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Does a matrix with all non-negative, real entries have all non-negative, real eigenvalues? Where might I find a proof of such?

Ideas: Perhaps we can multiply a prospective eigenvector so its biggest entries are positive, and then show that it is a contradiction for it to have a negative eigenvalue?

I am currently looking at the Perron-Frobenius theorem on Wikipedia, but it seems not to mention this issue. (I suspect my conjecture is not true.)

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  • $\begingroup$ Do you still need help with this? $\endgroup$ – Git Gud Jun 6 '14 at 19:44
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Hint: Not true. Think of a $2\times 2$ matrix with non-negative entries and negative determinant. It looks like $I_2$.

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Define

$$A = \left(\begin{array}{ccc} .8147 & .9134 & .2785 \\.9058 & .6324 & .5469 \\ .1270 & .0975 & .9575\end{array}\right)$$

This has a negative eigenvalue, approximately $-.1879$.


Source: Run

eig(rand(3, 3))

in Matlab without changing the seed for the random number generator.

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No, let $A = \left[\begin{array}{ccc} 1 & 2\\3 & 4\end{array}\right]$.

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No. The matrix: $A = \left[\begin{array}{ccc} 0 & 1\\1 & 0\end{array}\right]$.

is a reflection in $y=x$. It has an eigenvalue equal to $-1$.

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