha }n_{2}+d_{\alpha }n_{1}n_{3}&s_{\alpha }n_{1}+d_{\alpha }n_{2}n_{3}&c_{\alpha }+d_{\alpha }n_{3}^{2}\end{pmatrix}}}$

mit ${\displaystyle c_{\alpha }=\cos(\alpha ),\;d_{\alpha }=1-\cos(\alpha ),\;s_{\alpha }=\sin(\alpha )}$.

Euler-Rodrigues-Formel: ${\displaystyle a=\cos \left({\tfrac {\alpha }{2}}\right),b=\sin \left({\tfrac {\alpha }{2}}\right)n_{1},c=\sin \left({\tfrac {\alpha }{2}}\right)n_{2},d=\sin \left({\tfrac {\alpha }{2}}\right)n_{3}}$ also ${\displaystyle a^{2}+b^{2}+c^{2}+d^{2}=1}$:

${\displaystyle \mathbf {R} :={\begin{pmatrix}a^{2}+b^{2}-c^{2}-d^{2}&2(bc-ad)&2(bd+ac)\\2(bc+ad)&a^{2}+c^{2}-b^{2}-d^{2}&2(cd-ab)\\2(bd-ac)&2(cd+ab)&a^{2}+d^{2}-b^{2}-c^{2}\end{pmatrix}}}$

Formulierung mit Drehvektor:

Drehvektor		Orthogonaler Tensor
${\displaystyle {\vec {\alpha }}=\alpha {\vec {n}}}$	→	${\displaystyle \mathbf {R} =\mathbf {1} +{\frac {\sin(\alpha )}{\alpha }}{\vec {\alpha }}\times \mathbf {1} +{\frac {1-\cos(\alpha )}{\alpha ^{2}}}({\vec {\alpha }}\times \mathbf {1} )^{2}}$
${\displaystyle {\vec {\alpha }}=\tan(\alpha ){\vec {n}}}$	→	${\displaystyle \mathbf {R} =\mathbf {1} +\cos(\alpha ){\vec {\alpha }}\times \mathbf {1} +{\frac {\cos ^{2}(\alpha )}{1+\cos(\alpha )}}({\vec {\alpha }}\times \mathbf {1} )^{2}}$
${\displaystyle {\vec {\alpha }}=\tan \left({\frac {\alpha }{2}}\right)\;{\vec {n}}}$	→	${\displaystyle \mathbf {R} =\mathbf {1} +{\frac {2}{1+{\vec {\alpha }}\cdot {\vec {\alpha }}}}({\vec {\alpha }}\times \mathbf {1} +({\vec {\alpha }}\times \mathbf {1} )^{2})}$
${\displaystyle {\vec {\alpha }}=\sin(\alpha )\;{\vec {n}}}$	→	${\displaystyle \mathbf {R} =\mathbf {1} +{\vec {\alpha }}\times \mathbf {1} +{\dfrac {1}{1+\cos(\alpha )}}({\vec {\alpha }}\times \mathbf {1} )^{2}}$
${\displaystyle {\vec {\alpha }}=\sin \left({\frac {\alpha }{2}}\right)\;{\vec {n}}}$	→	${\displaystyle \mathbf {R} =\mathbf {1} +2\cos \left({\frac {\alpha }{2}}\right){\vec {\alpha }}\times \mathbf {1} +2({\vec {\alpha }}\times \mathbf {1} )^{2}}$
${\displaystyle {\vec {\alpha }}=\cos(\alpha )\;{\vec {n}}}$	→	${\displaystyle \mathbf {R} =\mathbf {1} +\tan(\alpha ){\vec {\alpha }}\times \mathbf {1} +{\frac {1-\cos(\alpha )}{{\vec {\alpha }}\cdot {\vec {\alpha }}}}({\vec {\alpha }}\times \mathbf {1} )^{2}}$
${\displaystyle {\vec {\alpha }}=\cos \left({\frac {\alpha }{2}}\right)\;{\vec {n}}}$	→	${\displaystyle \mathbf {R} =\mathbf {1} +2\sin \left({\frac {\alpha }{2}}\right){\vec {\alpha }}\times \mathbf {1} +2{\frac {1-{\vec {\alpha }}\cdot {\vec {\alpha }}}{{\vec {\alpha }}\cdot {\vec {\alpha }}}}({\vec {\alpha }}\times \mathbf {1} )^{2}}$

Darin ist ${\displaystyle ({\vec {\alpha }}\times \mathbf {1} )^{2}=({\vec {\alpha }}\times \mathbf {1} )\cdot ({\vec {\alpha }}\times \mathbf {1} )={\vec {\alpha }}\otimes {\vec {\alpha }}-({\vec {\alpha }}\cdot {\vec {\alpha }})\mathbf {1} }$

Beispiel für Drehspiegelung:

${\displaystyle \mathbf {Q} =-\mathbf {1} +\sin(\alpha ){\hat {n}}\times \mathbf {1} -(1+\cos(\alpha ))({\hat {n}}\times \mathbf {1} )^{2}}$

Drehung von Vektorraumbasis ${\displaystyle {\vec {u}}_{1,2,3}\;{\textsf {nach}}\;{\vec {v}}_{1,2,3}}$ mit Drehachse ${\displaystyle {\hat {n}}}$:

${\displaystyle \mathbf {Q} \cdot {\vec {u}}_{i}={\vec {v}}_{i}\,,\quad \mathbf {Q} \cdot {\vec {u}}^{i}={\vec {v}}^{i}\,,\quad \mathbf {Q} ={\vec {v}}_{i}\otimes {\vec {u}}^{i}={\vec {v}}^{i}\otimes {\vec {u}}_{i}}$

${\displaystyle {\hat {n}}\simeq {\vec {v}}_{i}\times {\vec {u}}^{i}={\vec {v}}^{i}\times {\vec {u}}_{i}=-2{\stackrel {A}{\overrightarrow {\mathbf {Q} }}}={\vec {\mathrm {i} }}(\mathbf {Q} )}$

mit #Dualer axialer Vektor ${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {Q} }}}}$ und #Vektorinvariante ${\displaystyle {\vec {\mathrm {i} }}(\mathbf {Q} )}$.

Gegeben Orthonormalbasis ${\displaystyle {\hat {v}}_{1,2,3}}$, Drehwinkel ${\displaystyle \alpha }$ und ${\displaystyle {\hat {v}}_{1}}$ ist Drehachse:

${\displaystyle {\begin{aligned}\mathbf {Q} =&{\color {red}\pm }{\hat {v}}_{1}\otimes {\hat {v}}_{1}+\cos(\alpha )({\hat {v}}_{2}\otimes {\hat {v}}_{2}+{\hat {v}}_{3}\otimes {\hat {v}}_{3})+\sin(\alpha )({\hat {v}}_{3}\otimes {\hat {v}}_{2}-{\hat {v}}_{2}\otimes {\hat {v}}_{3})\\=&{\begin{pmatrix}{\color {red}\pm 1}&0&0\\0&\cos(\alpha )&-\sin(\alpha )\\0&\sin(\alpha )&\cos(\alpha )\end{pmatrix}}_{{\hat {v}}_{i}\otimes {\hat {v}}_{j}}\end{aligned}}}$

${\displaystyle {\color {red}+1}}$: Drehung, ${\displaystyle {\color {red}-1}}$: Drehspiegelung um ${\displaystyle {\hat {v}}_{1}}$

Wenn ${\displaystyle {\hat {v}}_{1,2,3}}$ ein Rechtssystem (Mathematik) bilden, dann dreht Q gegen den Uhrzeigersinn, sonst im Uhrzeigersinn um die Drehachse.

#Eigensystem:

${\displaystyle {\begin{aligned}\lambda _{1}=&\mathrm {det} (\mathbf {Q} )\,,&{\vec {q}}_{1}=&{\hat {v}}_{1}\\\lambda _{2}=&e^{\mathrm {i} \alpha },&{\vec {q}}_{2}=&{\frac {1}{\sqrt {2}}}({\hat {v}}_{2}-\mathrm {i} {\hat {v}}_{3}).\\\lambda _{3}=&e^{-\mathrm {i} \alpha },&{\vec {q}}_{3}=&{\frac {1}{\sqrt {2}}}({\hat {v}}_{2}+\mathrm {i} {\hat {v}}_{3})\end{aligned}}}$

Drehwinkel:

${\displaystyle \cos(\alpha )={\frac {1}{2}}(\mathrm {Sp} (\mathbf {Q} )-\mathrm {det} (\mathbf {Q} ))}$

Drehachse ${\displaystyle {\hat {n}}}$ ist #Vektorinvariante:

${\displaystyle {\hat {n}}\simeq {\vec {\mathrm {i} }}(\mathbf {Q} )=\mathbf {1} \cdot \!\!\times \mathbf {Q} }$

${\displaystyle \mathbf {Q} ={\vec {s}}_{i}\otimes {\vec {e}}_{i}={\vec {e}}_{i}\otimes {\vec {z}}_{i}\quad \rightarrow \quad {\hat {n}}\simeq {\vec {s}}_{i}\times {\vec {e}}_{i}={\vec {e}}_{i}\times {\vec {z}}_{i}}$

${\displaystyle {\frac {1}{2}}(\mathbf {Q} -\mathbf {Q} ^{\top })=\sin(\alpha ){\hat {n}}\times \mathbf {1} =\sin(\alpha ){\begin{pmatrix}0&-n_{3}&n_{2}\\n_{3}&0&-n_{1}\\-n_{2}&n_{1}&0\end{pmatrix}},\quad |{\hat {n}}|=1}$

Definition

${\displaystyle \mathbf {A} :\quad {\vec {v}}\cdot \mathbf {A} \cdot {\vec {v}}>0\quad \forall \;{\vec {v}}\in \mathbb {V} \setminus \{{\vec {0}}\}}$

Kofaktor: ${\displaystyle \mathrm {cof} (\mathbf {A} )=\mathbf {A^{\top }\cdot A^{\top }} -\mathrm {I} _{1}(\mathbf {A} )\mathbf {A} ^{\top }+\mathrm {I} _{2}(\mathbf {A} )\mathbf {1} =\det(\mathbf {A} )\mathbf {A} ^{\top -1}}$

Notwendige Bedingungen für positive Definitheit:

${\displaystyle \mathrm {det} (\mathbf {A} )>0}$

${\displaystyle \mathbf {A} =A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}\quad \rightarrow \quad A_{11},\,A_{22},\,A_{33}>0}$

${\displaystyle \mathbf {A} =A_{j}^{i}{\vec {a}}_{i}\otimes {\vec {a}}^{j}\quad \rightarrow \quad A_{1}^{1},\,A_{2}^{2},\,A_{3}^{3}>0}$

Notwendige und hinreichende Bedingung für positive Definitheit: Alle #Eigenwerte von A sind größer als null.

