TITLE: The transpose of a permutation matrix is its inverse.
QUESTION [43 upvotes]: This is a question from the free Harvard online abstract algebra lectures.  I'm posting my solutions here to get some feedback on them.  For a fuller explanation, see this post.
This problem is from assignment 4.

Prove that the transpose of a permutation matrix $P$ is its inverse.

A permutation matrix $P$ has a single 1 in each row and a single 1 in each column, all other entries being 0.  So column $j$ has a single 1 at position $e_{i_jj}$.  $P$ acts by moving row $j$ to row $i_j$ for each column $j$.  Taking the transpose of $P$ moves each 1 entry from $e_{i_jj}$ to $e_{ji_j}$.  Then $P^t$ acts by moving row $i_j$ to row $j$ for each row $i_j$.  Since this is the inverse operation, $P^t=P^{-1}$.
Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem.
Thanks.

REPLY [5 votes]: Let $π$ be a permutation on $n$ objects and
\begin{equation}
\pi=\left(\begin{matrix}
    1 & 2 &\ldots& n \\
    \pi(1) & \pi(2) &\ldots& \pi(n)
    \end{matrix}
\right)
\end{equation}
Assume that $P_π$ be a permutation matrix. We need to prove that $P_π^T P_π=I$.
Note that, $π$ sends the $i$th row of the identity matrix to the $π(i)$th row, i.e.,
\begin{eqnarray*}
P_\pi=[P_{ij}]=\left\{
\begin{array}{ll}
       1; & i=\pi(j)\\
       0;  & i \ne \pi(j).
\end{array}
\right.
\end{eqnarray*}
The $ij$th component of $P_\pi^TP_\pi$ is
\begin{eqnarray}
(P_\pi^TP_\pi)_{ij}&=&\sum_{k=1}^n P^T_{ik}P_{kj}\\
&=&\sum_{k=1}^n P_{ki}P_{kj}\\
&=& P_{\pi(j)i}P_{\pi(j)j}\\
&=& P_{\pi(j)i}=\left\{
\begin{array}{ll}
       1; & i=j\\
       0;  & i \ne j.
\end{array}
\right.
\end{eqnarray}
Therefore, $P^T_\pi P_\pi=I$.