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 $0$?
Best Answer
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 $0$.