Immer positiv definit falls det(A) ≠ 0:

A·A^⊤ und A^⊤·A

Definition

${\displaystyle \mathbf {A} :\quad \mathbf {A} =\mathbf {A} ^{\top }}$

Kofaktor: ${\displaystyle \mathrm {cof} (\mathbf {A} )=\mathrm {adj} (\mathbf {A} )=\mathbf {A} ^{2}-\mathrm {Sp} (\mathbf {A} )\mathbf {A} +\mathrm {I} _{2}(\mathbf {A} )\mathbf {1} }$

#Betrag:

${\displaystyle \quad \parallel \mathbf {A} \parallel ={\sqrt {\mathrm {Sp} ^{2}(\mathbf {A} )-2\mathrm {I} _{2}(\mathbf {A} )}}={\sqrt {\mathrm {Sp} (\mathbf {A} ^{2})}}={\sqrt {\lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}}}}$

Bei Symmetrischen Tensoren verschwinden ihr #Dualer axialer Vektor und ihre #Vektorinvariante:

${\displaystyle {\stackrel {A}{\overrightarrow {\mathbf {A} ^{\mathrm {S} }}}}={\vec {\mathrm {i} }}(\mathbf {A} ^{\mathrm {S} })={\vec {0}}}$

Bilinearform:

${\displaystyle {\vec {u}}\cdot \mathbf {A} \cdot {\vec {v}}={\vec {v}}\cdot \mathbf {A} \cdot {\vec {u}}\quad \forall {\vec {u}},{\vec {v}}\in \mathbb {V} }$

Alle #Eigenwerte λ_1,2,3 sind reell. Alle #Eigenvektoren ${\displaystyle {\vec {a}}_{1,2,3}}$ sind reell und paarweise orthogonal zueinander oder orthogonalisierbar. Hauptachsentransformation:

${\displaystyle {\begin{aligned}\mathbf {A} =&\sum _{i=1}^{3}\lambda _{i}{\hat {a}}_{i}\otimes {\hat {a}}_{i}=({\hat {a}}_{i}\otimes {\hat {e}}_{i})\left(\sum _{i=1}^{3}\lambda _{j}{\hat {e}}_{j}\otimes {\hat {e}}_{j}\right)({\hat {e}}_{k}\otimes {\hat {a}}_{k})\\=&{\begin{pmatrix}{\hat {a}}_{1}&{\hat {a}}_{2}&{\hat {a}}_{3}\end{pmatrix}}\cdot {\begin{pmatrix}\lambda _{1}&0&0\\0&\lambda _{2}&0\\0&0&\lambda _{3}\end{pmatrix}}\cdot {\begin{pmatrix}{\hat {a}}_{1}&{\hat {a}}_{2}&{\hat {a}}_{3}\end{pmatrix}}^{\top }\end{aligned}}}$

Bezüglich der Standardbasis:

${\displaystyle \mathbf {A} =A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}={\begin{pmatrix}A_{11}&A_{12}&A_{13}\\A_{12}&A_{22}&A_{23}\\A_{13}&A_{23}&A_{33}\end{pmatrix}}}$

#Invarianten:

${\displaystyle \mathrm {Sp} (A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})=A_{11}+A_{22}+A_{33}}$

${\displaystyle \mathrm {I} _{2}(A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})=A_{11}A_{22}+A_{11}A_{33}+A_{22}A_{33}-A_{12}^{2}-A_{13}^{2}-A_{23}^{2}}$

${\displaystyle {\begin{aligned}\mathrm {det} (A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})=&A_{11}(A_{22}A_{33}-A_{23}^{2})+A_{12}(A_{23}A_{13}-A_{12}A_{33})\\&+A_{13}(A_{12}A_{23}-A_{13}A_{22})\end{aligned}}}$

${\displaystyle \parallel A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}\parallel ={\sqrt {A_{11}^{2}+A_{22}^{2}+A_{33}^{2}+2A_{12}^{2}+2A_{13}^{2}+2A_{23}^{2}}}}$

Definition

${\displaystyle \mathbf {A} :\quad \mathbf {A} =\mathbf {A} ^{\top }\quad {\text{und}}\quad {\vec {v}}\cdot \mathbf {A} \cdot {\vec {v}}>0\quad \forall \;{\vec {v}}\in \mathbb {V} \setminus \{{\vec {0}}\}}$

Kofaktor: ${\displaystyle \mathrm {cof} (\mathbf {A} )=\mathrm {adj} (\mathbf {A} )=\mathrm {det} (\mathbf {A} )\mathbf {A} ^{-1}=\mathbf {A} ^{2}-\mathrm {Sp} (\mathbf {A} )\mathbf {A} +\mathrm {I} _{2}(\mathbf {A} )\mathbf {1} }$

Mit den #Eigenwerten ${\displaystyle \lambda _{1},\,\lambda _{2},\,\lambda _{3}}$, den #Eigenvektoren ${\displaystyle {\hat {a}}_{1},\,{\hat {a}}_{2},\,{\hat {a}}_{3}}$ und einer reellwertigen Funktion ${\displaystyle f(x)\in \mathbb {R} }$ eines reellen Argumentes ${\displaystyle x\in \mathbb {R} }$ definiert man über das #Eigensystem symmetrischer Tensoren

${\displaystyle {\begin{aligned}\mathbf {A} =&\sum _{i=1}^{3}\lambda _{i}{\hat {a}}_{i}\otimes {\hat {a}}_{i}\\=&{\begin{pmatrix}{\hat {a}}_{1}&{\hat {a}}_{2}&{\hat {a}}_{3}\end{pmatrix}}\cdot {\begin{pmatrix}\lambda _{1}&0&0\\0&\lambda _{2}&0\\0&0&\lambda _{3}\end{pmatrix}}\cdot {\begin{pmatrix}{\hat {a}}_{1}&{\hat {a}}_{2}&{\hat {a}}_{3}\end{pmatrix}}^{\top }\end{aligned}}}$

den Funktionswert des Tensors:

${\displaystyle {\begin{aligned}f(\mathbf {A} ):=&\sum _{i=1}^{3}f(\lambda _{i}){\hat {a}}_{i}\otimes {\hat {a}}_{i}\\=&{\begin{pmatrix}{\hat {a}}_{1}&{\hat {a}}_{2}&{\hat {a}}_{3}\end{pmatrix}}\cdot {\begin{pmatrix}f(\lambda _{1})&0&0\\0&f(\lambda _{2})&0\\0&0&f(\lambda _{3})\end{pmatrix}}\cdot {\begin{pmatrix}{\hat {a}}_{1}&{\hat {a}}_{2}&{\hat {a}}_{3}\end{pmatrix}}^{\top }\end{aligned}}}$

Ist f eine mehrdeutige Funktion, wie die Wurzel (Mathematik), mit n alternativen Werten, dann steht f(A) mehrdeutig für n³ alternative Tensoren.

Insbesondere mit dem Deformationsgradient F:

Rechter Strecktensor

${\displaystyle \mathbf {U} =+{\sqrt {\mathbf {F} ^{\top }\cdot \mathbf {F} }}}$

Linker Strecktensor

${\displaystyle \mathbf {v} =+{\sqrt {\mathbf {F\cdot F} ^{\top }}}}$

Henky-Dehnung

${\displaystyle \mathbf {E} _{H}:=\ln(\mathbf {U} )={\frac {1}{2}}\ln(\mathbf {F} ^{\top }\cdot \mathbf {F} )}$

Die Tensoren

${\displaystyle {\begin{aligned}\mathbf {S} _{1}=&{\vec {e}}_{1}\otimes {\vec {e}}_{1}\\\mathbf {S} _{2}=&{\vec {e}}_{2}\otimes {\vec {e}}_{2}\\\mathbf {S} _{3}=&{\vec {e}}_{3}\otimes {\vec {e}}_{3}\\\mathbf {S} _{4}=&{\vec {e}}_{2}\otimes {\vec {e}}_{3}+{\vec {e}}_{3}\otimes {\vec {e}}_{2}\\\mathbf {S} _{5}=&{\vec {e}}_{1}\otimes {\vec {e}}_{3}+{\vec {e}}_{3}\otimes {\vec {e}}_{1}\\\mathbf {S} _{6}=&{\vec {e}}_{1}\otimes {\vec {e}}_{2}+{\vec {e}}_{2}\otimes {\vec {e}}_{1}\end{aligned}}}$

bilden eine Basis im Vektorraum ${\displaystyle \mathrm {sym} (\mathbb {V} ,\mathbb {V} )\subset {\mathcal {L}}}$ der symmetrischen Tensoren zweiter Stufe. Bezüglich dieser Basis können alle symmetrischen Tensoren zweiter Stufe in Voigt'scher Notation dargestellt werden:

${\displaystyle \mathbf {A} \in \mathrm {sym} (\mathbb {V} ,\mathbb {V} )\quad \rightarrow \quad \mathbf {A} =A_{r}\mathbf {S} _{r}{\hat {=}}{\begin{pmatrix}A_{1}\\A_{2}\\A_{3}\\A_{4}\\A_{5}\\A_{6}\end{pmatrix}}}$

Diese Vektoren dürfen addiert, subtrahiert und mit einem Skalar multipliziert werden. Beim Skalarprodukt muss

${\displaystyle \mathbf {A} :\mathbf {B} =A_{1}B_{1}+A_{2}B_{2}+A_{3}B_{3}+2A_{4}B_{4}+2A_{5}B_{5}+2A_{6}B_{6}}$

berücksichtigt werden. Siehe auch #Voigt'sche Notation von Tensoren vierter Stufe.

