I need the formal proof of this fact. I define the permutation matrix in this way
given $\pi$ a permutation of $n$ elements its permutation matrix is: $${P}_\pi=\begin{bmatrix} {e}_{\pi_{1}} \\ \vdots \\{e}_{\pi_{n}} \end{bmatrix},$$ where ${e}_i$ is the row vector with $1$ in the $i$-th position and $0$ in every other position. How can I prove that $${P}_{\sigma}^T{P}_{\pi}^T={{P}_{\sigma\circ\pi}}^T\quad?$$
thanks!
Observe that, for every $j=1,\ldots,n$, $$ P_\pi e_j=e_{\pi^{-1}(j)}. $$ Hence $$ P_\pi P_{\sigma}e_j=P_\pi e_{\sigma^{-1}(j)}=e_{\pi^{-1}(\sigma^{-1}(j))}=e_{(\sigma\circ\pi)^{-1}(j)}=P_{\sigma\circ \pi}e_j. $$ Thus $$ P_\pi P_{\sigma}=P_{\sigma\circ \pi}\quad\text{or}\quad P_\sigma^TP_{\pi}^T=P_{\sigma\circ \pi}^T. $$