Proof

Suppose that A is an \(n\times m\) matrix as follows:

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

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

Then we have

$$[A]_{ij}=a_{ij}\ \ \ \text{ and }\ \ \ [A^{t}]_{ij}=[A]_{ji}=a_{ji}$$

Therefore, we obtain

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