Definition

${\displaystyle \mathbf {A} :\quad \mathbf {A} =-\mathbf {A} ^{\top }}$

Kofaktor: ${\displaystyle \mathrm {cof} (\mathbf {A} )=\mathrm {adj} (\mathbf {A} )=\mathbf {A\cdot A} +\mathrm {I} _{2}(\mathbf {A} )\mathbf {1} =\mathbf {A\cdot A} -{\frac {1}{2}}\mathrm {Sp} (\mathbf {A} ^{2})\mathbf {1} }$

#Invarianten:

${\displaystyle \mathrm {Sp} (\mathbf {A} )=0}$

${\displaystyle \mathrm {I} _{2}(\mathbf {A} )=-{\frac {1}{2}}\mathrm {Sp} (\mathbf {A} ^{2})}$

${\displaystyle \mathrm {det} (\mathbf {A} )=0}$

${\displaystyle \quad \parallel \mathbf {A} \parallel ={\sqrt {2\mathrm {I} _{2}(\mathbf {A} )}}={\sqrt {-\mathrm {Sp} (\mathbf {A} ^{2})}}}$

In kartesischen Koordinaten:

${\displaystyle \mathbf {A} =A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}={\begin{pmatrix}0&A_{12}&A_{13}\\-A_{12}&0&A_{23}\\-A_{13}&-A_{23}&0\end{pmatrix}}}$

#Invarianten:

${\displaystyle \mathrm {Sp} (A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})=0}$

${\displaystyle \mathrm {I} _{2}(A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})=A_{12}^{2}+A_{13}^{2}+A_{23}^{2}}$

${\displaystyle \mathrm {det} (A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})=0}$

${\displaystyle \parallel A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}\parallel ={\sqrt {2}}{\sqrt {A_{12}^{2}+A_{13}^{2}+A_{23}^{2}}}}$

Bilinearform:

${\displaystyle {\vec {u}}\cdot \mathbf {A} \cdot {\vec {v}}=-{\vec {v}}\cdot \mathbf {A} \cdot {\vec {u}}\quad \forall {\vec {u}},{\vec {v}}\in \mathbb {V} }$

${\displaystyle {\vec {v}}\cdot \mathbf {A} \cdot {\vec {v}}=0\quad \forall {\vec {v}}\in \mathbb {V} }$

Ein Eigenwert ist null, zwei imaginär konjugiert komplex, siehe #Axialer Tensor oder Kreuzproduktmatrix.

#Dualer axialer Vektor:

${\displaystyle {\begin{aligned}&\mathbf {A} _{\times }:={\stackrel {A}{\overrightarrow {\mathbf {A} }}}:=-{\frac {1}{2}}\mathbf {1} \times \mathbf {A} ^{\top }=-{\frac {1}{2}}\mathbf {1} \cdot \!\!\times \mathbf {A} =-{\frac {1}{2}}{\vec {\mathrm {i} }}(\mathbf {A} )\\&\rightarrow \quad \mathbf {A} \cdot {\vec {v}}={\stackrel {A}{\overrightarrow {\mathbf {A} }}}\times {\vec {v}}\quad \forall {\vec {v}}\in \mathbb {V} \end{aligned}}}$

mit #Vektorinvariante ${\displaystyle {\vec {\mathrm {i} }}(\mathbf {A} )}$. Der zum Eigenwert null gehörende #Eigenvektor ist proportional zum dualen axialen Vektor ${\displaystyle \mathbf {A} _{\times }}$ denn

${\displaystyle \mathbf {A\cdot A} _{\times }=\mathbf {A} _{\times }\times \mathbf {A} _{\times }={\vec {0}}}$

${\displaystyle \mathbf {A} =A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}\quad \rightarrow \;\mathbf {A} _{\times }=-{\frac {1}{2}}A_{ij}{\hat {e}}_{i}\times {\hat {e}}_{j}={\begin{pmatrix}-A_{23}\\A_{13}\\-A_{12}\end{pmatrix}}}$

${\displaystyle \mathbf {A} =A_{ij}({\vec {a}}_{i}\otimes {\vec {b}}_{j}-{\vec {b}}_{j}\otimes {\vec {a}}_{i})\quad \rightarrow \;\mathbf {A} _{\times }=-A_{ij}{\vec {a}}_{i}\times {\vec {b}}_{j}}$

Kreuzproduktmatrix ${\displaystyle [{\vec {u}}]_{\times }}$ eines Vektors ${\displaystyle {\vec {u}}}$:

${\displaystyle {\begin{aligned}{\vec {u}}=u_{i}{\hat {e}}_{i}=&{\begin{pmatrix}u_{1}\\u_{2}\\u_{3}\end{pmatrix}}\\\rightarrow \;[{\vec {u}}]_{\times }=&{\vec {u}}\times \mathbf {1} ={\vec {u}}\times {\hat {e}}_{i}\otimes {\hat {e}}_{i}=-{\stackrel {3}{\mathbf {E} }}\cdot {\vec {u}}={\begin{pmatrix}0&-u_{3}&u_{2}\\u_{3}&0&-u_{1}\\-u_{2}&u_{1}&0\end{pmatrix}}\in {\mathcal {L}}\end{aligned}}}$

Kofaktor: ${\displaystyle \mathrm {cof} ([{\vec {u}}]_{\times })=\mathrm {cof} ({\vec {u}}\times \mathbf {1} )=\mathrm {adj} ({\vec {u}}\times \mathbf {1} )={\vec {u}}\otimes {\vec {u}}}$

#Invarianten:

${\displaystyle \mathrm {Sp} ([{\vec {u}}]_{\times })=0}$

${\displaystyle \mathrm {I} _{2}([{\vec {u}}]_{\times })={\vec {u}}\cdot {\vec {u}}=u_{1}^{2}+u_{2}^{2}+u_{3}^{2}}$

${\displaystyle \mathrm {det} ([{\vec {u}}]_{\times })=0}$

${\displaystyle \|[{\vec {u}}]_{\times }\|=\|{\vec {u}}\times \mathbf {1} \|={\sqrt {2{\vec {u}}\cdot {\vec {u}}}}={\sqrt {2}}{\sqrt {u_{1}^{2}+u_{2}^{2}+u_{3}^{2}}}}$

${\displaystyle {\stackrel {A}{\overrightarrow {{\vec {u}}\times \mathbf {1} }}}={\vec {u}}}$

#Eigensystem:

${\displaystyle {\begin{aligned}\lambda _{1}=&0\,,&{\vec {v}}_{1}=&{\vec {u}}\\\lambda _{2,3}=&\mp \mathrm {i} |{\vec {u}}|\,,&{\vec {v}}_{2,3}&\simeq &{\frac {u_{1}}{|{\vec {u}}|}}{\begin{pmatrix}u_{1}\\u_{2}\\u_{3}\end{pmatrix}}\pm \mathrm {i} {\begin{pmatrix}\pm \mathrm {i} |{\vec {u}}|\\-u_{3}\\u_{2}\end{pmatrix}}\end{aligned}}}$

Eigenschaften:

${\displaystyle {\vec {u}}\times {\vec {v}}=({\vec {u}}\times \mathbf {1} )\cdot {\vec {v}}={\vec {u}}\cdot ({\vec {v}}\times \mathbf {1} )}$

${\displaystyle {\vec {u}}\times \mathbf {1} =\mathbf {1} \times {\vec {u}}}$

${\displaystyle ({\vec {u}}\times \mathbf {1} )^{\top }=-{\vec {u}}\times \mathbf {1} }$

${\displaystyle {\vec {u}}=-{\frac {1}{2}}\mathbf {1} \cdot \!\!\times ({\vec {u}}\times \mathbf {1} )=-{\frac {1}{2}}(\mathbf {1} \times {\vec {u}})\times \mathbf {1} ={\frac {1}{2}}\mathbf {1} \times ({\vec {u}}\times \mathbf {1} )}$

${\displaystyle {\vec {u}}\times ({\vec {v}}\times \mathbf {1} )=({\vec {u}}\times \mathbf {1} )\cdot ({\vec {v}}\times \mathbf {1} )={\vec {v}}\otimes {\vec {u}}-({\vec {u}}\cdot {\vec {v}})\mathbf {1} }$

${\displaystyle {\vec {u}}\times ({\vec {v}}\times \mathbf {1} )\cdot {\vec {w}}={\vec {u}}\times ({\vec {v}}\times {\vec {w}})=({\vec {u}}\cdot {\vec {w}}){\vec {v}}-({\vec {u}}\cdot {\vec {v}}){\vec {w}}}$

Potenzen von ${\displaystyle [{\vec {u}}]_{\times }={\vec {u}}\times \mathbf {1} }$

