Let
and
be two second order tensors. Show that
![{\displaystyle {\boldsymbol {A}}:{\boldsymbol {B}}=({\boldsymbol {A}}^{T}\cdot {\boldsymbol {B}}):{\boldsymbol {\mathit {1}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/579557c043a192dea1a7484928994f21ec0f5773)
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}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77b7dd2881a7ff4c44cff4a7dfb18cd1319f8ccb)
Hence,
![{\displaystyle {{\boldsymbol {A}}:{\boldsymbol {B}}=({\boldsymbol {A}}^{T}\cdot {\boldsymbol {B}}):{\boldsymbol {\mathit {1}}}\qquad \square }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c8f511a48f7118d62d80339c309220882e009b9)
Let
be a second order tensor and let
and
be two
vectors. Show that
![{\displaystyle {\boldsymbol {A}}:(\mathbf {a} \otimes \mathbf {b} )=({\boldsymbol {A}}\cdot \mathbf {b} )\cdot \mathbf {a} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0eb418fca6d299c817a7f5981afe88be32ee70f4)
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} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/01fa2006da832d22e1bcebafa6bfb27aed1cb835)
Hence,
![{\displaystyle {{\boldsymbol {A}}:(\mathbf {a} \otimes \mathbf {b} )=({\boldsymbol {A}}\cdot \mathbf {b} )\cdot \mathbf {a} \qquad \square }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49573f17408d39495ba7c25aed1d940bdde1617c)
Let
and
be two second order tensors and let
and
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} )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f7292844e486a2297bc202e923f59fc40a222dc)
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} )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/55f7b4994deb0bb97a7058e0b18ddfad5547545a)
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 }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3efd451ba8eb9fdc7e806934fd352007b89739c7)
Let
be a second order tensors and let
and
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}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/710806475e4ee051f4431f9e5e5b0de8d8a69b3a)
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} )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d39188e07739cdf149684d7b1b185937b0b37b96)
Hence,
![{\displaystyle {({\boldsymbol {A}}\cdot \mathbf {a} )\otimes \mathbf {b} ={\boldsymbol {A}}\cdot (\mathbf {a} \otimes \mathbf {b} )\qquad \square }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b307ff9f0afbf0d6b115edb55394ac14299b261)
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}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6199dc11fecb5872deb65ded1fb794a25f8ec52a)
Therefore,
![{\displaystyle \mathbf {a} \otimes ({\boldsymbol {A}}\cdot \mathbf {b} )=(\mathbf {a} \otimes \mathbf {b} )\cdot {\boldsymbol {A}}^{T}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12610f0f631136ee9d24680b41b971e92830a091)
Now,
and
. 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}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6d8eb124ae751ed131ec0d29a3e21f94394ab89)
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 }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ba77b24d01959e18b48c066b7a20112a3e77d21)