$A=\begin{pmatrix}-5 & 0 & 0 \\ 2 & 2 & 1 \\ 3 & -5 & 4\end{pmatrix}$
Find the Gerschgorin-discs, where the eigenvalues of $A$ lie.
According to the formula if we have a $n\times n$-Matrix (where $i$:row, $j$:column)
$D_i=\{|x-(a_{ii})\le\sum\limits_{j=1,j\neq i}^{n} |a_{ij}| \}$
The eigenvalues must lie in $\cup_{i=1}^{n}D_i$
Hence;
$D_1=\{|x+5|\le 0\}\Rightarrow\lambda_1=-5$
$D_2=\{|x-2|\le 3\}$
$D_3=\{|x-4|\le 8\}$
The same for the transposed of $A$ (since they have smae eigenvalues)
$D'_2=\{|x-2|\le 5\}$
$D'_3=\{|x-4|\le 1\}$
Since $\lambda_1,\lambda_2\in(D_1\cup D_2)$ and $\lambda_1,\lambda_2\in(D'_1\cup D'_2)$
$\Rightarrow\lambda_1,\lambda_2\in(D_1\cup D_2)\cap(D'_1\cup D'_2)$
we can always choose the smaller circle of both $D_i$ or $D'_i$, is that correct ?
and there are no smaller discs for the eigenvalues, right ?
and the last question: If we have a matrix, whose i-th row entries are all zero except for the diagonal entry $a_{ii}$, then $a_{ii}$ is a eigenvalue ?
You get eigenvalues in $(D_2 \cup D_3) \cap (D_2' \cap D_3')$, so you can use the smaller discs for finding eigenvalues.
If the i-th row of the matrix contains all zeros except for $a_{ii}$, then $e_i^T$ - the i-th unit vector is a left eigenvector of $A$ to the eigenvalue $a_{ii}$.