After solving for eigenvalues, how do you solve for eigenvectors if your matrix has free variables? 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?
 A: As Eigenvectors are not unique, this will happen every time you want to calculate the eigenvectors.
What you can do is, in order to calculate this undetermined system is to assume that, $x_1 = s$, $x_2 = t$ and calculate $x_3$ depending on these two parameters. If you do this, you will end up with a vector depending on these two parameters, or if you separate the parameters, an eigenvector which is a linear combination of two linearly independent vectors where $s,t$ are the coefficients: $v= sv_1 + tv_2$, i.e. your Eigen-space is $span \{v_1,v_2\}$
A: 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
$\begin{bmatrix} 0 & 0 & 0 \\ -3 & 1 & 9 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}v_1\\0\\v_3\end{bmatrix} = -3v_1 + 9v_3 = 0$
$v_1 = 3v_3$
$v = \begin{bmatrix}3\\0\\1\end{bmatrix}$
A: Solve the linear system as usual. Intuitively, when you get a row of zeros, you know you have more than one solution. In this case, you have more than one eigenvector for the corresponding eigenvalue.
$$\begin{cases} -3x_1 +x_2+9x_3 = 0 \\ x_2 = t \\ x_3 = s \end{cases}$$
$$\implies x_1 = \frac{1}{3}t + 3s$$
Thus a basis for the eigenspace corresponding to the eigenvalue of $3$ is
$$B_{\lambda_3} = \Bigg \{ t\begin{bmatrix} \frac{1}{3} \\ 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix} \Bigg \}$$
This means all eigenvectors can be written as a linear combination of the vectors in this basis.
