Question

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

Short answer

Answer

The equation has been proven.

Step-by-step solution

Given Information

The tensor$X_i{A}_{ij}=B_j$.

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 $X_i{A}_{ij}=B_i$where B is a non-zero tensor.

Apply transformation law on A and B.

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

Let $S_{\alpha }=\left(X{\text{'}}_{\alpha }-a_{\alpha i}{X}_i\right)A{\text{'}}_{\alpha \beta }$

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

The equation mentioned above is true for any matrix A.

Thus, X is a tensor.