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Suppose $A$ is a doubly stochastic matrix, that is each row and column have the sum of their entries as $1$ and the entries are all nonnegative. Now, suppose $A=BC$ where $B$ is idempotent and $C$ is unitary. Then, it is easy to see that $B$ and $C$ have the row and column sum of entries in each row and/ or column as $1$. But, does this also imply that $B, C$ are also doubly stochastic? Any hints? Thanks beforehand.

vidyarthi
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2 Answers2

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No. Counterexample: $$ A=B=\frac13\pmatrix{1&1&1\\ 1&1&1\\ 1&1&1},\ C=\frac13\pmatrix{-1&2&2\\ 2&-1&2\\ 2&2&-1}. $$

user1551
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A remarkable question with a convincing counterexample answer.

You may consider the following as an abstract complement which states a weaker result:
The assumptions on idempotence $B^2=B$ and unitarity $C^T\!=C^{-1}$ force at least $B$ to be doubly-stochastic.
(which is satisfied in user$1551$'s counterexample)

Proof:
Since $B\neq 0$, one has $\|B\|\geqslant 1\,$ in view of $\,0\neq\|B\|=\|B^2\|\leqslant\|B\|^2$. Furthermore, $\|B\|=\|AC^{-1}\|\leqslant\|A\|\|C^{-1}\|=1$ because the double-stochastic $A$ and the unitary $C$ both have norm equal to $1$. Hence $\|B\|= 1$ and $B$ is necessarily symmetric, cf. that answer .
Then we get $A^T\!=C^{-1}B$, thus $\,B=BCC^{-1}\!B=AA^T$ is a doubly-stochastic matrix, being a product of doubly-stochastic ones.

Hanno
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