Trouble with dependent matrix solution Im determining the eigenvector for $\lambda = 6$. Here is the following matrix $A - 6*I$:
0  0  0  0
0  1  0  1
0  0  0  0 
0  0  0  0

Thus the corresponding equation: $x_2 = -x_4$
Thus $x_1,x_3,x_4$ are free.
How do I express this in terms of an eigenvector, a little confused..?

A simpler example to show what can happen.
$$\left[\begin{array}{cc}2&1\\0&2\end{array}\right]$$
You will get $(2-\lambda)^2 = 0$ to solve for eigenvalues. So two eigenvalues at $2$.
$$\left[\begin{array}{cc|c}
0&1&0\\
0&0&0
\end{array}\right]$$
Wee see that the only requirement is $x_2 = 0$ and $x_1$ can be what it wants. Only one eigenvector!
