How to show that product of two matrices $\mathbf{A}$ and $\mathbf{B}$ of convex combinations as $\mathbf{C=A*B}$ is also a matrix of convex combinations.
Convex combinations: entries of each column of matrix are non-negative and they sum to 1.
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1$\begingroup$ "Convex combination" is what's usually called "stochastic", right? $\endgroup$– Gerry MyersonCommented Jan 8, 2018 at 16:37
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$\begingroup$ @GerryMyerson I guess the word "stochastic" is used in reference to a square probability/transition matrix. In my example none of the matrices are square. I would probably use the term "convex combination" or a combination that is "conic & affine". May be I am not missing anything? $\endgroup$– AstroCommented Jan 9, 2018 at 5:11
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$\begingroup$ OK, I guess I assumed the matrices were square. Anyway, Sebastian seems to have settled things. $\endgroup$– Gerry MyersonCommented Jan 11, 2018 at 9:33
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Just turn being a convex combination into vector arithmetic: $(1, \ldots, 1) \cdot C = (1, \ldots, 1) \cdot A \cdot B = (1, \ldots, 1) \cdot B = (1, \ldots, 1)$. The non-negativity is clear I suppose.
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$\begingroup$ Hi, can you provide more details. What about sum to one constraint.? $\endgroup$– AstroCommented Jan 8, 2018 at 12:10
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2$\begingroup$ That's exactly what this answer shows. Also it seems more common to call these matrices left stochastic or column stochastic rather than "being convex combinations". $\endgroup$– DirkCommented Jan 8, 2018 at 12:28
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$\begingroup$ @Dirk Thanks all for the help. Cheers! $\endgroup$– AstroCommented Jan 9, 2018 at 5:04