Given the matrix
$$\begin{pmatrix} 3 & 0& 0 \\ -3& 4& 9 \\ 0 & 0& 3 \end{pmatrix}$$
you get eigenvalues $3$ (twice) and $4$. However, when solving for the eigenvector of $3$, you end up with
$$\begin{pmatrix} 0 & 0 & 0 \\ -3 & 1 & 9 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$
How do you solve for the eigenvectors with the free variables?
Best Answer
Your caracteristic equation is $(3-\lambda)^2(4-\lambda)$ so we need to find two vectors such that
$\begin{bmatrix} 0 & 0 & 0 \\ -3 & 1 & 9 \\ 0 & 0 & 0 \end{bmatrix}v = 0$
Fortunately, you have the freedom do to that.
I think the easiest way to do these is to plug one of your entries equal to 0, and then plug a different entry equal to 0
$\begin{bmatrix} 0 & 0 & 0 \\ -3 & 1 & 9 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}v_1\\v_2\\0\end{bmatrix} = -3u_1 + u_2 = 0$
$u_2 = 3u_1$
$u = \begin{bmatrix}1\\3\\0\end{bmatrix}$
then
$\begin{bmatrix} 0 & 0 & 0 \\ -3 & 1 & 9 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}v_1\\0\\v_3\end{bmatrix} = -3v_1 + 9v_3 = 0$
$v_1 = 3v_3$
$v = \begin{bmatrix}3\\0\\1\end{bmatrix}$