<s>[INST] 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??
 [/INST] (Too long for a comment.)
This looks like a trick question and I don't know the trick, but you may try the following approach. Let $B=A-I$. The characteristic polynomial of $B$ is $x^4 + 15x^2 - 14x + 16$. As $15x^2-14x+16>0$ over $\mathbb R$, $B$ has no real eigenvalues. Since $B$ has zero trace and nonreal eigenvalues of a real matrix must occur in conjugate pairs, it follows that
$$
x^4+15x^2-14x+16=(x^2-2ax+r^2)(x^2+2ax+R^2)
$$
where $a\ne0$ is the real part of one of the eigenvalues of $B$ and $r,R$ ( with $(0<|a|<r\le R$) are the moduli of the eigenvalues of $B$. By comparing coefficients of both sides, we get
\begin{cases}
R^2 + r^2 - 4a^2 = 15,\\
2a(R^2-r^2) = 14,\\
r^2R^2 = 16.
\end{cases}
Substitute the last two equations into the first, we get
$$
\left(\frac{16}{r^2} + r^2 - 15\right)\left(\frac{16}{r^2} - r^2\right)^2 = 196.
$$
If you can show that $r>1$, then we may conclude from $r^2R^2=16$ that $R<4$ and hence $(3)$ is the correct answer.
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