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Given a $3\times3$ matrix is there a criterion capable of telling whether the matrix has a positive eigenvalue?

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The trace of a matrix is the sum of the eigenvalues and the determinant is the product. So in the $3\times 3$ case, if either the trace or determinant is positive, there is at least one positive eigenvalue. The converse is not true though. A matrix with -10, 1, 1 on the diagonal, zeros elsewhere, has a negative trace and determinant but two positive eigenvalues. – axblount Aug 9 '12 at 18:07
Probably not what you are looking for, but you can actually calculate the eigenvalues using the cubic formula. – N. S. Aug 9 '12 at 18:09
Thanks, @N.S., I have been working with this approach but thought there may be better ways and, it would seem, there are. – Richard Aug 9 '12 at 20:10
As it was pointed, $\det(A)< 0$ always guarantees a positive eigenvalue, anyhow it is not necessarily a necessary condition. And in my opinion, if this doesn't happen, most of the methods posted are not really much easier than simply writing the solutions. – N. S. Aug 9 '12 at 20:39

Let $M$ be the matrix (which I assume has real entries), and $p(x) = x^3 + a_2 x^2 + a_1 x + a_0$ its characteristic polynomial $\det(xI - M)$. If $a_0 < 0$, i.e. $\det(M) > 0$, there is always a positive eigenvalue.

Now suppose $a_0 \ge 0$. If $a_2^2 < 3 a_1$, the roots of $p'$ are complex, so $p(x)$ is increasing and there are no positive eigenvalues. If $a_2^2 \ge 3 a_1$, let $r = (\sqrt{a_2^2-3a_1}-a_2)/3$ which is the greatest root of $p'(x)$. In order for there to be a positive eigenvalue, we need $r > 0$ and $p(r) \le 0$.

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Just to avoid misunderstandings: The second paragraph relies on the determinant test in the first paragraph; some of its statements wouldn't hold if the determinant were positive. – joriki Aug 9 '12 at 18:43
@joriki: That's why I started the paragraph with "Now suppose $a_0 \ge 0$". – Robert Israel Aug 9 '12 at 19:02
I know, I just thought that someone not familiar with all this might not immediately see which statements depend on that -- e.g. in concluding from $p$ increasing that there are no positive eigenvalues you're using the fact that if there were one negative and two conjugate complex eigenvalues the product of the latter would be positive and the determinant negative -- that's not entirely obvious. – joriki Aug 9 '12 at 19:15
@joriki: Actually I'm just using the fact that if $p$ is increasing, then $p(x) > p(0) = a_0 \ge 0$ for $x > 0$. – Robert Israel Aug 9 '12 at 21:04
Ah, I see -- then I guess it was obvious after all ;-) – joriki Aug 9 '12 at 21:25

The characteristic polynomial of the matrix $M$ is the following function of $\lambda$: $$ \det(\lambda I - M) $$ where $I$ is the identity matrix of the same size. The values of $\lambda$ for which the characteristic polynomial are $0$ are the eigenvalues. For a $3\times3$ matrix, you get a third-degree polynomial. So the question is whether a specified third-degree equation has a positive root. Now try looking at Descartes' rule of signs. Certainly there is no positive root if all of the coefficients of the polynomial are positive.

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If the determinant is positive, there is (at least) one (or three) positive eigenvalues.

If not, the characteristic polynomial is $$P(x)=d-c.x+b.x^2-x^3$$

with $d$ as the determinant, $b$ the trace, and $c$ the sum of principal minors.

If there is a positive root, the maximum of this polynomial on $\mathbb R^+$ is positive. So look at the derivate polynomial $$P'(x)=-c+2bx-3x^2$$ If this polynomial has his largest root $r>0$ , just compute $P(r)$. If $P(r)>0$, then $P$ has a positive root.

So, in the worst case, you need to compute the characteristic polynomial, derivate it, and find the roots of a degree 2 polynomial.

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could you please clarify your first sentence - I think there's a word missing and it seems important. – Richard Aug 9 '12 at 20:11

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