Find Jordan Decomposition without calculating $\ker(A - \lambda E)$ 
Without calculating the null space of $(A - \lambda E)$, find the Jordan decomposition of 
  $$
A = \begin{pmatrix}
0 & 1 & 1 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
$$

My ideas


*

*The characteristic polynomial of $A$ ist $P_A(t) = -t^5$, implying that $\lambda = 0$ is the only eigenvalue.

*$F_A(x) := Ax$ describes a nilpotent endomorphism, since $A^3 = 0$.
Therefore the minimal polynomial of $A$ ist $M_A(t) = t^3$.


My question


*

*How can I derive from the minimal polynomial, if I have two (2 x 2 and 3 x 3) or three (1 x 1, 1 x 1, 3 x 3) Jordan blocks?

 A: As the matrix is nilpotent, you have already noticed that the diagonal will be 0 in Jordan form. The fact that $A^2 \neq 0$ but $A^3 = 0$ tells you that one block will be $3 \times 3$. The question then is wether the remainder is a $2\times 2$ block or two $1\times 1$ Jordan blocks.
To answer the last question (without computing the kernel), just notice that the matrix is rank 2, then by rank-nullity, the dimension of the kernel is $3$. (dim Kernel = n-dim Rank). This states there are 3 linearly independent eigenvectors in the 0 eigenspace. This says there are three Jordan blocks. So you have two $1\times 1$ blocks.

EDIT: Just to extend the explanation, this is because the number of Jordan blocks corresponds exactly with the number of linearly independent eigenvectors, the first column of each block being one such vector. The $k$th column of the block is what is referred to normally as a "generalized" eigenvector, belonging to $\ker(A-\lambda E)^k$.

These are in fact easy to find without computation as well. Columns 1 and 4, respectively will correspond to eigenvectors
$$
\left\{\begin{pmatrix}
1 \\ 0 \\ 0 \\ 0 \\ 0
\end{pmatrix},
\begin{pmatrix}
0 \\ 0 \\ 0 \\ 1 \\ 0
\end{pmatrix}\right\}
$$
(No computation necessary to see this.)
Neither of these will be associated with the three-by-three block, although it is also obvious that another linearly independent eigenvector
$$
\begin{pmatrix}
0 \\ 0 \\ 1 \\ 0 \\ -1
\end{pmatrix}.
$$
Cheers.