${\displaystyle [{\vec {u}}]_{\times }^{2}=[{\vec {u}}]_{\times }\cdot [{\vec {u}}]_{\times }={\vec {u}}\otimes {\vec {u}}-({\vec {u}}\cdot {\vec {u}})\mathbf {1} }$

${\displaystyle [{\vec {u}}]_{\times }^{3}=-({\vec {u}}\cdot {\vec {u}})[{\vec {u}}]_{\times }}$

Definition

${\displaystyle \mathbf {A} :\quad \mathrm {Sp} (\mathbf {A} )=0}$

Kofaktor: ${\displaystyle \mathrm {cof} (\mathbf {A} )=\left(\mathbf {A} ^{2}\right)^{\top }+\mathrm {I} _{2}(\mathbf {A} )\mathbf {1} =\left(\mathbf {A} ^{2}\right)^{\top }-{\frac {1}{2}}\mathrm {Sp} (\mathbf {A} ^{2})\mathbf {1} }$

#Hauptinvarianten:

${\displaystyle \mathrm {Sp} (\mathbf {A} ):=0}$

${\displaystyle \mathrm {I} _{2}(\mathbf {A} )=-{\frac {1}{2}}\mathrm {Sp} (\mathbf {A} ^{2})}$

${\displaystyle \mathrm {det} (\mathbf {A} )={\frac {1}{3}}\mathrm {Sp} (\mathbf {A} ^{3})}$

Bezüglich der Standardbasis:

${\displaystyle \mathbf {A} =A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}={\begin{pmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&-A_{11}-A_{22}\end{pmatrix}}}$

${\displaystyle \mathrm {Sp} (A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})=0}$

${\displaystyle \mathrm {I} _{2}(A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})=-A_{11}^{2}-A_{22}^{2}-A_{11}A_{22}-A_{12}A_{21}-A_{13}A_{31}-A_{23}A_{32}}$

${\displaystyle {\begin{aligned}\mathrm {det} (A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j})=&-A_{11}(A_{11}A_{22}+A_{22}^{2}+A_{23}A_{32})\\&+A_{12}(A_{23}A_{31}+A_{21}A_{11}+A_{21}A_{22})\\&+A_{13}(A_{21}A_{32}-A_{22}A_{31})\end{aligned}}}$

${\displaystyle \parallel A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}\parallel ={\sqrt {\begin{array}{r}2A_{11}^{2}+2A_{22}^{2}+2A_{11}A_{22}+A_{12}^{2}+A_{21}^{2}+\ldots \\\ldots +A_{13}^{2}+A_{31}^{2}+A_{23}^{2}+A_{32}^{2}\end{array}}}}$

Definition

${\displaystyle \mathbf {A} :\quad \mathbf {A} =a\mathbf {1} ={\begin{pmatrix}a&0&0\\0&a&0\\0&0&a\end{pmatrix}}}$

Kofaktor: ${\displaystyle \mathrm {cof} (\mathbf {A} )=\mathrm {adj} (\mathbf {A} )=a^{2}\mathbf {1} }$

${\displaystyle \mathrm {Sp} (\mathbf {A} )=3a}$

${\displaystyle \mathrm {I} _{2}(\mathbf {A} )=3a^{2}}$

${\displaystyle \mathrm {det} (\mathbf {A} )=a^{3}}$

${\displaystyle \parallel \mathbf {A} \parallel ={\sqrt {3}}|a|}$

Gegeben ein beliebiger Tensor ${\displaystyle \mathbf {A} =A_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}}$

${\displaystyle \mathbf {A} ^{\mathrm {S} }=\mathrm {sym} (\mathbf {A} ):={\frac {1}{2}}(\mathbf {A} +\mathbf {A} ^{\top })}$

${\displaystyle \mathbf {A} ^{\mathrm {S} }={\frac {1}{2}}{\begin{pmatrix}2A_{11}&A_{12}+A_{21}&A_{13}+A_{31}\\A_{12}+A_{21}&2A_{22}&A_{23}+A_{32}\\A_{13}+A_{31}&A_{23}+A_{32}&2A_{33}\end{pmatrix}}}$

${\displaystyle \mathrm {Sp} (\mathbf {A} ^{\mathrm {S} })=\mathrm {Sp} (\mathbf {A} )=A_{11}+A_{22}+A_{33}}$

${\displaystyle {\begin{aligned}\mathrm {I} _{2}(\mathbf {A} ^{\mathrm {S} })=&{\frac {1}{2}}\mathrm {I} _{2}(\mathbf {A} )+{\frac {1}{4}}\mathrm {Sp} ^{2}(\mathbf {A} )-{\frac {1}{4}}\mathbf {A:A} \\=&A_{11}A_{22}+A_{11}A_{33}+A_{22}A_{33}\\&-{\frac {1}{4}}\left((A_{12}+A_{21})^{2}+(A_{13}+A_{31})^{2}+(A_{23}+A_{32})^{2}\right)\end{aligned}}}$

${\displaystyle {\begin{aligned}\mathrm {det} (\mathbf {A} ^{\mathrm {S} })=&{\frac {1}{4}}\mathrm {det} (\mathbf {A} )+{\frac {1}{4}}\mathbf {A} :\mathrm {adj} (\mathbf {A} )\\=&A_{11}A_{22}A_{33}+{\frac {1}{4}}(A_{12}+A_{21})(A_{23}+A_{32})(A_{13}+A_{31})\\&-{\frac {1}{4}}\left(A_{11}(A_{23}+A_{32})^{2}+A_{22}(A_{13}+A_{31})^{2}+A_{33}(A_{12}+A_{21})^{2}\right)\end{aligned}}}$

${\displaystyle {\begin{aligned}\parallel (\mathbf {A} ^{\mathrm {S} })\parallel =&{\sqrt {\mathbf {A:A} ^{\mathrm {S} }}}\\=&{\sqrt {A_{11}^{2}+A_{22}^{2}+A_{33}^{2}+{\frac {1}{2}}{\big (}(A_{12}+A_{21})^{2}+(A_{13}+A_{31})^{2}+(A_{23}+A_{32})^{2}{\big )}}}\end{aligned}}}$

${\displaystyle \mathbf {A} ^{\mathrm {A} }=\mathrm {skw} (\mathbf {A} ):={\frac {1}{2}}(\mathbf {A} -\mathbf {A} ^{\top })}$

${\displaystyle \mathbf {A} ^{\mathrm {A} }={\frac {1}{2}}{\begin{pmatrix}0&A_{12}-A_{21}&A_{13}-A_{31}\\A_{21}-A_{12}&0&A_{23}-A_{32}\\A_{31}-A_{13}&A_{32}-A_{23}&0\end{pmatrix}}}$

${\displaystyle \mathrm {Sp} (\mathbf {A} ^{\mathrm {A} })=0}$

${\displaystyle {\begin{aligned}\mathrm {I} _{2}(\mathbf {A} ^{\mathrm {A} })=&{\frac {1}{4}}\left(\mathbf {A:A} -\mathrm {Sp} (\mathbf {A} ^{2})\right)\\=&{\frac {1}{4}}\left((A_{12}-A_{21})^{2}+(A_{13}-A_{31})^{2}+(A_{23}-A_{32})^{2}\right)\end{aligned}}}$

${\displaystyle \mathrm {det} (\mathbf {A} ^{\mathrm {A} })=0}$

${\displaystyle \parallel \mathbf {A} ^{\mathrm {A} }\parallel ={\sqrt {\mathbf {A:A} ^{\mathrm {A} }}}={\sqrt {\frac {1}{2}}}{\sqrt {(A_{12}-A_{21})^{2}+(A_{13}-A_{31})^{2}+(A_{32}-A_{23})^{2}}}}$

${\displaystyle \mathbf {A} ^{\mathrm {D} }=\mathrm {dev} (\mathbf {A} ):=\mathbf {A} -{\frac {1}{3}}\mathrm {Sp} (\mathbf {A} )\mathbf {1} }$

${\displaystyle \mathbf {A} ^{\mathrm {D} }={\begin{pmatrix}{\frac {2A_{11}-A_{22}-A_{33}}{3}}&A_{12}&A_{13}\\A_{21}&{\frac {2A_{22}-A_{11}-A_{33}}{3}}&A_{23}\\A_{31}&A_{32}&{\frac {2A_{33}-A_{11}-A_{22}}{3}}\end{pmatrix}}}$

${\displaystyle \mathrm {Sp} (\mathbf {A} ^{\mathrm {D} })=0}$

${\displaystyle {\begin{aligned}\mathrm {I} _{2}(\mathbf {A} ^{\mathrm {D} })=&\mathrm {I} _{2}(\mathbf {A} )-{\frac {1}{3}}\mathrm {Sp} ^{2}(\mathbf {A} )={\frac {1}{6}}\mathrm {Sp} ^{2}(\mathbf {A} )-{\frac {1}{2}}\mathrm {Sp} (\mathbf {A} ^{2})\\=&{\frac {1}{3}}(A_{11}A_{22}+A_{11}A_{33}+A_{22}A_{33}-A_{11}^{2}-A_{22}^{2}-A_{33}^{2})\\&-A_{12}A_{21}-A_{13}A_{31}-A_{23}A_{32}\end{aligned}}}$

