Determine the smallest disc in which all the eigen values of a given matrix lie Let , $$A=\left[\begin{matrix}1&-2&3&-2\\1&1&0&3\\-1&1&1&-1\\0&-3&1&1\end{matrix}\right]$$Which of the following is the smallest disc in $\mathbb C$  which contains all eigen values of $A$.


*

*$|z-1|\le 7$

*$|z-1|\le 6$

*$|z-1|\le 4$.
Characteristic polynomial of $A$ is $x^4-4x^3+21x^2-48x+46$. From this computing eigen values is very difficult in hand. Without computing eigen values how we can detect the required interval ?
Does there any other process??

Yes it's an application of the Gershgorin Theorem:
The theorem states that the eigenvalues are bounded in the union of the regions:
\begin{equation}
K_i=\left\{ |z-a_{ii}|<\underset{{\scriptscriptstyle i\neq j}}{\sum|a_{ij}|}\right\} 
\end{equation}
Here we have:
$$A=\left[\begin{matrix}1&-2&3&-2\\1&1&0&3\\-1&1&1&-1\\0&-3&1&1\end{matrix}\right]$$
The circles are all centered on value 1 since all $a_{ii}=1$ so summing horizontally line by line you would obtain the following restriction on the eigenvalues:
$$K_1=\left\{|z-1|< |-2|+|3|+|-2|=7 \right\} $$
$$K_2=\left\{|z-1|<  4 \right\}$$
$$K_3=\left\{|z-1|<  3\right\}$$
$$K_4=\left\{|z-1|<  4\right\}$$
The union of all $K_i$ is $K_1$. So $K=\left\{|z-1|<7 \right\}$ seems to be the minimum disc including all the eigenvalues. But since the eigenvalues are the same for the transpose of the matrix you can calculate again watching for the columns instead of the rows and obtain
$$|z-a_{ii}|< {\sum}|a_{ji}|$$ with $i \neq j$ and obtain as maximum radius
$$|z-1|< |-2|+|-3|+|1|=6$$
So the correct answer is the $\bf (2)$ and all eigenvalues are included in $K=\left\{|z-1|<6 \right\}$
