Here's the question:
$\lambda$
The typical formula I've seen is $(A-\lambda I)v = 0$ where A is the starting matrix, $\lambda$ is the eigenvalue, I is:
$$\begin{bmatrix}
1&0&0 \\
0&1&0 \\
0&0&1 \end{bmatrix}$$
and $v$ is:
$$\begin{bmatrix}
v_{1} \\
v_{2} \\
v_{3} \end{bmatrix}$$
When I plug in I get the matrix:
$$
\begin{bmatrix}3&-2&3 \\0&-2&3 \\-1&2&-3 \end{bmatrix}
\begin{bmatrix}v_{1} \\v_{2} \\v_{3} \end{bmatrix}
=
\begin{bmatrix}0 \\0 \\0 \end{bmatrix}
$$
I'm not so sure what to do after this.
Best Answer
The characteristic polynomial is given by $|A - \lambda I| = 0$, hence:
$$-\lambda ^3+\lambda ^2+13 \lambda -13 = -(\lambda -1) \left(\lambda ^2-13\right) = 0$$
This shows one eigenvalue is $\lambda_1 = 1$.
To find the eigenvector, we solve $[A-\lambda_i I]v_i = [A - I]v_1=0$.
We perform Row-Reduced-Echelon-Form (RREF), yielding:
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 0 \\ \end{bmatrix}v_1 = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$$
Choose the eigenvector as $a = 0, c = 2 , b = 3$, so
$$ v_1 = \begin{bmatrix} 0 \\ 3 \\ 2\end{bmatrix}$$