${\displaystyle {\begin{aligned}\mathrm {det} (\mathbf {A} ^{\mathrm {D} })=&\mathrm {det} (\mathbf {A} )+{\frac {2}{27}}\mathrm {Sp} ^{3}(\mathbf {A} )-{\frac {1}{3}}\mathrm {Sp} (\mathbf {A} )\mathrm {I} _{2}(\mathbf {A} )\\=&{\frac {1}{27}}{\Big (}12A_{11}A_{22}A_{33}+2(A_{11}^{3}+A_{22}^{3}+A_{33}^{3})\ldots \\&\qquad \ldots -3A_{11}^{2}(A_{22}+A_{33})-3A_{22}^{2}(A_{11}+A_{33})-3A_{33}^{2}(A_{11}+A_{22}){\Big )}\\&-{\frac {1}{3}}{\Big (}(2A_{11}-A_{22}-A_{33})A_{23}A_{32}+(2A_{22}-A_{11}-A_{33})A_{13}A_{31}+\ldots \\&\qquad \ldots +(2A_{33}-A_{11}-A_{22})A_{12}A_{21}{\Big )}\\&+A_{13}A_{32}A_{21}+A_{12}A_{23}A_{31}\end{aligned}}}$

${\displaystyle \parallel \mathbf {A} ^{\mathrm {D} }\parallel ={\sqrt {\begin{array}{r}{\frac {2}{3}}(A_{11}^{2}+A_{22}^{2}+A_{33}^{2}-A_{11}A_{22}-A_{11}A_{33}-A_{22}A_{33})+\ldots \\\ldots +A_{12}^{2}+A_{21}^{2}+A_{13}^{2}+A_{31}^{2}+A_{23}^{2}+A_{32}^{2}\end{array}}}}$

${\displaystyle \mathbf {A} ^{\mathrm {K} }=\mathrm {sph} (\mathbf {A} ):={\frac {1}{3}}\mathrm {Sp} (\mathbf {A} )\mathbf {1} }$

${\displaystyle \mathbf {A} ^{\mathrm {K} }={\frac {1}{3}}(A_{11}+A_{22}+A_{33}){\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}}$

${\displaystyle \mathrm {Sp} (\mathbf {A} ^{\mathrm {K} })=\mathrm {Sp} (\mathbf {A} )=A_{11}+A_{22}+A_{33}}$

${\displaystyle \mathrm {I} _{2}(\mathbf {A} ^{\mathrm {K} })={\frac {1}{3}}\mathrm {Sp} ^{2}(\mathbf {A} )={\frac {1}{3}}(A_{11}+A_{22}+A_{33})^{2}}$

${\displaystyle \mathrm {det} (\mathbf {A} ^{\mathrm {K} })={\frac {1}{27}}\mathrm {Sp} ^{3}(\mathbf {A} )={\frac {1}{27}}(A_{11}+A_{22}+A_{33})^{3}}$

${\displaystyle \parallel \mathbf {A} ^{\mathrm {K} }\parallel ={\frac {1}{\sqrt {3}}}|\mathrm {Sp} (\mathbf {A} )|={\frac {1}{\sqrt {3}}}|A_{11}+A_{22}+A_{33}|}$

${\displaystyle \mathbf {A} =\mathbf {A} ^{\mathrm {S} }+\mathbf {A} ^{\mathrm {A} }=\mathbf {A} ^{\mathrm {D} }+\mathbf {A} ^{\mathrm {K} }}$

Symmetrische und schiefsymmetrische Tensoren sind orthogonal zueinander:

${\displaystyle \mathbf {A} ^{\mathrm {S} }:\mathbf {B} ^{\mathrm {A} }=0}$

Deviatoren und Kugeltensoren sind orthogonal zueinander:

${\displaystyle \mathbf {A} ^{\mathrm {D} }:\mathbf {B} ^{\mathrm {K} }=0}$

Für jeden Tensor F mit #Determinante ≠ 0 gibt es #Orthogonale Tensoren Q und #Symmetrische und positiv definite Tensoren U in eindeutiger Weise, sodass

F = Q·U

Im Fall des Deformationsgradienten ist U der rechte Strecktensor, siehe #Symmetrische und positiv definite Tensoren. Der Anteil U berechnet sich wie dort angegeben aus

${\displaystyle \mathbf {U} =+{\sqrt {\mathbf {F^{\top }\cdot F} }}}$

Dann ist U·U = F^⊤·F und

${\displaystyle \mathbf {Q} =\mathbf {F\cdot U} ^{-1}}$

Bei det(F)=0 ergeben sich U sowie Q aus der Singulärwertzerlegung von F und U ist nur noch symmetrisch positiv semidefinit.

Gegeben sei die Gerade durch den Punkt ${\displaystyle {\vec {x}}}$ mit Richtungsvektor ${\displaystyle {\vec {g}}}$ und ein beliebiger anderer Punkt ${\displaystyle {\vec {p}}}$.

Dann ist

${\displaystyle {\begin{aligned}{\vec {p}}=&{\vec {x}}+{\vec {a}}+{\vec {b}}\quad {\textsf {mit}}\quad {\vec {a}}\|{\vec {g}}\quad {\text{und}}\quad {\vec {b}}\bot {\vec {g}}\\\mathbf {G} =&{\frac {{\vec {g}}\otimes {\vec {g}}}{{\vec {g}}\cdot {\vec {g}}}}\quad \rightarrow \quad \mathbf {G} \cdot {\vec {g}}={\vec {g}}\,,\quad (\mathbf {1} -\mathbf {G} )\cdot {\vec {g}}={\vec {0}}\\&{\vec {n}}\cdot {\vec {g}}=0\quad \rightarrow \quad \mathbf {G} \cdot {\vec {n}}={\vec {0}}\,,\quad (\mathbf {1} -\mathbf {G} )\cdot {\vec {n}}={\vec {n}}\\{\vec {a}}=&\mathbf {G} \cdot ({\vec {p}}-{\vec {x}})={\frac {{\vec {g}}\cdot ({\vec {p}}-{\vec {x}})}{{\vec {g}}\cdot {\vec {g}}}}{\vec {g}}\\{\vec {b}}=&\left(\mathbf {1} -\mathbf {G} \right)\cdot ({\vec {p}}-{\vec {x}})={\vec {p}}-{\vec {x}}-{\vec {a}}\end{aligned}}}$

Der Punkt ${\displaystyle {\vec {x}}+{\vec {a}}}$ ist die senkrechte Projektion von ${\displaystyle {\vec {p}}}$ auf die Gerade. Der Tensor G extrahiert den Anteil eines Vektors in Richtung von ${\displaystyle {\vec {g}}}$ und 1-G den Anteil senkrecht dazu.

Gegeben sei die Ebene durch den Punkt ${\displaystyle {\vec {x}}}$ und zwei die Ebene aufspannende Vektoren ${\displaystyle {\vec {u}}}$ und ${\displaystyle {\vec {v}}\not \!\|{\vec {u}}}$ sowie ein beliebiger anderer Punkt ${\displaystyle {\vec {p}}}$. Dann verschwindet die Normale

${\displaystyle {\hat {n}}={\frac {{\vec {u}}\times {\vec {v}}}{|{\vec {u}}\times {\vec {v}}|}}}$

nicht. Dann ist

${\displaystyle {\begin{aligned}{\vec {p}}=&{\vec {x}}+{\vec {a}}+{\vec {b}}\quad {\textsf {mit}}\quad {\vec {a}}\bot {\hat {n}}\quad {\text{und}}\quad {\vec {b}}\|{\hat {n}}\\\mathbf {P} =&{\frac {({\vec {v}}\cdot {\vec {v}}){\vec {u}}\otimes {\vec {u}}-({\vec {u}}\cdot {\vec {v}})({\vec {u}}\otimes {\vec {v}}+{\vec {v}}\otimes {\vec {u}})+({\vec {u}}\cdot {\vec {u}}){\vec {v}}\otimes {\vec {v}}}{({\vec {u}}\cdot {\vec {u}})({\vec {v}}\cdot {\vec {v}})-({\vec {u}}\cdot {\vec {v}})^{2}}}=\mathbf {1} -{\hat {n}}\otimes {\hat {n}}\\&\rightarrow \mathbf {P} \cdot {\vec {u}}={\vec {u}}\,,\quad \mathbf {P} \cdot {\vec {v}}={\vec {v}}\,,\quad \mathbf {P} \cdot {\hat {n}}={\vec {0}}\,,\quad (\mathbf {1} -\mathbf {P} )\cdot {\hat {n}}={\hat {n}}\\&\rightarrow \mathbf {P} \cdot (x{\vec {u}}+y{\vec {v}})=x{\vec {u}}+y{\vec {v}}\quad {\text{und}}\quad (\mathbf {1} -\mathbf {P} )\cdot (x{\vec {u}}+y{\vec {v}})={\vec {0}}\quad \forall x,y\in \mathbb {R} \\{\vec {a}}=&\mathbf {P} \cdot ({\vec {p}}-{\vec {x}})\\{\vec {b}}=&(\mathbf {1} -\mathbf {P} )\cdot ({\vec {p}}-{\vec {x}})={\vec {p}}-{\vec {x}}-{\vec {a}}\end{aligned}}}$

Der Punkt ${\displaystyle {\vec {x}}+{\vec {a}}}$ ist die senkrechte Projektion von ${\displaystyle {\vec {p}}}$ auf die Ebene.^[2] Der Tensor P extrahiert den Anteil eines Vektors in der Ebene und 1-P den Anteil senkrecht dazu.

Die Projektion der Geraden, die durch die Punkte ${\displaystyle {\vec {x}}}$ und ${\displaystyle {\vec {p}}}$ verläuft, liegt in der Ebene in Richtung des Vektors ${\displaystyle {\vec {a}}}$.

