How to prove a tighter bound $|\lambda_3-1| \leq \epsilon^2$ for an eigenvalue of $A$ with Gerschgorin's theorem and similar matrices?

Question

Given the following matrix $$A = \begin{bmatrix} 8 & 1 & 0\\ 1 & 4 & \epsilon\\ 0 & \epsilon & 1\\ \end{bmatrix}, |\epsilon| < 1.$$ Gerschgorin's theorem states that each of the $\lambda_i$ eigenvalues of A will be placed inside a circular disk with center $a_{i,i}$ and radius $\sum^n_{j = 1, j\neq i} |a_{i,j}|$. As a direct consequence, the lowest eigenvalue $\lambda_3$ of A is such that $|\lambda_3 -1| < \epsilon$.

However I'm now asked to prove the tighter bound $|\lambda_3-1| < \epsilon^2$ using diagonal similarity transformations. My initial idea was to use the first step of the QR algorithm for Hessemberg diagonalization as usual to obtain a similar matrix $A_1 = Q^*AQ$ (with $Q$ computed by Householder transformations) preserving the eigenvalues and hopefully the tighter bound appears using Gerschgorin's theorem once again. I've made some numerical tests using Octave for some particular $\epsilon$ and it seems to be indeed the case that $A_1$ satisfies the bound. The problem is that the exact computation of $Q$ and $A_1$ have been absolutely painful, specially when dealing with the unknown symbol $\epsilon$. I also tried to use a shift of $\rho = 1$ with no better results. So now I'm suspicious there is a simpler similarity transformation that could lead to the desired result. Any suggestions would be highly appreciated.

Answer 1

Subtract off the identity matrix so the upper left term is now $7.$ The characteristic polynomial is now $$ x^3 - 10 x^2 + (20 -\epsilon^2)x + 7 \epsilon^2 $$

We wish to show that this has a root near zero, between $- \epsilon^2$ and $\epsilon^2.$ I guess we take $\epsilon \neq 0$ and treat that as a separate case.

The value of the characteristic polynomial when $x = \epsilon^2$ is $$ \epsilon^2 \left( 27 - 11 \epsilon^2 + \epsilon^4 \right) $$ which is positive, as $\epsilon^2 < 1.$

The value of the characteristic polynomial when $x = - \epsilon^2$ is $$ -\epsilon^2 \left( 13 +9 \epsilon^2 + \epsilon^4 \right) $$ which is negative, as $\epsilon^2 < 1.$

Te characteristic polynomial is continuous in $x,$ thus it has a root with $$ - \epsilon^2 < x < \epsilon^2 $$

The original matrix has an eigenvalue $x$ with $$ 1- \epsilon^2 < x < 1 +\epsilon^2 $$

ADDED: As Robert points out, we can simply evaluate the shifted characteristic polynomial $ x^3 - 10 x^2 + (20 -\epsilon^2)x + 7 \epsilon^2 $ at $x = t \epsilon^2,$ with $t$ real. When $t=0$ we get (positive) $7 \epsilon^2.$ There is some cancellation available if we take $t = -\frac{7}{20},$ and the polynomial comes out as $$ \epsilon^2 \left( - \frac{350}{400} \epsilon^2 - \frac{343}{8000} \epsilon^4\right) $$ which is negative. Thus the shifted eigenvalue is between $-\frac{7}{20} \epsilon^2$ and $0,$ the original eigenvalue between $1-\frac{7}{20} \epsilon^2$ and $1.$

Answer 2

One of the purposes of the question was to apply similarity transformations (which preserves eigenvalues) and Gerschgorin's theorem, so I'm posting an alternative to the accepted answer. As suggested by @user8675309, the problem may also be solved using a similarity transformation $A \mapsto D^{-1}AD$ with $D$ being a diagonal matrix.

If we let $c$ be a free parameter in $D = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & c\\ \end{bmatrix}$, then $D^{-1}AD = \begin{bmatrix} 8 & 1 & 0\\ 1 & 4 & c\epsilon\\ 0 & \frac{\epsilon}{c} & 1\\ \end{bmatrix}$.

To make the bound as requested, take $c = \epsilon^{-1}$ to get the 3 following Gerschgorin's theorem disks $$D_1 = \{x \in \mathbb{C} : |x-8| \leq 1\}, D_2 = \{x \in \mathbb{C} : |x-4| \leq 2\}, D_3 = \{x \in \mathbb{C} : |x-1| \leq \epsilon^2\}$$ Since they are disjoint under $|\epsilon| < 1$, each eigenvalue belongs to exactly one of this sets and therefore the lowest one is restricted to $|\lambda_3 - 1| \leq \epsilon^2.$