Real Eigen Values: Proving/Disproving Matrix AB

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The discussion centers on the proposition that if A is an m by n row-stochastic matrix with all positive real entries and B is an n by m column-stochastic matrix with the same properties, then all eigenvalues of the m by m matrix AB are real. Participants seek to either prove or disprove this statement through counterexamples. Notably, the matrix AB is not necessarily symmetric (Hermitian), which adds complexity to the analysis. A reference to positive-definite matrices is provided as a potential starting point for understanding the implications of the proposition.

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asisbanerjee
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I need to either prove or disprove by a counterexample the following proposition:
" Let A be an m by n row-stochastic matrix in which all entries are positive real numbers and let B be an n by m column-stochastic matrix with the same feature. Then all eigen values of the m by m matrix AB are real."
Can anyone help? It may be noted that AB is not necessarily symmetric (Hermitian).
 
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asisbanerjee said:
I need to either prove or disprove by a counterexample the following proposition:
" Let A be an m by n row-stochastic matrix in which all entries are positive and let B be an n by m column-stochastic matrix with the same feature. Then all eigen values of the m by m matrix AB are real."
Can anyone help?

Hello asisbanerjee and welcome to the forums.

Maybe this will give you a headstart:

http://en.wikipedia.org/wiki/Positive-definite_matrix
 
asisbanerjee said:
I need to either prove or disprove by a counterexample the following proposition:
" Let A be an m by n row-stochastic matrix in which all entries are positive real numbers and let B be an n by m column-stochastic matrix with the same feature. Then all eigen values of the m by m matrix AB are real."
Can anyone help? It may be noted that AB is not necessarily symmetric (Hermitian).
Do you know an example of a (square) stochastic matrix A whose entries are positive but which has a nonreal eigenvalue?

That's enough to find a counterexample, because you can then take B=...
 
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