Proof

Suppose that A is \(n\times m\) matrix. Then we have

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

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

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

Therefore, we obtain

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