Help finding eigenvectors? The given matrix is:
$$
\begin{pmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{pmatrix}\qquad
$$
I got the characteristic polynomial of $$x^3 - x^2 - 5x - 3 = 0$$
which factors down to $$(x+1)^2 * (x-3) = 0$$
I see that it  has eigenvalues of -1 and 3.
I know I'm almost there, I plugged in the eigenvalues to $A-\lambda I$ but completely forgot how to find the eigenvectors after this.
When $\lambda$ = 3, I got:
$$\begin{pmatrix}0 & 1 & 6 \\ 2 & -2 & 0 \\ -1 & 0 & 0\end{pmatrix} \begin{pmatrix}x
_1 \\ x_2 \\ x_3\end{pmatrix}=0\qquad$$
and when $\lambda$ = -1, I got:
$$\begin{pmatrix}4 & 1 & 6 \\ 2 & 0 & 0 \\ -1 & 0 & 4\end{pmatrix}\begin{pmatrix}x
_1 \\ x_2 \\ x_3\end{pmatrix}=0\qquad$$
Where do I go from here? Row-reduce the $3x3$ matrices to solve?
 A: I am going to use the approach you are using so you can see your issues.
We are given: $A = \begin{bmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{bmatrix}$
We set up and and solve: $|A - \lambda I| = 0$, which yields:
$$\left|\begin{matrix}3-\lambda & 1 & 6 \\ 2 & 1-\lambda & 0 \\ -1 & 0 & -3-\lambda\end{matrix}\right| = 0$$
This yields a characteristic polynomial and eigenvalues as:
$$-\lambda^3+\lambda^2+5 \lambda+3 = -(\lambda-3) (\lambda+1)^2 = 0 ~~~\rightarrow ~~~ \lambda_1 = 3, \lambda_{2,3} = -1$$
We have multiplicities of $1$ and $2$ for those eigenvalues.
To find the eigenvectors, we generally solve $[ A - \lambda_i I]v_i = 0$, but since we have a repeated eigenvalue, we may need to change that strategy and find a generalized eigenvalue (I'll let you deal with the details of this and geometric multiplicities).
So, for $\lambda_1 = 3$, we have:
$[A- 3I]v_1 = \begin{bmatrix}0 & 1 & 6 \\ 2 & -2 & 0 \\ -1 & 0 & -6\end{bmatrix}v_1 = 0$
Doing row-reduced-echelon-form (RREF), yields:
$\begin{bmatrix}1 & 0 & 6 \\ 0 & 1 & 6 \\ 0 & 0 & 0\end{bmatrix}v_1 = 0$
Thus, $b = -6c, a = -6c \rightarrow ~~\text{let}~~ c = 1 \rightarrow a = b = -6, v_1 = (-6,-6,1)$.
Repeating this same process for the second eigenvalue, we have as RREF:
$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 0\end{bmatrix}v_1 = 0$
This yields an eigenvector of $v_2 = (-2, 2, 1)$.
Unfortunately, we cannot get another linearly independent eigenvector, so need to get a generalized one, by doing $[A - \lambda I]v_3 = v_2$, so we have:
$\begin{bmatrix}4 & 1 & 6 \\ 2 & 2 & 0 \\-1 & 0 & 2\end{bmatrix}v_3 = \begin{bmatrix} -2  \\ 2  \\ 1 \end{bmatrix}$
After RREF, we arrive at:
$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 0\end{bmatrix}v_3 = \begin{bmatrix} -1  \\ 2  \\ 0 \end{bmatrix}$
So, we have: $a = -1 -2c, b = 2 + 2c \rightarrow ~~ \text{let} ~~ c = 0 \rightarrow a = -1, b = 2$, thus $v_3 = (-1,2,0)$
Putting all of this together, we have the eigenvalue/eigenvector pairs:


*

*$\lambda_1 = 3, v_1 = (-6, -6, 1)$   

*$\lambda_2 = -1, v_2 = (-2, 2, 1)$

*$\lambda_3 = -1, v_3 = (-1,2,0)$

A: it is enough you find:
$$\left(\begin{bmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{bmatrix}- 3I\right)\begin{bmatrix}x
_1 \\ x_2 \\ x_3\end{bmatrix}=0\qquad
\text{and} \qquad 
\left(\begin{bmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{bmatrix}+ I\right)\begin{bmatrix}x
_1 \\ x_2 \\ x_3\end{bmatrix}=0 \ .$$
A: Remember an eigenvector $v_\lambda$ of the matrix $M$ associated to the eigenvalue $\lambda$ is a vector such $Mv_\lambda=\lambda v_\lambda$ or equivalently $(M-\lambda I)v_\lambda=0$. So, you're trying to find a vector (or vectors depending on multiplicity) $v_\lambda$ which spans the nullspace of the matrix $M-\lambda I$. Do you know how to find the nullspace of a matrix?
