A theorem related to the Gershgorin's theorem 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!
 A: 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. 
