Let $A$ be a $3$ by $3$ symmetric matrix with trace being zero, that is, $a_{11}+a_{22}+a_{33}=0$. And let $\lambda_1\leq \lambda_2\leq \lambda_3$ be real eigenvalues of $A$.
We have $\lambda_1+ \lambda_2+ \lambda_3=0$. It is easy to see $\lambda_1\leq 0\leq \lambda_3$. $|\lambda_2|\leq |\lambda_1|$, $|\lambda_2|\leq |\lambda_3|$. I am wondering can we estimate $\lambda_2$, $|\lambda_2|$ or $\max(\lambda_2,0)$ by entries $a_{i,j}$ of $A$?
Best Answer
The characteristic equation is a depressed cubic, of the form $\lambda^3-p\lambda-q$ (Wikipedia). Mind the minus signs.
The roots of the derivative are $$\lambda=\pm\sqrt{\frac p3}$$ and they separate the roots of the cubic. This gives an upper bound.
As we are in the casus irreductibilis, if you want exact roots, you can't spare the trigonometric solution.
An approximate value can be obtained by linear interpolation between the two extrema,
$$\left(-\sqrt{\frac p3},\frac{2p}{3}-q\right),\left(\sqrt{\frac p3},-\frac{2p}{3}-q\right).$$