Finding the Jordan Canonical Form of a Matrix

Question

Let $J_5$ be the $5x5$ Jordan block, with eigenvalue zero, and define $B=J^2$. Find the Jordan form of $B$.

I found that the $J_5$ is a matrix with $0s$ on the diagonal and $1s$ on top. I squared it and tried to find the Jordan canonical form.

However, is there an easier way of doing this?

Thanks!

Answer 1

Note that the kernel (nullspace) of $B$ has dimension $2$, so $B$ contains two Jordan blocks.

Note that since $J^5 = 0$ but $J^4 \neq 0$, we have $B^3 = 0$ but $B^2 \neq 0$. Conclude that the largest Jordan block of $B$ has size $3$.

Conclude that $B$ has Jordan form $J_3 \oplus J_2$.