Can a doubly stochastic matrix be asymmetric?

Question

It's a fairly simple question but I cannot find the answer to it anywhere.

It’d be great if you could include the definitions of ‘doubly stochastic’ and ‘asymmetric’. :) — Haskell Curry, Mar 17 '13 at 23:17
@Haskell, "doubly stochastic" means all the row sums and all the column sums are 1; "asymmetric" means not symmetric, that is, not equal to its transpose. — Gerry Myerson, Apr 06 '17 at 22:59

score 6 · Accepted Answer · answered Mar 17 '13 at 22:57

6

$${1\over15}\pmatrix{8&1&6\cr3&5&7\cr4&9&2\cr}$$

answered Mar 17 '13 at 22:57

Gerry Myerson

179,216

4

It occurs to me that a simpler answer would be $$\pmatrix{0&1&0\cr0&0&1\cr1&0&0\cr}$$ – Gerry Myerson Apr 06 '17 at 23:01

score 1 · Answer 2 · answered Mar 19 '18 at 18:14

For more examples, one can do no better than $n$-by-$n$ circulant matrices; for instance, if $a$, $b$, and $c$ are nonnegative numbers such that $a+b+c=1$, then the matrix $$ \begin{bmatrix} a & b & c \\ c & a & b \\ b & c & a \end{bmatrix} $$ is doubly stochastic and not symmetric.

Can a doubly stochastic matrix be asymmetric?

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