Generalized Eigenvectors when algebraic multiplicity greater than 1 Find the Generalized Eigenvectors of 
$$
\begin{bmatrix}
1 & 0 & 0 & 0 & 0\\
1 & -1 & 0 & 0 & -1\\
1 & -1 & 0 & 0 & -1\\
0 & 0 & 0 & 0 & -1\\
-1 & 1 & 0 & 0 & 1
\end{bmatrix}
$$
Since for $\lambda = 0, null(A-\lambda I) = 2.$ I have 2 eigenvectors = 
$$
\begin{bmatrix}
0\\0\\0\\1\\0
\end{bmatrix}
and
\begin{bmatrix}
0\\0\\1\\0\\0
\end{bmatrix}
$$
I know to use $(A - \lambda I)v_2 = v_1$ to get all eigenvectors corresponding to a $\lambda$. But this one has 2 blocks corresponding to $\lambda = 0$ in the Jordan Cannonical form.
$$
J = 
\begin{bmatrix}
1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
$$
So, how do I associate generalised eigenvectors to each Jordan block?
 A: The first Jordan block is the one-by-one matrix [1].  An associated eigenvector is x = [–1, –1, –1, –1, 1]$^{\mathrm T}.$  The multiplicity of that eigenvalue is one, so there are no complications here.
Dealing with eigenvalue 0 is not as easy.  The remainder of my answer is based on Wikipedia subtopic "Computation of generalized eigenvectors."
Because rank(A – 0 I) = rank A = 3, rank A$^2$ = 2, and rank A$^3$ = 1, there will be three linearly-independent generalized eigenvectors—z$_3$ of order 3, z$_2$ of order 2, and z$_1$ of order 1—that form a chain. To find z$_3,$ we solve A$^3$z$_3$ = 0 such that A$^2$z$_3$ ≠ 0.  The simplest solution is z$_3$ = [0, 0, 0, 0, 1]$^{\mathrm T}.$  Then z$_2$ = Az$_3$ = [0, –1, –1, –1, 1]$^{\mathrm T},$ and z$_1$ = Az$_2$ = [0, 0, 0, –1, 0]$^{\mathrm T}.$  Those three generalized eigenvectors are associated with the third, three-by-three Jordan block.
Finally, note that z$_1$ is a multiple of the first eigenvector that you give.  Hence, you can associate the other eigenvector y = [0, 0, 1, 0, 0]$^{\mathrm T}$ with the second, one-by-one Jordan block [0].
FYI, the Jordan decomposition A = S J S$^{-1}$ has S = [x, y, z$_1,$ z$_2,$ z$_3]$ where the generalized eigenvectors are the columns of S.
