Proof

Suppose that A and B are \(n\times m\) matrices as follows

$$A=\begin{bmatrix}a_{11} &a_{12}&\cdots&a_{1m}\\a_{21} &a_{22}&\cdots&a_{2m}\\\vdots &\vdots & \ddots & \vdots\\a_{n1} &a_{n2}&\cdots&a_{nm}\\\end{bmatrix},\ \ \ B=\begin{bmatrix}b_{11} &b_{12}&\cdots&a_{1m}\\b_{21} &b_{22}&\cdots&a_{2m}\\\vdots &\vdots & \ddots & \vdots\\b_{n1} &b_{n2}&\cdots&b_{nm}\\\end{bmatrix},$$

Let us denote the \((i,j)\)-entry in A and B by \([A]_{ij}\) and \([B]_{ij}\).

Then we have

\begin{eqnarray*}&&[A]_{ij}=a_{ij}\ \ \text{and }\ \ [A^{t}]_{ij}=a_{ji}\\&&[B]_{ij}=b_{ij}\ \ \text{and }\ \ [B^{t}]_{ij}=b_{ji}\\&&[A+B]_{ij}=a_{ij}+b_{ij}\ \ \text{and }\ \ \ [(A+B)^{t}]_{ij}=a_{ji}+b_{ji}\end{eqnarray*}

Therefore, we obtain

$$[A^{t}+B^{t}]_{ij}=a_{ji}+b_{ji}=[(A+B)^{t}]_{ij}$$