Find the smallest disk in complex plane containing the eigenvalues of a matrix (NBHM 2012) Consider the given matrix
$$\left[\begin{matrix}1& -2& 3& -2 \\1& 1& 0& 3\\-1& 1& 1& -1\\0& -3& 1& 1&\end{matrix}\right]$$
Find out the smallest disk like ($|z-1| < r$ ) in the complex plane containing the eigenvalues of the given matrix.
One easiest method is to find out the eigenvalues from the characteristic equation of the matrix. I can do it so no problem with it. 
I am looking for a method to solve the problem without the above method that can be evaluated in very short time. This type of method is possible, I believe.
Thank you for your help

For this particular problem, the matrix (say $A$) turns out to have two pairs of conjugate eigenvalues $\lambda,\bar{\lambda},\mu,\bar{\mu}$. The center of the required disc is therefore $\frac14\operatorname{trace}(A)$ and the radius is $|\lambda-\frac14\operatorname{trace}(A)|$. So, even if you can infer by some clever argument that $A$ has two pairs of conjugate eigenvalues, you still have to calculate one eigenvalue of $A$. So, I don't think you can solve the problem in a simple manner without calculating the spectrum of $A$.
