Determine whether a $3\times3$ matrix has a positive eigenvalue? Given a $3\times3$ matrix is there a criterion capable of telling whether the matrix has a positive eigenvalue?
 A: 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.
A: 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 $p(r) \le 0$.
A: 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.
