# Finding Spectral Radius of Matrix

Find the Spectral Radius of $A=$ $\mbox{} \left[ \begin{array}{cc} 1 & 0 & 0 & 0 \\ 0 & 0 & c & 0 \\ 0 & -c & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]$

After going through the motions, I got that the $-x^{3}-c^{2}x=0$ $\Rightarrow$ $-x(x^{2}+c^{2})$

Now C is supposed to be in $\mathbb{R}$ while $A \in \mathbb{C}$

When I use the quadratic formula, I get that 2iC=0, and by the definition the spectral radius is the largest absolute value of the eigenvalues.

So what am I to make of the results of the quadratic formula? On the right path?

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Care to explain how you get $2iC = 0$ rather than $x = something$ from the quadratic formula? – Erick Wong Sep 26 '12 at 4:05
Okay, totally understand mistake. Thank you. – Edgar Aroutiounian Sep 26 '12 at 4:20
Then would $i c$ be the largest possible eigenvalue? – Edgar Aroutiounian Sep 26 '12 at 4:27

$\mathrm{det}(A-xI) =\mathrm{det}\left[ \begin{array}{cccc} 1-x & 0 & 0 & 0 \\ 0 & -x & c & 0 \\ 0 & -c & -x & 0 \\ 0 & 0 & 0 & -x \end{array} \right].$ Once you get that, you can factor it into linear factors over $\mathbb{C}$ to find the eigenvalues.
Note that $1$ is always an eigenvalue, since $\left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & c & 0 \\ 0 & -c & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \left[ \begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right]=\left[ \begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right]$ so your spectral radius can never be less than $1$.