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I was reading an old paper by Wilf: http://www.math.upenn.edu/~wilf/website/Eigenvalues%20of%20a%20graph.pdf

In short, he has made some claim about a lower bound for the largest eigenvalue of a non-negative matrix. (Which, by the Perron–Frobenius theorem is non-negative, and it has an all positive eigenvector)

And it made great sense to me until I got to one claim in the end, where he discussed another matrix, which was identical to the old one, except he replaced some of its entries by 0s, and claimed that the new matrix' highest eigenvalue can't be higher than the original's, and after spending some time trying to prove that fact, I haven't been able to.

Am I missing something?

Is it true in general that if a non-negative matrix A dominates a second non-negative matrix B, all of B's positive eigenvalues are no greater than A's?

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migrated from mathoverflow.net Jun 18 '14 at 22:07

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Wilf's claim follows from the Perron-Frobenius theorems, see the wikipedia article. (The Perron-Frobenius theory tells us much more than what you have quoted.)

I do not understand what you mean by "all of B's positive eigenvalues are no greater than A's", but at this level of generality there's very little that can said about any eigenvalue than the largest.

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Probably what is meant is that if $A \geq B$ (entrywise), then the largest eigenvalue of $A$ is at least as large as that of $B$ (when $A$ and $B$ are nonnegative matrices). If $B$ is primitive (that is, some power is strictly positive), then an easy argument with the H-transform shows that the spectral radius of $A$ is strictly greater than that of $B$ if $A \geq B$ and $A \neq B$.

A problem with the statement about other positive eigenvalues, presumably corresponding to different blocks, is that adding stuff can reduce the number of irreducible blocks, thereby in principle reducing the number of positive eigenvalues, e.g., if the perturbed matrix is primitive.

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