Determine size or number of jordan blocks

Question

I have done most of the work but I struggle to put this matrix into Jordan Normal Form.

$$C=\begin{bmatrix}1 & 0 & 0 & 0&0\\1 & -1 & 0 & 0&-1 \\1 & -1 & 0 & 0&-1 \\ 0 & 0 & 0 & 0&-1\\-1&1&0&0&1\end{bmatrix}$$

The characteristic polynomial is $(x-1)x^4$, so the eigenvalues are $1,0,0,0,0$. The rank of $C$ is 3. The rank of $C^2=2$.

I know there is only $1$ Jordan block of size $1$ for the eigenvalue $1$, $J_1(1)$. But I am stuck to determine the blocks for the eigenvalue $0$. I know the block sizes must add up to $4$. Can someone help me finish finding the Jordan blocks for $0$?

Answer 1

If $B$ is a Jordan block of sike $k$ for the eigenvalue $0$, then $B$ has rank $k-1$, and if $k > 1$, then $B^{2}$ has rank $k-2$. Since going from $C$ to $C^{2}$ the rank decreases by one, there must only one Jordan block of size $> 1$ for the eigenvalue $0$, and it must have size $3$ by the above, so you have a further block of size $1$ for $0$.

Answer 2

As at Jordan normal form (Basis)
we see here the method of finding some column vector $r_5,$ such that $C^3 r_5 = 0$ but $C^2 r_5 \neq 0.$ Once we choose that, we are forced to take $r_4 = C r_5,$ then column $r_3 = C r_4.$ This becomes an eigenvector because $Cr_3 = C^2 r_4 = C^3 r_5 = 0.$

I have picked $r_5 = (0,0,0,0,-1)^T$ and finished up. The other null vector $r_2$ can be anything in the two- dimensional subspace of genuine zero eigenvectors, so more choices are possible.

$$ \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 0&-1&1&0&0 \\ 0&-1&0&1&0 \\ -1&1&0&0&0 \\ 0&-1&0&0&-1 \\ \end{array} \right) \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 1&0&0&1&0 \\ 1&1&0&1&0 \\ 1&0&1&1&0 \\ -1&0&0&-1&-1 \\ \end{array} \right) = \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 0&1&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1& 0\\ 0&0&0&0&1 \\ \end{array} \right) $$

$$ \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 0&-1&1&0&0 \\ 0&-1&0&1&0 \\ -1&1&0&0&0 \\ 0&-1&0&0&-1 \\ \end{array} \right) \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 1&-1&0&0&-1 \\ 1&-1&0&0&-1 \\ 0&0&0&0&-1 \\ -1&1&0&0&1 \\ \end{array} \right) \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 1&0&0&1&0 \\ 1&1&0&1&0 \\ 1&0&1&1&0 \\ -1&0&0&-1&-1 \\ \end{array} \right) = \left( \begin{array}{c|c|ccc} 1&0&0&0&0 \\ \hline 0&0&0&0&0 \\ \hline 0&0&0&1&0 \\ 0&0&0&0& 1\\ 0&0&0&0&0 \\ \end{array} \right) $$

$$ \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 1&0&0&1&0 \\ 1&1&0&1&0 \\ 1&0&1&1&0 \\ -1&0&0&-1&-1 \\ \end{array} \right) \left( \begin{array}{c|c|ccc} 1&0&0&0&0 \\ \hline 0&0&0&0&0 \\ \hline 0&0&0&1&0 \\ 0&0&0&0& 1\\ 0&0&0&0&0 \\ \end{array} \right) \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 0&-1&1&0&0 \\ 0&-1&0&1&0 \\ -1&1&0&0&0 \\ 0&-1&0&0&-1 \\ \end{array} \right) = \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 1&-1&0&0&-1 \\ 1&-1&0&0&-1 \\ 0&0&0&0&-1 \\ -1&1&0&0&1 \\ \end{array} \right) $$