I'm working on LU factorization that involves pivoting rows and I'm still trying to work my head around how to obtain the permutation matrices with ease.
Say I want to interchange two rows. If I have something as simple as $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ then I can left multiply by $P = \begin{bmatrix} p_1 & p_2 \\ p_3 & p_4 \end{bmatrix}$, form a system under these matrices so that Gaussian elimination gives me $P = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, satisfying \begin{equation*} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} c & d \\ a & b \end{bmatrix} \end{equation*}
But how do I do this for larger matrices between any arbitrary rows? Is there a fast algorithm that allows me to do this?
We say that $P$ is a permutation matrix if any row and any column have exactly one element equals 1 while the others are equals 0. Take, for example: $$ P = \begin{bmatrix} 1&0&0\\0&0&1\\0&1&0\end{bmatrix}$$
Let $A$ be a $3\times 3$ matrix. If $PA = A'$, $A'$ will be equal the following row swapping of $A$:
The first row of $A'$ is equal the first row of $A$, because the first row of $P$ has $1$ in the 1st column.
The second row of $A'$ is equal the third row of $A$, because the second row of $P$ has $1$ in the 3rd column.
The third row of $A'$ is equal the second row of $A$, because the second row of $P$ has $1$ in the 2rd column.
I hope that the explanation was clear!