Solution

A second-order tensor ${\displaystyle \mathbf {T} }$ transforms a vector ${\displaystyle \mathbf {u} }$ into another vector ${\displaystyle \mathbf {v} }$. Thus,

${\displaystyle \mathbf {v} =\mathbf {T} \mathbf {u} =\mathbf {T} \bullet \mathbf {u} }$

In index and matrix notation,

${\displaystyle {\text{(1)}}\qquad v_{i}=T_{ij}u_{i}\leftrightarrow v_{p}=T_{pq}u_{q}~{\text{or,}}~\left[v\right]=\left[T\right]\left[u\right]}$

Let us determine the change in the components of ${\displaystyle \mathbf {T} }$ with change the basis from (${\displaystyle \mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}}$) to (${\displaystyle \mathbf {e} _{1}^{'},\mathbf {e} _{2}^{'},\mathbf {e} _{3}^{'}}$). The vectors ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$, and the tensor ${\displaystyle \mathbf {T} }$ remain the same. What changes are the components with respect to a given basis. Therefore, we can write

${\displaystyle {\text{(2)}}\qquad v_{i}^{'}=T_{ij}^{'}u_{i}^{'}~{\text{or,}}~\left[v\right]^{'}=\left[T\right]^{'}\left[u\right]^{'}}$

Now, using the vector transformation rule,

${\displaystyle {\begin{aligned}{\text{(3)}}\qquad v_{i}^{'}&=l_{ip}v_{p}~;~u_{i}^{'}=l_{ip}u_{p}~{\text{or,}}~\left[v\right]^{'}=\left[L\right]\left[v\right]~;\left[u\right]^{'}=\left[L\right]\left[u\right]\\v_{q}&=l_{iq}v_{i}^{'}~;~u_{q}=l_{iq}u_{i}^{'}~{\text{or,}}~\left[v\right]=\left[L\right]^{T}\left[v\right]^{'}~;\left[u\right]=\left[L\right]^{T}\left[u\right]^{'}\end{aligned}}}$

Plugging the first of equation (3) into equation (2) we get,

${\displaystyle {\text{(4)}}\qquad l_{ip}v_{p}=T_{ij}^{'}u_{i}^{'}~{\text{or,}}~\left[L\right]\left[v\right]=\left[T\right]^{'}\left[u\right]^{'}}$

Substituting for ${\displaystyle v_{p}}$ in equation~(4) using equation~(1),

${\displaystyle {\text{(5)}}\qquad l_{ip}T_{pq}u_{q}=T_{ij}^{'}u_{i}^{'}~{\text{or,}}~\left[L\right]\left[T\right]\left[u\right]=\left[T\right]^{'}\left[u\right]^{'}}$

Substituting for ${\displaystyle u_{q}}$ in equation (5) using equation (3),

${\displaystyle {\text{(6)}}\qquad l_{ip}T_{pq}l_{iq}u_{i}^{'}=T_{ij}^{'}u_{i}^{'}~{\text{or,}}~\left[L\right]\left[T\right]\left[L\right]^{T}\left[u\right]^{'}=\left[T\right]^{'}\left[u\right]^{'}}$

Therefore, if ${\displaystyle \mathbf {u} \equiv \left[u\right]}$ is an arbitrary vector,

${\displaystyle l_{ip}T_{pq}l_{iq}=T_{ij}^{'}\Rightarrow T_{ij}^{'}=l_{ip}l_{jq}T_{pq}~{\text{or,}}~\left[T\right]^{'}=\left[L\right]\left[T\right]\left[L\right]^{T}}$

which is the transformation rule for second order tensors.

Therefore, in matrix notation, the transformation rule can be written as

${\displaystyle \left[T\right]^{'}=\left[L\right]\left[T\right]\left[L\right]^{T}}$