Find the eigenvectors of $$A = \left[ {\begin{array}{ccc} 2 & 0 & 0 \\ 4 & 2 & 3 \\ 3 & 0 & -1 \end{array} } \right]$$
Step 1. Find the eigenvalues.
$$ \left| {\begin{array}{ccc} 2-\lambda & 0 & 0 \\ 4 & 2-\lambda & 3 \\ 3 & 0 & -1-\lambda \end{array} } \right| = (2-\lambda)^2(-1-\lambda) $$
So the eigenvalues $\lambda = 2, -1$, with repeated multiplicity for $2$.
Step 2. The eigenvector for $\lambda = 2$ is
$$\left[ {\begin{array}{ccc} 0 & 0 & 0 \\ 4 & 0 & 3 \\ 3 & 0 & -3 \end{array} } \right] \left[ {\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} } \right] = \left[ {\begin{array}{c} 0 \\ 0 \\ 0 \end{array} } \right] $$
This gives the system of equations
$$\begin{align*} 4x_1 + 3x_3 &= 0 \\ 3x_1 - 3x_3 &= 0 \end{align*}$$
Which gives $(x_1, x_2, x_2)$ as an "eigenvector." However, this seems extremely defective and I don't know how to proceed.
The equations imply $x_1=x_3$ and $x_1=-3/4x_3$ So $x_3 = 0, x_1=0$ and an eigenvector is $(0,1,0)$.
To find another generalized eigenvector for $\lambda =2$ you can set $(A-2I)^2x = 0$ and do the same thing.