Question

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

Short answer

Answer

The equation has been proven.

Step-by-step solution

Given Information

The tensor$X_i{A}_j=B_{ij}$.

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

Apply transformation law on A and B.

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

A is an arbitrary vector.

Hence $\left(X{\text{'}}_{\alpha }-a_{\alpha i}{X}_i\right)=0$

Thus, X is a tensor.