Falls ${\displaystyle |{\vec {u}}|=|{\vec {v}}|=1}$ und ${\displaystyle {\vec {u}}\bot {\vec {v}}}$ folgt:

${\displaystyle {\hat {n}}={\vec {u}}\times {\vec {v}}\quad {\text{mit}}\quad |{\hat {n}}|=1}$

${\displaystyle \mathbf {P} ={\vec {u}}\otimes {\vec {u}}+{\vec {v}}\otimes {\vec {v}}=\mathbf {1} -{\hat {n}}\otimes {\hat {n}}}$

${\displaystyle {\vec {a}}=({\vec {u}}\otimes {\vec {u}}+{\vec {v}}\otimes {\vec {v}})\cdot ({\vec {p}}-{\vec {x}})=(\mathbf {1} -{\hat {n}}\otimes {\hat {n}})\cdot ({\vec {p}}-{\vec {x}})}$

${\displaystyle {\vec {b}}=(\mathbf {1} -{\vec {u}}\otimes {\vec {u}}-{\vec {v}}\otimes {\vec {v}})\cdot ({\vec {p}}-{\vec {x}})=({\hat {n}}\otimes {\hat {n}})\cdot ({\vec {p}}-{\vec {x}})}$

Definition:

${\displaystyle {\begin{aligned}{\stackrel {3}{\mathbf {E} }}:=&\epsilon _{ijk}\,{\hat {e}}_{i}\otimes {\hat {e}}_{j}\otimes {\hat {e}}_{k}\\=&({\hat {e}}_{j}\times {\hat {e}}_{k})\otimes {\hat {e}}_{j}\otimes {\hat {e}}_{k}\\=&{\hat {e}}_{i}\otimes ({\hat {e}}_{k}\times {\hat {e}}_{i})\otimes {\hat {e}}_{k}\\=&{\hat {e}}_{i}\otimes {\hat {e}}_{j}\otimes ({\hat {e}}_{i}\times {\hat {e}}_{j})\end{aligned}}}$

Kreuzprodukt von Vektoren:

${\displaystyle {\vec {u}}\times {\vec {v}}={\stackrel {3}{\mathbf {E} }}:({\vec {u}}\otimes {\vec {v}})={\vec {v}}\cdot {\stackrel {3}{\mathbf {E} }}\cdot {\vec {u}}=-{\vec {u}}\cdot {\stackrel {3}{\mathbf {E} }}\cdot {\vec {v}}=-{\stackrel {3}{\mathbf {E} }}:({\vec {v}}\otimes {\vec {u}})=-{\vec {v}}\times {\vec {u}}}$

${\displaystyle {\vec {e}}_{i}\times {\vec {e}}_{j}=\epsilon _{ijk}\,{\hat {e}}_{k}}$

#Kreuzprodukt von Tensoren, #Skalarkreuzprodukt von Tensoren:

${\displaystyle {\begin{aligned}{\stackrel {3}{\mathbf {E} }}:\mathbf {A} =&\mathbf {A} :{\stackrel {3}{\mathbf {E} }}=-{\stackrel {3}{\mathbf {E} }}:(\mathbf {A} ^{\top })=-(\mathbf {A} ^{\top }):{\stackrel {3}{\mathbf {E} }}\\=&\mathbf {1} \times \mathbf {A} ^{\top }=\mathbf {1} \cdot \!\!\times \mathbf {A} \end{aligned}}}$

#Dualer axialer Vektor und #Vektorinvariante:

${\displaystyle {\stackrel {3}{\mathbf {E} }}:\mathbf {A} =-2{\stackrel {A}{\overrightarrow {\mathbf {A} }}}={\vec {\mathrm {i} }}(\mathbf {A} )}$

#Kreuzprodukt von Tensoren:

${\displaystyle \mathbf {A} \times \mathbf {B} ={\stackrel {3}{\mathbf {E} }}:(\mathbf {A\cdot B} ^{\top })}$

${\displaystyle (A_{ik}{\vec {e}}_{i}\otimes {\vec {e}}_{k})\times (B_{jl}{\vec {e}}_{j}\otimes {\vec {e}}_{l})=A_{ik}B_{jk}{\vec {e}}_{i}\times {\vec {e}}_{j}=\epsilon _{ijk}A_{jl}B_{kl}{\vec {e}}_{i}}$

#Skalarkreuzprodukt von Tensoren:

${\displaystyle \mathbf {A} \cdot \!\!\times \mathbf {B} ={\stackrel {3}{\mathbf {E} }}:(\mathbf {A\cdot B} )}$

${\displaystyle (A_{ik}{\vec {e}}_{i}\otimes {\vec {e}}_{k})\cdot \!\!\times (B_{lj}{\vec {e}}_{l}\otimes {\vec {e}}_{j})=A_{ik}B_{kj}{\vec {e}}_{i}\times {\vec {e}}_{j}=\epsilon _{ijk}A_{jl}B_{lk}{\vec {e}}_{i}}$

#Axialer Tensor oder Kreuzproduktmatrix:

${\displaystyle {\stackrel {3}{\mathbf {E} }}\cdot {\vec {u}}={\vec {u}}\cdot {\stackrel {3}{\mathbf {E} }}=-{\vec {u}}\times \mathbf {1} =-\mathbf {1} \times {\vec {u}}}$

Tensoren zweiter Stufe sind ebenfalls Elemente eines Vektorraums ${\displaystyle {\mathcal {L}}}$ wie im Abschnitt #Tensoren als Elemente eines Vektorraumes dargestellt. Daher kann man Tensoren vierter Stufe definieren, indem man in dem Kapitel formal die Tensoren zweiter Stufe durch Tensoren vierter Stufe und die Vektoren durch Tensoren zweiter Stufe ersetzt, z. B.:

${\displaystyle {\stackrel {4}{\mathbf {A} }}=A_{pq}(\mathbf {A} _{p}\otimes \mathbf {G} _{q})}$

mit Komponenten ${\displaystyle A_{pq}}$ und die Tensoren ${\displaystyle \mathbf {A} _{1},\mathbf {A} _{2},\ldots ,\mathbf {A} _{9}\in {\mathcal {L}}}$ sowie ${\displaystyle \mathbf {G} _{1},\mathbf {G} _{2},\ldots ,\mathbf {G} _{9}\in {\mathcal {L}}}$ bilden eine Basis von ${\displaystyle {\mathcal {L}}}$.

Standardbasis in ${\displaystyle {\mathcal {L}}}$:

${\displaystyle \mathbf {E} _{1}={\vec {e}}_{1}\otimes {\vec {e}}_{1},\mathbf {E} _{2}={\vec {e}}_{1}\otimes {\vec {e}}_{2},\mathbf {E} _{3}={\vec {e}}_{1}\otimes {\vec {e}}_{3},\mathbf {E} _{4}={\vec {e}}_{2}\otimes {\vec {e}}_{1},\ldots ,\mathbf {E} _{9}={\vec {e}}_{3}\otimes {\vec {e}}_{3}}$

Tensortransformation:

${\displaystyle {\stackrel {4}{\mathbf {A} }}:\mathbf {H} =A_{pq}(\mathbf {A} _{p}\otimes \mathbf {G} _{q}):\mathbf {H} :=A_{pq}(\mathbf {G} _{q}:\mathbf {H} )\mathbf {A} _{p}}$

Tensorprodukt:

${\displaystyle {\big (}A_{pq}(\mathbf {A} _{p}\otimes \mathbf {G} _{q}){\big )}:{\big (}B_{rs}(\mathbf {H} _{r}\otimes \mathbf {U} _{s}){\big )}:=A_{pq}(\mathbf {G} _{q}:\mathbf {H} _{r})B_{rs}\mathbf {A} _{p}\otimes \mathbf {U} _{s}}$

Übliche Schreibweisen für Tensoren vierter Stufe:

${\displaystyle {\stackrel {4}{\mathbf {A} }}=\mathbb {A} =A_{ijkl}\;{\vec {e}}_{i}\otimes {\vec {e}}_{j}\otimes {\vec {e}}_{k}\otimes {\vec {e}}_{l}}$

Transposition:

${\displaystyle (\mathbf {A} \otimes \mathbf {B} )^{\top }=\mathbf {B} \otimes \mathbf {A} }$

${\displaystyle (A_{ijkl}\;{\vec {e}}_{i}\otimes {\vec {e}}_{j}\otimes {\vec {e}}_{k}\otimes {\vec {e}}_{l})^{\top }:=A_{ijkl}\;{\vec {e}}_{k}\otimes {\vec {e}}_{l}\otimes {\vec {e}}_{i}\otimes {\vec {e}}_{j}}$

Spezielle Transposition ${\displaystyle {\stackrel {4}{\mathbf {A} }}{}^{\stackrel {mn}{\top }}}$ vertauscht ${\displaystyle m}$-tes mit ${\displaystyle n}$-tem Basissystem.

