How do I find the characteristic polynomial and eigenvalues?

Question

For the following matrix, compute

1. its characteristic polynomial
2. its eigenvalues

$$A = \begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & -5 & 4\end{bmatrix}$$

So I think I know to find the characteristic polynomial, I have to compute det(A-$\lambda$I) = 0. Which gives...

$$A-\lambda I = \begin{bmatrix}-\lambda & 1 & 0 \\ 0 & -\lambda & 1 \\ 2 & -5 & 4-\lambda\end{bmatrix}$$

$$-\lambda\begin{vmatrix}-\lambda & 1 \\ -5 & 4-\lambda\end{vmatrix} - 1 \begin{vmatrix}0 & 1 \\2 & 4-\lambda\end{vmatrix} $$

After calculating the determinant I get, $-\lambda(-4\lambda + \lambda^{2}+5) +2 = 0 $

What do I do next? I'm completely stuck.

Answer 1

The sum of the entries on each row is always $1$, so $\left(1,\begin{bmatrix} 1\\1\\1\end{bmatrix}\right)$ is an eigenpair.

Since $\det(A)=2$ (easily seen by Laplace expansion on the first row) and $\text{tr}(A)=4$, the other 'two' eigenvalues follow easily.

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Actually answering your question, since $$-\lambda(-4\lambda + \lambda^{2}+5) +2 = 0\iff \lambda^3-4\lambda ^2+5\lambda -2=0$$

(I'd rather work with monic polynomials, this equivalence is unnecessary), the rational root theorem says the rational roots, if any, are in $\left\{-2-1,1,2\right\}$, so you can try these and hope for the best.

Answer 2

Visual inspection tells us that matrix $\rm A$ is a companion matrix and that $1$ is an eigenvalue of $\rm A$. Hence, the characteristic polynomial of $\rm A$ is

$$q (s) := s^3 - 4 s^2 + 5 s - 2 = (s - 1) (s^2 - 3 s + 2) = \color{blue}{(s-1)^2 (s-2)}$$