There is a question/theorem in the book Matrix Analysis by R. Horn, page 351 says that:
Let $A \in M_{n}$. Then $\sigma(A) = \bigcap\limits_{S} G(S^{-1}AS)$ if the intersection is taken over all nonsingular $S$. (Here, $G(\cdot)$ represents the Gershgorin's region).
I'd like to prove it and I need a hint to start. Any comment/hint/answer is appreciated.
We know that:
$\sigma(A) \subseteq \bigcap_{S} G(S^{-1}AS)$:
this is because $\sigma(A) = \sigma(S^{-1}AS) \subseteq G(S^{-1}AS)$ for any invertible $S$ (by the Gerschgorin's theorem) and hence $\sigma(A) \subseteq \bigcap_{S} G(S^{-1}AS)$.
About equality I need a hint!
Inspired by comments of @achillehui:
We know that for any matrix $A \in M_{n}$ there is a nonsingular matrix $Q$ such that $$ Q^{-1}AQ = J = \begin{bmatrix} J(\lambda_{1}) & & \\ & \ddots & \\ & & J(\lambda_{m}) \end{bmatrix}, $$ where each Jordan block $$ J(\lambda_{i}) = \begin{bmatrix} \lambda_{i} & 1 & & \\ & \lambda_{i} & \ddots & \\ & & \ddots & 1\\ & & & \lambda_{i} \end{bmatrix}_{n_{i}\times n_{i}}, \quad i = 1, \ldots, m. $$ Let $D_{i}(t)= diag\{1,t,t^2, \ldots, t^{n_{i}}\}$. Then we have
\begin{align} D_{i}(t)^{-1}J(\lambda_{i})D_{i}{}(t) &= \begin{bmatrix} 1 & & & \\ & t^{-1} & & \\ & & & \ddots & \\ & & & & t^{-n_{i}} \end{bmatrix} \begin{bmatrix} \lambda_{i} & 1 & & \\ & \lambda_{i} & \ddots & \\ & & \ddots & 1\\ & & & \lambda_{i} \end{bmatrix} \begin{bmatrix} 1 & & & \\ & t & & \\ & & & \ddots & \\ & & & & t^{n_{i}} \end{bmatrix} \\ &= \begin{bmatrix} \lambda_{i} & t & & \\ & \lambda_{i} & \ddots & \\ & & \ddots & t\\ & & & \lambda_{i} \end{bmatrix}. \end{align}
Define $D = diag\{D_{1}(t), \ldots, D_{m}(t)\}$. Then we can write $$ D^{-1}JD = D^{-1}Q^{-1}AQD = S^{-1}AS = \begin{bmatrix} D_{1}(t)^{-1} J(\lambda_{1}) D_{1}(t) & & \\ & & \\ & \ddots & \\ & & D_{m}(t)^{-1} J(\lambda_{m}) D_{m}(t) \end{bmatrix}, $$ where $S = QD$. This implies $S^{-1}AS$ is a matrix with entries $\lambda_{i}$ along the main diagonal, $t$ along the superdiagonal and $0$ otherwise. Then $$ G(S^{-1}AS) = \bigcup_{i=1}^{n} \{z: |z-\lambda_{i}| \leq r \in \{t,0\}\} = \sigma(A), \quad \text{as} \quad t\rightarrow 0, $$ which means that $G(S^{-1}AS)$ is the union of circles of radius at most $t$ centered at points of $\sigma(A)$. Therefore, we can write $\bigcap_{S} G(S^{-1}AS) = \sigma(A)$ and the proof is completed.