Beispielsweise:

${\displaystyle {\stackrel {4}{\mathbf {A} }}{}^{\stackrel {13}{\top }}:=A_{ijkl}\;{\vec {e}}_{k}\otimes {\vec {e}}_{j}\otimes {\vec {e}}_{i}\otimes {\vec {e}}_{l}}$

${\displaystyle {\stackrel {4}{\mathbf {A} }}{}^{\stackrel {24}{\top }}:=A_{ijkl}\;{\vec {e}}_{i}\otimes {\vec {e}}_{l}\otimes {\vec {e}}_{k}\otimes {\vec {e}}_{j}}$

${\displaystyle {\stackrel {4}{\mathbf {A} }}\,^{\top }=\left({\stackrel {4}{\mathbf {A} }}{}^{\stackrel {13}{\top }}\right){}^{\stackrel {24}{\top }}=A_{ijkl}\;{\vec {e}}_{k}\otimes {\vec {e}}_{l}\otimes {\vec {e}}_{i}\otimes {\vec {e}}_{j}}$

Definition: ${\displaystyle {\stackrel {4}{\mathbf {A} }}={\stackrel {4}{\mathbf {A} }}{}^{\top }}$

Dann gilt: ${\displaystyle {\stackrel {4}{\mathbf {A} }}:\mathbf {B} =\mathbf {B} :{\stackrel {4}{\mathbf {A} }}}$

${\displaystyle {\begin{aligned}{\stackrel {4}{\mathbf {1} }}:=&\mathbf {E} _{p}\otimes \mathbf {E} _{p}={\stackrel {4}{\mathbf {1} }}{}^{\top }=(\mathbf {1} \otimes \mathbf {1} )\,^{\stackrel {23}{\top }}\\=&{\vec {e}}_{i}\otimes {\vec {e}}_{j}\otimes {\vec {e}}_{i}\otimes {\vec {e}}_{j}=\delta _{ik}\delta _{jl}({\vec {e}}_{i}\otimes {\vec {e}}_{j}\otimes {\vec {e}}_{k}\otimes {\vec {e}}_{l})\\=&{\vec {g}}_{i}\otimes {\vec {g}}_{j}\otimes {\vec {g}}^{i}\otimes {\vec {g}}^{j}\end{aligned}}}$

Für beliebige Tensoren zweiter Stufe A gilt:

${\displaystyle {\stackrel {4}{\mathbf {T} }}=\mathbf {E} _{p}^{\top }\otimes \mathbf {E} _{p}=\delta _{il}\delta _{jk}({\vec {e}}_{i}\otimes {\vec {e}}_{j}\otimes {\vec {e}}_{k}\otimes {\vec {e}}_{l})\rightarrow {\stackrel {4}{\mathbf {T} }}:\mathbf {A} =\mathbf {A} ^{\top }}$

${\displaystyle {\stackrel {4}{\mathbf {S} }}={\frac {1}{2}}\left({\stackrel {4}{\mathbf {1} }}+\mathbf {E} _{p}^{\top }\otimes \mathbf {E} _{p}\right)={\frac {1}{2}}(\delta _{ik}\delta _{jl}+\delta _{il}\delta _{jk})({\vec {e}}_{i}\otimes {\vec {e}}_{j}\otimes {\vec {e}}_{k}\otimes {\vec {e}}_{l})\rightarrow {\stackrel {4}{\mathbf {S} }}:\mathbf {A} =\mathbf {A} ^{\mathrm {S} }}$

${\displaystyle {\stackrel {4}{\mathbf {A} }}={\frac {1}{2}}\left({\stackrel {4}{\mathbf {1} }}-\mathbf {E} _{p}^{\top }\otimes \mathbf {E} _{p}\right)={\frac {1}{2}}(\delta _{ik}\delta _{jl}-\delta _{il}\delta _{jk})({\vec {e}}_{i}\otimes {\vec {e}}_{j}\otimes {\vec {e}}_{k}\otimes {\vec {e}}_{l})\rightarrow {\stackrel {4}{\mathbf {A} }}:\mathbf {A} =\mathbf {A} ^{\mathrm {A} }}$

${\displaystyle {\stackrel {4}{\mathbf {K} }}={\frac {1}{3}}\mathbf {1} \otimes \mathbf {1} ={\frac {1}{3}}\delta _{ij}\delta _{kl}({\vec {e}}_{i}\otimes {\vec {e}}_{j}\otimes {\vec {e}}_{k}\otimes {\vec {e}}_{l})\rightarrow {\stackrel {4}{\mathbf {K} }}:\mathbf {A} =\mathbf {A} ^{\mathrm {K} }}$

${\displaystyle {\stackrel {4}{\mathbf {D} }}={\stackrel {4}{\mathbf {1} }}-{\frac {1}{3}}\mathbf {1} \otimes \mathbf {1} =(\delta _{ik}\delta _{jl}-{\frac {1}{3}}\delta _{ij}\delta _{kl})({\vec {e}}_{i}\otimes {\vec {e}}_{j}\otimes {\vec {e}}_{k}\otimes {\vec {e}}_{l})\rightarrow {\stackrel {4}{\mathbf {D} }}:\mathbf {A} =\mathbf {A} ^{\mathrm {D} }}$

Diese fünf Tensoren sind sämtlich symmetrisch und es gilt:

${\displaystyle {\stackrel {4}{\mathbf {C} }}\in \left\{{\stackrel {4}{\mathbf {A} }},\,{\stackrel {4}{\mathbf {D} }},\,{\stackrel {4}{\mathbf {K} }},\,{\stackrel {4}{\mathbf {S} }}\right\}\;\rightarrow \;{\stackrel {4}{\mathbf {C} }}:{\stackrel {4}{\mathbf {C} }}={\stackrel {4}{\mathbf {C} }}}$

${\displaystyle {\stackrel {4}{\mathbf {K} }}+{\stackrel {4}{\mathbf {D} }}={\stackrel {4}{\mathbf {S} }}+{\stackrel {4}{\mathbf {A} }}={\stackrel {4}{\mathbf {T} }}:{\stackrel {4}{\mathbf {T} }}={\stackrel {4}{\mathbf {1} }},\quad {\stackrel {4}{\mathbf {K} }}:{\stackrel {4}{\mathbf {D} }}={\stackrel {4}{\mathbf {S} }}:{\stackrel {4}{\mathbf {A} }}={\stackrel {4}{\mathbf {0} }}}$

Mit beliebigen Tensoren zweiter Stufe A, B und G gilt:

${\displaystyle {\stackrel {4}{\mathbf {C} }}=(\mathbf {A} \otimes \mathbf {B} )^{\stackrel {23}{\top }}=A_{ik}B_{jl}({\vec {e}}_{i}\otimes {\vec {e}}_{j}\otimes {\vec {e}}_{k}\otimes {\vec {e}}_{l})\rightarrow {\stackrel {4}{\mathbf {C} }}:\mathbf {G} =\mathbf {A\cdot G\cdot B} ^{\top }}$

${\displaystyle {\stackrel {4}{\mathbf {C} }}=(\mathbf {A} \otimes \mathbf {B} )^{\stackrel {24}{\top }}=A_{il}B_{kj}({\vec {e}}_{i}\otimes {\vec {e}}_{j}\otimes {\vec {e}}_{k}\otimes {\vec {e}}_{l})\rightarrow {\stackrel {4}{\mathbf {C} }}:\mathbf {G} =\mathbf {A\cdot G} ^{\top }\cdot \mathbf {B} }$

Sei Q ein #Orthogonaler Tensor und

${\displaystyle {\begin{aligned}{\stackrel {4}{\mathbf {Q} }}=&(\mathbf {Q} \otimes \mathbf {Q} )^{\stackrel {23}{\top }}=(\mathbf {Q} \cdot {\vec {e}}_{i})\otimes (\mathbf {Q} \cdot {\vec {e}}_{j})\otimes {\vec {e}}_{i}\otimes {\vec {e}}_{j}\\=&Q_{ik}\,Q_{jl}\,{\vec {e}}_{i}\otimes {\vec {e}}_{j}\otimes {\vec {e}}_{k}\otimes {\vec {e}}_{l}\end{aligned}}}$

Dann ist speziell

${\displaystyle {\stackrel {4}{\mathbf {Q} }}\,:\,{\stackrel {4}{\mathbf {Q} }}{\!}^{\top }={\stackrel {4}{\mathbf {Q} }}{\!}^{\top }:\,{\stackrel {4}{\mathbf {Q} }}={\stackrel {4}{\mathbf {1} }}}$

${\displaystyle {\stackrel {4}{\mathbf {Q} }}\,:\mathbf {G} =\mathbf {Q\cdot G\cdot Q} ^{\top }}$

${\displaystyle {\stackrel {4}{\mathbf {Q} }}\,:\mathbf {A} \otimes \mathbf {B} :\,{\stackrel {4}{\mathbf {Q} }}{\!}^{\top }=(\mathbf {Q\cdot A\cdot Q} ^{\top })\otimes (\mathbf {Q\cdot B\cdot Q} ^{\top })}$

${\displaystyle {\stackrel {4}{\mathbf {Q} }}\,:(T_{ijkl}{\vec {a}}_{i}\otimes {\vec {b}}_{j}\otimes {\vec {c}}_{k}\otimes {\vec {d}}_{l}):\,{\stackrel {4}{\mathbf {Q} }}{\!}^{\top }=T_{ijkl}(\mathbf {Q} \cdot {\vec {a}}_{i})\otimes (\mathbf {Q} \cdot {\vec {b}}_{j})\otimes (\mathbf {Q} \cdot {\vec {c}}_{k})\otimes (\mathbf {Q} \cdot {\vec {d}}_{l})}$

${\displaystyle \left(a{\stackrel {4}{\mathbf {1} }}+\mathbf {B} \otimes \mathbf {C} \right)^{-1}={\frac {1}{a}}\left({\stackrel {4}{\mathbf {1} }}-{\frac {1}{a+\mathbf {B} :\mathbf {C} }}\mathbf {B} \otimes \mathbf {C} \right)}$

Mit ${\displaystyle {\stackrel {4}{\mathbf {S} }}}$ und ${\displaystyle {\stackrel {4}{\mathbf {A} }}}$ aus #Spezielle Tensoren vierter Stufe:

${\displaystyle {\begin{aligned}&\left(a{\stackrel {4}{\mathbf {S} }}+\mathbf {B} ^{\mathrm {S} }\otimes \mathbf {C} ^{\mathrm {S} }+b{\stackrel {4}{\mathbf {A} }}+\mathbf {D} ^{\mathrm {A} }\otimes \mathbf {E} ^{\mathrm {A} }\right)^{-1}=\dots \\&\qquad \dots ={\frac {1}{a}}\left({\stackrel {4}{\mathbf {S} }}-{\frac {1}{a+\mathbf {B} ^{\mathrm {S} }:\mathbf {C} ^{\mathrm {S} }}}\mathbf {B} ^{\mathrm {S} }\otimes \mathbf {C} ^{\mathrm {S} }\right)+{\frac {1}{b}}\left({\stackrel {4}{\mathbf {A} }}-{\frac {1}{b+\mathbf {D} ^{\mathrm {A} }:\mathbf {E} ^{\mathrm {A} }}}\mathbf {D} ^{\mathrm {A} }\otimes \mathbf {E} ^{\mathrm {A} }\right)\end{aligned}}}$

Mit den Spannungen ${\displaystyle \mathbf {T} }$ und den Dehnungen ${\displaystyle \mathbf {E} }$ im Hookeschen Gesetz gilt:

${\displaystyle {\stackrel {4}{\mathbf {C} }}:=2\mu {\stackrel {4}{\mathbf {1} }}+\lambda \mathbf {1} \otimes \mathbf {1} \quad \rightarrow \quad {\stackrel {4}{\mathbf {C} }}:\mathbf {E} =\mathbf {T} }$

mit den Lamé-Konstanten ${\displaystyle \lambda }$ und ${\displaystyle \mu }$. Dieser Elastizitätstensor ist symmetrisch.

