Continuum mechanics/Tensor algebra identities: Wikis

Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.

Study guide

Up to date as of January 14, 2010

Identity 1

Let $\boldsymbol{A}$ and $\boldsymbol{B}$ be two second order tensors. Show that

$\boldsymbol{A}:\boldsymbol{B} = (\boldsymbol{A}^T\cdot\boldsymbol{B}):\boldsymbol{\mathit{1}}~.$

Proof:

Using index notation,

$\boldsymbol{A}:\boldsymbol{B} = A_{ij}~B_{ij} = A^T_{ji}~B_{ij} = A^T_{ji}~B_{ik}~\delta_{jk} = [\boldsymbol{A}^T\cdot\boldsymbol{B}]_{jk}~\delta_{jk} = (\boldsymbol{A}^T\cdot\boldsymbol{B}):\boldsymbol{\mathit{1}} ~.$

Hence,

${ \boldsymbol{A}:\boldsymbol{B} = (\boldsymbol{A}^T\cdot\boldsymbol{B}):\boldsymbol{\mathit{1}} \qquad \square }$

Identity 2

Let $\boldsymbol{A}$ be a second order tensor and let $\mathbf{a}$ and $\mathbf{b}$ be two vectors. Show that

$\boldsymbol{A}:(\mathbf{a}\otimes\mathbf{b}) = (\boldsymbol{A}\cdot\mathbf{b})\cdot\mathbf{a} ~.$

Proof:

It is convenient to use index notation for this. We have

$\boldsymbol{A}:(\mathbf{a}\otimes\mathbf{b}) = A_{ij}~a_i~b_j = (A_{ij}~b_j)~a_i = (\boldsymbol{A}\cdot\mathbf{b})\cdot\mathbf{a} ~.$

Hence,

${ \boldsymbol{A}:(\mathbf{a}\otimes\mathbf{b}) = (\boldsymbol{A}\cdot\mathbf{b})\cdot\mathbf{a} \qquad \square }$

Identity 3

Let $\boldsymbol{A}$ and $\boldsymbol{B}$ be two second order tensors and let $\mathbf{a}$ and $\mathbf{b}$ be two vectors. Show that

$(\boldsymbol{A}\cdot\mathbf{a})\cdot(\boldsymbol{B}\cdot\mathbf{b}) = (\boldsymbol{A}^T\cdot\boldsymbol{B}):(\mathbf{a}\otimes\mathbf{b}) ~.$

Proof:

Using index notation,

$(\boldsymbol{A}\cdot\mathbf{a})\cdot(\boldsymbol{B}\cdot\mathbf{b}) = (A_{ij}~a_j)(B_{ik}~b_k) = (A_{ij}~B_{ik})(a_j~b_k) = (A^T_{ji}~B_{ik})(a_j~b_k) = (\boldsymbol{A}^T\cdot\boldsymbol{B}):(\mathbf{a}\otimes\mathbf{b}) ~.$

Hence,

${ (\boldsymbol{A}\cdot\mathbf{a})\cdot(\boldsymbol{B}\cdot\mathbf{b}) = (\boldsymbol{A}^T\cdot\boldsymbol{B}):(\mathbf{a}\otimes\mathbf{b}) \qquad \square }$

Identity 4

Let $\boldsymbol{A}$ be a second order tensors and let $\mathbf{a}$ and $\mathbf{b}$ be two vectors. Show that

$(\boldsymbol{A}\cdot\mathbf{a})\otimes\mathbf{b} = \boldsymbol{A}\cdot(\mathbf{a}\otimes\mathbf{b}) \qquad \text{and} \qquad \mathbf{a}\otimes(\boldsymbol{A}\cdot\mathbf{b}) = [\boldsymbol{A}\cdot(\mathbf{b}\otimes\mathbf{a})]^T = (\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T ~.$

Proof:

For the first identity, using index notation, we have

$[(\boldsymbol{A}\cdot\mathbf{a})\otimes\mathbf{b}]_{ik} = (A_{ij}~a_j)~b_k = A_{ij}~(a_j~b_k) = A_{ij}~[\mathbf{a}\otimes\mathbf{b}]_{jk} = \boldsymbol{A}\cdot(\mathbf{a}\otimes\mathbf{b}) ~.$

Hence,

${ (\boldsymbol{A}\cdot\mathbf{a})\otimes\mathbf{b} = \boldsymbol{A}\cdot(\mathbf{a}\otimes\mathbf{b}) \qquad \square }$

For the second identity, we have

$[\mathbf{a}\otimes(\boldsymbol{A}\cdot\mathbf{b})]_{ij} = a_i~(A_{jk}~b_k) = (a_i~b_k)~A_{jk} = (a_i~b_k)~A^T_{kj} = [(\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T]_{ij} ~.$

Therefore,

$\mathbf{a}\otimes(\boldsymbol{A}\cdot\mathbf{b}) = (\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T ~.$

Now, $\mathbf{a}\otimes\mathbf{b} = [\mathbf{b}\otimes\mathbf{a}]^T$ and $(\boldsymbol{A}\cdot\boldsymbol{B})^T = \boldsymbol{B}^T\cdot\boldsymbol{A}^T$. Hence,

$(\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T = (\mathbf{b}\otimes\mathbf{a})^T\cdot\boldsymbol{A}^T = [\boldsymbol{A}\cdot(\mathbf{b}\otimes\mathbf{a})]^T ~.$

Therefore,

${ \mathbf{a}\otimes(\boldsymbol{A}\cdot\mathbf{b}) = [\boldsymbol{A}\cdot(\mathbf{b}\otimes\mathbf{a})]^T = (\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T \qquad \square }$