Question

${\mathit{X}}_{\mathbf{i}\mathbf{j}\mathbf{k}\mathbf{l}}{\mathit{A}}_{\mathbf{k}\mathbf{l}}\mathbf{=}{\mathit{B}}_{\mathbf{i}\mathbf{j}}$.

Short answer

Answer

The equation has been proven.

Step-by-step solution

Given Information

The tensor$X_{ijkl}{A}_{kl}=B_{ij}$.

Definition of a cartesian tensor.

The first rank tensor is just a vector. A tensor of the second rank has nine components (in three dimensions) in every rectangular coordinate system.

Prove that X is a tensor.

Let $X_{ijkl}{A}_{kl}=B_{ij}$where B is a non-zero tensor.

Apply transformation law on A and B.

$$\begin{gathered} 0=X{\text{'}}_{\alpha \beta \omega }A{\text{'}}_{\omega }-B{\text{'}}_{\alpha \beta } \\ 0=X{\text{'}}_{\alpha \beta \omega }A{\text{'}}_{\omega }-a_{\alpha i}{a}_{\beta j}{B}_{ij} \\ 0=X{\text{'}}_{\alpha \beta \omega }A{\text{'}}_{\omega }-a_{\alpha i}{a}_{\beta j}{X}_{ijkl}{A}_{kl} \\ 0=X{\text{'}}_{\alpha \beta \omega }A{\text{'}}_{\omega }-a_{\alpha i}{a}_{\beta j}{a}_{\gamma k}{a}_{nl}{X}_{ijkl}A{\text{'}}_{\omega } \\ 0=\left(X{\text{'}}_{\alpha \beta \omega }-a_{\alpha i}{a}_{\beta j}{a}_{\gamma k}{a}_{nl}{X}_{ijkl}\right)A{\text{'}}_{\omega } \end{gathered}$$

Let $S_{\alpha }=\left(X{\text{'}}_{\alpha \beta \omega }-a_{\alpha i}{a}_{\beta j}{a}_{\gamma k}{a}_{nl}{X}_{ijkl}\right)A{\text{'}}_{\alpha \beta }$

Hence, $S^{T}A=0$

The equation mentioned above is true for any matrix A.

Thus, X is a tensor.