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Suppose I have a matrix $A$ with real entries such that the off-diagonal entries of $A$ are positive or zero. (The diagonal entries may be positive, negative or zero.)

Is this a sufficient condition for the eigenvalues to all be real? (Maybe there is a theorem to this effect, but if so I haven't managed to identify it...)

It seems (to my ill-formed intuition!) like this might follow from the answer to this question:

Exponential of matrix with negative entries only on the diagonal

(The intuition is that if I think of the matrix as defining an ODE system, it seems like it then follows that a solution for any initial vector that is positive will stay positive; so no cyclic or spiral solutions, so no imaginary eigenvalues ... but as I say, my intuition may be right off.)

I think I can prove this straightforwardly for the $2\times 2$ case (essentially the discriminant in the solution of the characteristic equation is $((a-d)^2 + 4bc)$ so a sufficient condition for this to be non-negative is that $bc$ be non-negative, which follows if both off-diagonal terms, $b$ and $c$, are so...); but not sure if this can be generalised...

If the intuition is wrong, could someone offer a counter example (some specific such matrix that has imaginary eigenvalues - presumably bigger than $2\times 2$)?

Tks.

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Here is an counter-example

$ \left( \begin{array}{ccc} 0 & 1 & 3 & 4\\ 1 & -2 & 0.6 & 2\\ 1 & 0.8 & -1 & 1\\ 3 & 2 & 1 & -0.6 \end{array} \right)$

which has 2 complex eigenvalues (approximately):

$-2.04195 + 0.154793i, -2.04195 - 0.154793i$

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  • $\begingroup$ OK, thanks @Mr.T Pity. Back to the drawing board. ;-) $\endgroup$ – autofac Jul 9 '14 at 21:05

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