I have been given the following exercise
Use the Gerschgorin cirlce theorem to find an upper bound for the spectral condition of a matrix $A$ which is real, symmetric and diagonally dominant.
I guess we can also assume, that $A$ is invertible, because we defined the spectral condition only for invertible matrices.
I tried the following: $$ \kappa_2(A) \overset{\text{def.}}{=} \Vert A \Vert_2 \Vert A^{-1} \Vert_2 = \varrho(A)\varrho(A^{-1}) $$ where $\varrho(M)$ is the spectral radius of $M$.
There is an easy way to bound $\varrho(A)$: Using Gerschgorin and the diagonal dominance we get $$ \vert \lambda - a_{jj} \vert \leq \sum_{l \neq j} a_{jl} \leq \vert a_{jj} \vert\\ \Rightarrow \vert \lambda - a_{jj} \vert + \vert a_{jj} \vert \leq 2\vert a_{jj} \vert $$ and by the triangle inequality $$ \vert \lambda \vert \leq \vert \lambda - a_{jj} \vert + \vert a_{jj} \vert \leq 2\vert a_{jj} \vert $$ then we choose $\vert a_{jj} \vert$ maximal and so got $\varrho(A)$ bounded.
But bounding $\varrho(A^{-1})$ is rather difficult: I wanted to use $\varrho(A^{-1}) = 1/\min( \vert \text{eig}(A) \vert)$, but I have not found a way to lower bound $\min( \vert \text{eig}(A) \vert)$, which also seems intuitively impossible to me; Just imagine the circles we get by using the inequality which we derived by using the diagonal dominance $$ \vert \lambda - a_{jj} \vert \leq \vert a_{jj} \vert $$ all of thous circles include the origin. I guess you see my problem.
I could lower bound $\min( \vert \text{eig}(A) \vert$ by not using the diagonal dominance, but I guess this would just ruin the whole point of this exercise.
Maybe I am overlooking something, or maybe there is an entirely different method.
Thanks for your help!
By Gerschgorin disc theorem, every eigenvalue of $A$ lies inside the union of the open Gerschgorin discs $\cup_{i=1}^n B(a_{ii},\sum_{j\ne i}|a_{ij}|)$. Since the $A$ in question is symmetric and diagonally dominant (I suppose the question means that $A$ is strictly diagonally dominant, otherwise the condition number of $A$ may not exist, such as when $A=0$), we have either $$ 0<a_{ii}-\sum_{j\ne i}|a_{ij}|\le\lambda\le a_{ii}+\sum_{j\ne i}|a_{ij}| $$ or $$ a_{ii}-\sum_{j\ne i}|a_{ij}|\le\lambda\le a_{ii}+\sum_{j\ne i}|a_{ij}|<0 $$ for every eigenvalue $\lambda$ of $A$ in the disc (say, the $i$-th one) it belongs to. It follows that $$ 0<|a_{ii}|-\sum_{j\ne i}|a_{ij}|\le|\lambda|\le |a_{ii}|+\sum_{j\ne i}|a_{ij}| $$ in the disc that $\lambda$ belongs to. Consequently, $$ \kappa_2(A)=\frac{|\lambda|_\max(A)}{|\lambda|_\min(A)}\le\frac{ \max_i\left\{|a_{ii}|+\sum_{j\ne i}|a_{ij}|\right\}, }{ \min_i\left\{|a_{ii}|-\sum_{j\ne i}|a_{ij}|\right\} }. $$