# Product of permutation matrices

I want to prove that the product of two permutation matrices is itself a permutation matrix. But I don't know how. Please help!

• How have you had a permutation matrix defined? Where are you stuck (when multiplying them, which of the properties are you having trouble verifying?) – Tobias Kildetoft May 20 '13 at 14:22
• Unless you want to explicitly do this with matrix algebra, the most elegant proof would be to show that applying a permutation matrix to a vector permutes the vector's components, and then to use the facts that applying two permutations after each other is equivalent to applying a third permutation, and that applying two matrices to a vector after each other is equivalent to applying their product to the vector. – joriki May 20 '13 at 14:28

Method 1 - Using Matrices:

Suppose you have two permutation matrices $X$ and $Y$. So by definition both $X$ and $Y$ are $n\times n$ matrices having precisely one non-zero entry in each row and column, and these entries all take the value 1.

If the non-zero entry in the first row of $X$ is in the $i$-th column, then what will the first row of the matrix $X\cdot Y$ be? By definition of matrix multiplication it will be

$\left(\sum_{j=1}^{n}X_{1,j}Y_{j,1}, \sum_{j=1}^{n}X_{1,j}Y_{j,2},\ldots,\sum_{j=1}^{n}X_{1,j}Y_{j,n} \right)$

However, we know that $X_{1,j}=\delta_{i,j}$ (where $\delta_{i,j}$ is the Kronecker delta taking the value $1$ when $i=j$ and $0$ otherwise). Hence the first row of $X\cdot Y$ will be

$\left(X_{1,i}Y_{i,1}, X_{1,i}Y_{i,2},\ldots,X_{1,i}Y_{i,n} \right)$

which since $X_{1,i}=1$ will be

$\left(Y_{i,1}, Y_{i,2},\ldots,Y_{i,n} \right).$

So the first row of $X\cdot Y$ corresponds to the $i$-th row of $Y$. Continuing in this way, you see that the $m$-th row of $X\cdot Y$ will be the $l$-th row of $Y$, where the $1$ in the $m$-th row of $X$ appears in the $l$-th position (I hope that makes sense!)

Thus when you multiply two permutation matrices together, you are just rearranging the rows of one of the matrices.

Method 2 - Using Permutations:

A slightly easier approach is to use permutations. Suppose $X$ and $Y$ are $n\times n$ permutation matrices as above. Then define $\sigma_{X}:\{1,\ldots, n\}\rightarrow\{1,\ldots,n\}$ as follows:

$\sigma_{X}(i)=j$ where the $1$ in the $i$-th column of $X$ occurs in the $j$-th position. Similarly you may define $\sigma_{Y}$.

Then you can check that $\sigma:X\mapsto\sigma_{X}$ respects multiplication (in a similar way as above).

Method 3 - Using Vectors:

You can also show this using vectors (see joriki's comment above for more details).

Note: All of these methods are essentially the same, but are just different ways of viewing the problem.

Hint:

If $P$ is a permutation matrix and $A$ any matrix then $AP$ is a matrix consisting of the columns of $A$ permuted. Similarly $PA$ consists the rows of $A$ permuted. Therefore if $A$ is also a permutation matrix ...