# the absolute value smallnest Eigenvalue estimate of a symmetric matrix

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 $$A$$?

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,

$$\left(-\sqrt{\frac p3},\frac{2p}{3}-q\right),\left(\sqrt{\frac p3},-\frac{2p}{3}-q\right).$$

• It is also worth noting that the roots are nicely expressible in terms of the entries of $A$, since we have $p = \frac 12 \operatorname{tr}(A^2)$. With the symmetry of the matrix, the roots of the derivative are $$\lambda = \pm \frac{\|A\|_F}{\sqrt{6}}$$ where $\|\cdot \|_F$ denotes the Frobenius norm. Feb 19, 2020 at 9:44

You might find the Cauchy interlacing theorem useful in this regard.

Let $$B$$ denote a $$2 \times 2$$ principal submatrix of $$A$$. For instance, we can take the leading principal submatrix $$B = \pmatrix{a_{11} & a_{12}\\ a_{21} & a_{22}}.$$ Let $$\mu_1 \leq \mu_2$$ denote the eigenvalues of $$B$$. The interlacing theorem sates that we have $$\lambda_1 \leq \mu_1 \leq \lambda_2 \leq \mu_2 \leq \lambda_3.$$ In other words, $$\lambda_2$$ lies between the eigenvalues of $$B$$. This leads to a nice inequality since the eigenvalues of a $$2 \times 2$$ matrix are easy to compute. In particular, $$B$$ has characteristic polynomial $$x^2 - \operatorname{trace}(B)x + \det(B)$$ so that $$\mu_1 = \frac{\operatorname{trace}(B) - \sqrt{\operatorname{trace}(B)^2 - 4\det(B)}}{2},\\ \mu_2 = \frac{\operatorname{trace}(B) + \sqrt{\operatorname{trace}(B)^2 - 4\det(B)}}{2}.$$

This inequality becomes tight if we consider submatrices of the matrix $$U^TAU$$ for $$3 \times 3$$ orthogonal matrices $$U$$. That is, there necessarily exists a $$U$$ for which a submatrix $$B$$ of $$U^TAU$$ satisfies $$\mu_1 = \lambda_2$$, and there necessarily exists another $$U$$ for which a submatrix $$B$$ of $$U^TAU$$ satisfies $$\mu_2 = \lambda_2$$.