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I am reading from Linear Algebra in Action by Dym and am working through the proof of the Birkhoff-von Neumann Theorem. For the sake of clarity, ill write up everything until the point where I get lost.

Theorem: Let $P \in \mathbb{R}^{n\times n}$ be a doubly stochastic matrix.Then $P$ is a convex combination of finitely many permutation matrices.

Proof: If $P$ is a permutation matrix, then the assertion is self-evident. If $P$ is not a permutation matrix, them, in the view of Lemma 23.13

Lemma 23.13: Let $A \in \mathbb{R}^{n \times n}$ be a doubly stochastic matrix. Then $\operatorname{per}(A)>0$.

and the fact that $P$ is doubly stochastic, there exists a permutation $\sigma$ of the integers $\{1,\dots,n\}$ such that $1 > p_{1\sigma(1)}p_{2\sigma(2)} \cdots p_{n\sigma(n)} > 0.$ Let $$\lambda_1=\min\{p_{1\sigma(1)},p_{2\sigma(2)},\dots, p_{n\sigma(n)}\}$$ and let $\Pi_1$ be the permutation matrix with 1's in the $i\sigma(i)$ position for $i=1,\dots,n$. Then it is readily checked that $$P_1 = \frac{P-\lambda_1\Pi_1}{1-\lambda_1}$$ is a doubly stochastic matrix with at least one more zero entry than $P$ and that ...

I do not see how that is clearly the case? Can someone explain that to me?

Rodrigo de Azevedo
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Alex Sampson
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1 Answers1

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When we subtract $P-\lambda_1\Pi_1$ we are only subtracting $\lambda_1$ from entries which are $\ge\lambda_1$, so the result still has nonnegative entries. Furthermore every row and column has $\lambda_1$ taken from its sum total, i.e. all rows and columns sum to $1-\lambda_1$. When we divide by $1-\lambda_1$ (which is possible since $0<\lambda_1<1$) the result is thus a matrix whose rows and columns simply sum to $1$, and whose entries are still nonnegative, i.e. the result is a doubly stochastic matrix. Any matrix entry which was already $0$ remains $0$, but out of the entries corresponding to the $1$s in $\Pi_1$ the minimal entry, i.e. where $p_{r\sigma(r)}=\lambda_1$, is now $0$ where it wasn't before.

anon
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