Prove that if A and B are doubly-stochastic matrices of order n, then AB is also a doubly-stochastic matrix.
My attempt:
Since the size of A is nxn and the size of B is nxn, then the size of AB is also nxn.
Let A = ($a_{ij}$) and B = ($b_{ij}$). Consider the sum of the entries in the $i$th row of AB:
$a_{i1}b_{11} + a_{i2}b_{21} + a_{i3}b_{31} + … + a_{in}b_{n1}$
+ $a_{i1}b_{12} + a_{i2}b_{22} + a_{i3}b_{32} + … + a_{in}b_{n2}$
+ $…a_{in}b_{nn}$
= $a_{i1}(b_{11} + b_{12} + … + b_{1n}) + a_{i2}(b_{21} + b_{22} + … + b_{2n}) + …+ a_{in}(b_{n1} + b_{n2} + … + b_{nn})$
= $a_{i1} + a_{i2} + … + a_{in}$
= $1$
Next, I would consider the sum of the entries in the jth column of AB. However, at this point, I'm stuck. So, I'm wondering if my method is possible. Is there also a better way to prove this?
Best Answer
Your method could work. You might find it easier to consider the $i$th row of $(AB)^T$. There's a quick approach, though.
Hint: Let $x$ denote the column-vector of $1$s. Then a matrix $M$ is row-stochastic iff $Mx = x$ and column-stochastic iff $x^TM = x^T$ (or if you prefer, $M^Tx = x$).