# Continuum mechanics/Tensor algebra identities

## Identity 1

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

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

Proof:

Using index notation,

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

Hence,

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

## Identity 2

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

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

Proof:

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

${\displaystyle {\boldsymbol {A}}:(\mathbf {a} \otimes \mathbf {b} )=A_{ij}~a_{i}~b_{j}=(A_{ij}~b_{j})~a_{i}=({\boldsymbol {A}}\cdot \mathbf {b} )\cdot \mathbf {a} ~.}$

Hence,

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

## Identity 3

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

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

Proof:

Using index notation,

${\displaystyle ({\boldsymbol {A}}\cdot \mathbf {a} )\cdot ({\boldsymbol {B}}\cdot \mathbf {b} )=(A_{ij}~a_{j})(B_{ik}~b_{k})=(A_{ij}~B_{ik})(a_{j}~b_{k})=(A_{ji}^{T}~B_{ik})(a_{j}~b_{k})=({\boldsymbol {A}}^{T}\cdot {\boldsymbol {B}}):(\mathbf {a} \otimes \mathbf {b} )~.}$

Hence,

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

## Identity 4

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

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

Proof:

For the first identity, using index notation, we have

${\displaystyle [({\boldsymbol {A}}\cdot \mathbf {a} )\otimes \mathbf {b} ]_{ik}=(A_{ij}~a_{j})~b_{k}=A_{ij}~(a_{j}~b_{k})=A_{ij}~[\mathbf {a} \otimes \mathbf {b} ]_{jk}={\boldsymbol {A}}\cdot (\mathbf {a} \otimes \mathbf {b} )~.}$

Hence,

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

For the second identity, we have

${\displaystyle [\mathbf {a} \otimes ({\boldsymbol {A}}\cdot \mathbf {b} )]_{ij}=a_{i}~(A_{jk}~b_{k})=(a_{i}~b_{k})~A_{jk}=(a_{i}~b_{k})~A_{kj}^{T}=[(\mathbf {a} \otimes \mathbf {b} )\cdot {\boldsymbol {A}}^{T}]_{ij}~.}$

Therefore,

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

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

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

Therefore,

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