Invertierungsformel mit ${\displaystyle a=2\mu }$, ${\displaystyle \mathbf {B} =\lambda \mathbf {1} }$ und ${\displaystyle \mathbf {C} =\mathbf {1} }$:

${\displaystyle {\begin{aligned}&{\stackrel {4}{\mathbf {B} }}:={\stackrel {4}{\mathbf {C} }}{}^{-1}={\frac {1}{2\mu }}\left({\stackrel {4}{\mathbf {1} }}-{\frac {\lambda }{2\mu +3\lambda }}\mathbf {1} \otimes \mathbf {1} \right)={\frac {1}{2\mu }}{\stackrel {4}{\mathbf {1} }}-{\frac {\nu }{E}}\mathbf {1} \otimes \mathbf {1} \\&\rightarrow \quad {\stackrel {4}{\mathbf {B} }}:\mathbf {T} =\mathbf {E} \end{aligned}}}$

mit der Querdehnzahl ${\displaystyle \nu }$ und dem Elastizitätsmodul ${\displaystyle E}$.

Aus der Basis ${\displaystyle \mathbf {S} _{1},\ldots ,\mathbf {S} _{6}}$ des Vektorraums ${\displaystyle {\mathcal {S}}=\mathrm {sym} (\mathbb {V} ,\mathbb {V} )}$ der symmetrischen Tensoren zweiter Stufe, siehe #Voigt-Notation symmetrischer Tensoren zweiter Stufe, kann eine Basis des Vektorraums ${\displaystyle {\stackrel {4}{\mathcal {S}}}=\mathrm {Lin} ({\mathcal {S}},{\mathcal {S}})}$ der linearen Abbildungen von symmetrischen Tensoren auf symmetrische Tensoren konstruiert werden. Der #Einheitstensor vierter Stufe ist kein Element dieses Raumes, aber der Symmetrisierer ${\displaystyle {\stackrel {4}{\mathbf {S} }}}$ aus dem Abschnitt #Spezielle Tensoren vierter Stufe ist es. Die 36 Komponenten der Tensoren vierter Stufe aus ${\displaystyle {\stackrel {4}{\mathcal {S}}}}$ können als Voigt'scher Notation in eine 6×6-Matrix einsortiert werden:

${\displaystyle {\stackrel {4}{\mathbf {A} }}=A_{uv}\mathbf {S} _{u}\otimes \mathbf {S} _{v}\quad \leftrightarrow \quad \left[{\stackrel {4}{\mathbf {A} }}\right]={\begin{pmatrix}A_{11}&A_{12}&A_{13}&A_{14}&A_{15}&A_{16}\\A_{21}&A_{22}&A_{23}&A_{24}&A_{25}&A_{26}\\A_{31}&A_{32}&A_{33}&A_{34}&A_{35}&A_{36}\\A_{41}&A_{42}&A_{43}&A_{44}&A_{45}&A_{46}\\A_{51}&A_{52}&A_{53}&A_{54}&A_{55}&A_{56}\\A_{61}&A_{62}&A_{63}&A_{64}&A_{65}&A_{66}\end{pmatrix}}}$

Darin steht [x] für die Voigt-Notation von x. Diese Tensoren sind sämtlich singulär, denn sie bilden #Schiefsymmetrische Tensoren auf Nulltensoren ab:

${\displaystyle {\stackrel {4}{\mathbf {A} }}\,:\mathbf {T} ^{\mathrm {A} }={\stackrel {4}{\mathbf {S} }}\,:\mathbf {T} ^{\mathrm {A} }=\mathbf {0} }$

Die Vektoren und Matrizen in Voigt'scher Notation können addiert, subtrahiert und mit einem Skalar multipliziert werden. Beim Matrizenprodukt in Voigt'scher Notation muss eine Diagonalmatrix

${\displaystyle L=\mathrm {diag} (1,1,1,2,2,2)}$

mit den Einträgen ${\displaystyle L_{uv}=\mathbf {S} _{u}:\mathbf {S} _{v}}$ zwischengeschaltet werden:

${\displaystyle \mathbf {A} :\mathbf {B} =[\mathbf {A} ]^{\top }L[\mathbf {B} ]=A_{1}B_{1}+A_{2}B_{2}+A_{3}B_{3}+2A_{4}B_{4}+2A_{5}B_{5}+2A_{6}B_{6}}$

${\displaystyle \left[{\stackrel {4}{\mathbf {A} }}:\mathbf {T} \right]=\left[{\stackrel {4}{\mathbf {A} }}\right]L[\mathbf {T} ]}$

${\displaystyle \left[{\stackrel {4}{\mathbf {A} }}:{\stackrel {4}{\mathbf {B} }}\right]=\left[{\stackrel {4}{\mathbf {A} }}\right]L\left[{\stackrel {4}{\mathbf {B} }}\right]}$

Inverse Matrix bei #Determinante ungleich null in Voigt-Notation:

${\displaystyle {\begin{aligned}S:=&L^{-1}=\mathrm {diag} \left(1,1,1,{\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}}\right)\\\left[{\stackrel {4}{\mathbf {B} }}\right]=&S\left[{\stackrel {4}{\mathbf {A} }}\right]^{-1}S\;\leftrightarrow \;\left[{\stackrel {4}{\mathbf {A} }}\right]=S\left[{\stackrel {4}{\mathbf {B} }}\right]^{-1}S\\\rightarrow \left[{\stackrel {4}{\mathbf {S} }}\right]=&\left[{\stackrel {4}{\mathbf {A} }}\right]L\left[{\stackrel {4}{\mathbf {B} }}\right]=\left[{\stackrel {4}{\mathbf {B} }}\right]L\left[{\stackrel {4}{\mathbf {A} }}\right]=S\\\rightarrow {\stackrel {4}{\mathbf {S} }}=&{\stackrel {4}{\mathbf {A} }}\,:\,{\stackrel {4}{\mathbf {B} }}={\stackrel {4}{\mathbf {B} }}\,:\,{\stackrel {4}{\mathbf {A} }}\end{aligned}}}$

#Hookesches Gesetz in Voigt-Notation entspricht

${\displaystyle {\begin{aligned}&{\stackrel {4}{\mathbf {C} }}=2\mu {\stackrel {4}{\mathbf {S} }}+\lambda \mathbf {1} \otimes \mathbf {1} ,\quad {\stackrel {4}{\mathbf {B} }}={\frac {1}{2\mu }}{\stackrel {4}{\mathbf {S} }}-{\frac {\nu }{E}}\mathbf {1} \otimes \mathbf {1} ,\quad {\stackrel {4}{\mathbf {C} }}\,:\,{\stackrel {4}{\mathbf {B} }}={\stackrel {4}{\mathbf {B} }}\,:\,{\stackrel {4}{\mathbf {C} }}={\stackrel {4}{\mathbf {S} }}\end{aligned}}}$

Sei T ein #symmetrischer und ${\displaystyle \mathbf {Q} =Q_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}}$ ein #orthogonaler Tensor. Dann gilt:

${\displaystyle \left[\mathbf {Q\cdot T\cdot Q} ^{\top }\right]=RL[\mathbf {T} ]}$

${\displaystyle RLR^{\top }=R^{\top }LR=S}$

mit

${\displaystyle {\begin{aligned}R=&{\begin{pmatrix}R_{1111}&R_{1122}&R_{1133}&R_{1123}&R_{1113}&R_{1112}\\R_{2211}&R_{2222}&R_{2233}&R_{2223}&R_{2213}&R_{2212}\\R_{3311}&R_{3322}&R_{3333}&R_{3323}&R_{3313}&R_{3312}\\R_{2311}&R_{2322}&R_{2333}&R_{2323}&R_{2313}&R_{2312}\\R_{1311}&R_{1322}&R_{1333}&R_{1323}&R_{1313}&R_{1312}\\R_{1211}&R_{1222}&R_{1233}&R_{1223}&R_{1213}&R_{1212}\\\end{pmatrix}}\\R_{ijkl}:=&{\frac {1}{2}}(Q_{ik}\,Q_{jl}+Q_{il}\,Q_{jk})\end{aligned}}}$

1. ↑ P. B. Denton, S. J. Parke, T. Tao, X. Zhang: Eigenvectors from Eigenvalues. (PDF) 10. August 2019, S. 1–3, abgerufen am 29. November 2019 (englisch). 
2. ↑ J. Hanson: Rotations in three, four, and five dimensions. S. 4 f., arxiv:1103.5263. 

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