I'm trying to find a way to estimate the condition $\kappa(A)$ of a symmetric, real matrix $A$ based on the norm $\lVert\cdot\rVert_2$ by using the Gershgorin circle theorem.
For this, I calculated: $$\kappa(A) = \lVert A^tA\rVert_2 \lVert {A^{-1}}^tA^{-1}\rVert_2 = \lVert A^2\rVert_2 \lVert {A^{-1}}^2\rVert_2 = \rho(A)\rho(A^{-1})$$
where $\rho(A)$ is the biggest absolute eigenvalue of $A$.
Now, I can easily calculate a conservative estimate for $\rho(A)$ by finding the biggest eigenvalue possible using the Gershgorin circles, based on the fact that all eigenvalues will be real, since $A$ is real and symmetric, thus rendering the circles into intervals on the real number line.
The problems start with $\rho(A^{-1})$. Since it holds true, that $$\lambda \:\text{EV of }A \Leftrightarrow \lambda^{-1} \:\text{EV of }A^{-1}$$ I don't have to calculate $A^{-1}$, but instead approximate $\rho(A^{-1})$ with the biggest $\frac{1}{\lambda}$, where $\lambda$ is the smallest possible absolute EV of $A$. So far so good, and this approach gives good results for many matrices I have encountered.
But for some matrices, this approach creates problems. Consider the following matrix:
$$A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 3 \end{pmatrix}$$
This matrix is clearly symmetric and real. But consider the Gershgorin circles (now intervals):
- $D(1, 5) \rightarrow \lambda \in \left[-4, 6\right]$
- $D(4, 7)\rightarrow \lambda \in \left[-3, 11\right]$
- $D(3, 8)\rightarrow \lambda \in \left[-5, 11\right]$
Some of these intervals are crossing the zero point, which means that if I want to maximize $\frac{1}{\lambda}$, it becomes $\infty$, since I can find arbitrarily small $\lambda$ as the smallest possible EV, rendering my approach completely useless!
Can this be avoided without calculating $A^{-1}$? Can I choose some $\frac{1}{\lambda}$ to estimate this? Can I somehow "save" my approach?
P.S: The context is being able to solve problems in an exam, since the midterm had problems exactly like this, that required using the Gershgorin circles to provide an estimate.