I want to ask if this is a valid proof that $P * P^T = I$
Let have permutation matrix $P^{m\times m}$
We can write it as :
$ P_{(a_1..a_n)} = \begin{pmatrix} I_{a_1,*} \\ \vdots \\ I_{a_n,*} \end{pmatrix} $
(Edit: where $P_{(a_1..a_n)}$ is any permutation matrix with permutation given as it's 'argument' And $I_{x, *}$ is a row number $x$ of identity matrix )
and his transposed matrix as:
$ {P_{(a_1..a_n)}}^T = \begin{pmatrix} {(I_{a_1,*})}^T & \cdots & {(I_{a_n,*})}^T \end{pmatrix} $
then we can $P \cdot P^T$ express as : \begin{equation*} P_{(a_1..a_n)} \cdot {P_{(a_1..a_n)}}^T = \begin{pmatrix} I_{a_1,*} \cdot {(I_{a_1,*})}^T & \cdots & I_{a_1,*} \cdot {(I_{a_n,*})}^T \\ \vdots & \ddots & \vdots \\ I_{a_n,*} \cdot {(I_{a_1,*})}^T & \cdots & I_{a_n,*} \cdot {(I_{a_n,*})}^T \end{pmatrix} \end{equation*}
We know that :
$I_{n,x} = 1\ if \ x=n \ else \ 0 $
$ {(I_{n,x})}^T = 1\ if \ x=n \ else \ 0$
Also we know that:
$(P_{(a_1..a_n)} \cdot {P_{(a_1..a_n)}}^T)_{n,m} = {I_n \cdot (I_m)^T} = \sum_{i=1}^{m}I_{n,i}\cdot ((I_m)^T)_i$
and we see that this is equal to 1 only if $n = m$ which give us I matrix
I want to ask if this is a valid proof that $P * P^T = I$ ??
Some answers...
Using coordinates. This is what you did which looks correct.
Using orthogonal matrix. A permutation matrix... permutes the elements of an orthonormal basis. Therefore it is an orthogonal matrix. Therefore its transpose is equal to its inverse.