2
$\begingroup$

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?

  • 0
    This follows from the Birkhoff–von Neumann theorem, which states that the $n\times n$ doubly stochastic matrices are precisely the convex combinations of the $n\times n$ permutation matrices. Multiplying two permutation matrices yields another permutation matrix, from which it is straightforward to show that the multiplication of two stochastic matrices is a convex combination of permutation matrices, from which we conclude. – 2017-01-30
  • 1
    @Math1000 that seems like overkill. – 2017-01-30
  • 0
    I was trying to come up with an intuitive reason as to why the statement should be true and found this very interesting theorem. I wrote it as a comment instead of an answer because it is indeed "overkill." – 2017-01-30
  • 0
    @Math1000 fair enough. Your approach doesn't appeal directly to my intuition, but I can see why you like it. – 2017-01-30

2 Answers 2

1

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, $M^Tx = x$).

  • 0
    nice one. Didn't know that identity yet. – 2017-01-30
  • 0
    So, if matrix $M$ is both row-stochastic and column-stochastic, then it is doubly-stochastic? – 2017-01-31
  • 0
    Yes, that's exactly it – 2017-01-31
1

Let $C$ be the $n \times n$ matrix defined as $C = AB$. Then, \begin{align} \sum_jc_{ij} & = \sum_j \left(\sum_k a_{ik}b_{kj}\right) \\ & = \sum_j \sum_k a_{ik}b_{kj} \\ & = \sum_k \sum_j a_{ik}b_{kj} \\ & = \sum_k a_{ik} \left( \sum_j b_{kj} \right) \\ & = \sum_k a_{ik}(1) \\ & = 1. \end{align}

The proof for a column of $C$ is similar.

  • 0
    For the proof for a column of $C$, just replace the $j$ with $i$ and $i$ with $j$? – 2017-02-01