I have a matrix
$$A= \begin{pmatrix} 2 & 2 & -2 \\ 5 & 1 & -3 \\ 1 & 5 & -3 \end{pmatrix} $$
I want to find the eigenvalues and corresponding eigenvectors. I've tried this question multiple times but I end up with the same answer which is $\lambda^3 = 0$.
Q1) Can I have 3 repeated eigenvalues and can they all be zero?
Q2) Is my working out correct? Do you end up with $\lambda^3 = 0$?
Q3) If indeed we can have 3 repeated eigenvalues at zero, how do I find the eigenvectors?
Thank you
$\lambda^3 = 0$ is indeed characteristic equation.
You have only one true eiegnvector.
$v_1 = \begin{bmatrix}1\\1\\2\end{bmatrix}$
You have 2 "generalized eigenvectors.
Find $v_2$ such that
$Av_2 = \lambda v_2 + v_1$
In which case:
$(A-\lambda I)v_2 = v_1\\ (A-\lambda I)^2v_2 = (A-\lambda I)v_1 = 0$
And since $\lambda = 0$
$A^2 v_2 = 0$
$A^2 = \begin{bmatrix}12&-4&-4\\12&-4&-4\\24&-8&-8\end{bmatrix}$
And $v_2 = \begin{bmatrix}0\\1\\-1\end{bmatrix}$ would be a candidate.
Except $Av_2 = 4v_1$
now you need $v_3$ such that $Av_3 = v_2$
$A = \begin{bmatrix}1&0&\frac {1}{16}\\1&\frac 14 &-\frac 1{16}\\2&-\frac 14&0\end{bmatrix}\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}\begin{bmatrix}\frac 14&\frac14&\frac 14\\2&2&-2\\12&-4&-4\end{bmatrix}$