Q:

Finding eigenvectors of a 3x3 matrix 2

I am trying to find the eigenvectors of the following 3x3 matrix:
\begin{bmatrix}
        -9 & 4 & 4 \\
        -8 & 3 & 4 \\
        -16 & 8 & 7 \\
        \end{bmatrix}
The eigenvalues for the matrix are $\lambda_1=3$, $\lambda_2=-1$, $\lambda_3=-1$.
$\lambda_1=3$ :
$$\lambda I_3-A=\begin{bmatrix}
        (3)+9 & -4 & -4 \\
        8 & (3)-3 & -4 \\
        16 & -8 & (3)-7 \\
        \end{bmatrix}=\begin{bmatrix}
        12 & -4 & -4 \\
        8 & 0 & -4 \\
        16 & -8 & -4 \\
        \end{bmatrix}$$
$$8x-4z=0$$
Let $z=t$
$$8x=4t$$
$$x=\frac 12t$$
$$-4y=4t-12(\frac 12t)$$
$$y=\frac 12t$$
When $t=2$$$v_1=\begin{bmatrix}
        1 \\
        1 \\
        2 \\
        \end{bmatrix}$$
But now I'm struggling to find the eigenvectors when $\lambda_2=\lambda_3=-1$
$\lambda_2=\lambda_3=-1$ :
$$\lambda I_3-A=\begin{bmatrix}
        (-1)+9 & -4 & -4 \\
        8 & (-1)-3 & -4 \\
        16 & -8 & (-1)-7 \\
        \end{bmatrix}=\begin{bmatrix}
        8 & -4 & -4 \\
        8 & -4 & -4 \\
     16 & -8 & -8 \\
        \end{bmatrix}$$
This matrix gives me 3 linear equations that are all multiples of each other so I effectively only have one equation:
$$2x-y-z=0$$
According to WolframAlpha there should be two more eigenvectors:
$$v_2=\begin{bmatrix}
        1 \\
        0 \\
        2 \\
        \end{bmatrix}$$ and $$v_3=\begin{bmatrix}
        1 \\
        2 \\
        0 \\
        \end{bmatrix}$$
but I have no idea how to find these. I'm pretty new to matrices so this is probably a really simple problem but any help would be much appreciated.

A:

From the equation $2x-y-z=0$, you can finish your problem. All eigenvectors are of the form $(x,y,z)$ that satisfy that equation. That is, all vectors of the form $$\begin{bmatrix}x\\y\\2x-y\end{bmatrix}
=x\begin{bmatrix}1\\0\\2\end{bmatrix} + y \begin{bmatrix}0\\1\\-1\end{bmatrix}.$$
Essentially, anything in the span of $(1,0,2)$ and $(0,1,-1)$ is an eigenvector.

