Given a $3\times3$ matrix is there a criterion capable of telling whether the matrix has a positive eigenvalue?
Determine whether a $3\times3$ matrix has a positive eigenvalue?
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0As 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. – 2012-08-09
3 Answers
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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0Ah, I see -- then I guess [it was obvious after all](http://www.thescienceforum.com/mathematics/10833-math-joke.html) ;-) – 2012-08-09
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.
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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0could you please clari$f$y your first sentence - I think there's a word missing and it seems important. – 2012-08-09