Question

Following what we did in equations (2.14) to (2.17), show that the direct product of a vector and a ${\mathbf{3}}^{\mathbf{r}\mathbf{d}}$-rank tensor is a ${\mathbf{4}}^{\mathbf{r}\mathbf{h}}$-rank tensor. Also show that the direct product of two ${\mathbf{2}}^{\mathbf{n}\mathbf{d}}$-rank tensors is a ${\mathbf{4}}^{\mathbf{rh}}$-rank tensor. Generalize this to show that the direct product of two tensors of ranks m and n is a tensor of rank m + n .

Short answer

Answer

The statement has been proven.

Step-by-step solution

Given Information

A $2^{nd}$- rank tensor,$3^{rd}$-rank tensor and a $4^{th}$- rank tensor.

Prove the statement.

The formula that ${\left(S\otimes T\right)}_{i_1,i_2,j_1,j_2,j_m}=S_{i_1,i_2,,i_n}{T}_{j_1,j_2,,j_m}$

Solve the tensor(direct) product mentioned in the formula.

The formula becomes as follows.

$$\begin{gathered} {\left(S\otimes T\right)}_{i{\text{'}}_1,i{\text{'}}_2,,i{\text{'}}_n,j{\text{'}}_1,j{\text{'}}_2,j{\text{'}}_m}=S_{j{\text{'}}_{1}i{\text{'}}_2,,j{\text{'}}_n}{T}_{j{\text{'}}_1{j^{\text{'}}}_2,,j_m} \\ {\left(S\otimes T\right)}_{i{\text{'}}_1,i{\text{'}}_2,,i{\text{'}}_n,j{\text{'}}_1,j{\text{'}}_2,j{\text{'}}_m}=\sum _{i_1-i_{n/1},,i_m}{a}_{i{\text{'}}_1{i}_1}{a}_{i{\text{'}}_n{i}_n}{a}_{j{\text{'}}_1{j}_1}..a_{j{\text{'}}_m{j}_m}{S}_{i_1,i_2}..T_{j_1,j_2..j_m} \\ {\left(S\otimes T\right)}_{i{\text{'}}_1,i{\text{'}}_2,,i{\text{'}}_n,j{\text{'}}_1,j{\text{'}}_2,j{\text{'}}_m}=\sum _{i_1-i_{n/1},,i_m}{a}_{i{\text{'}}_1{i}_1}{a}_{i{\text{'}}_n{i}_n}{a}_{j{\text{'}}_1{j}_1}..a_{j{\text{'}}_m{j}_m}{\left(S\otimes T\right)}_{i_1,i_2}.{.}_{i_1,i_2,i_2..j_m} \end{gathered}$$

This proves that $\left(S\otimes T\right)$is a tensor of rank $\left(n++m\right)$.

Hence, the statement has been proven.