Show that $J^{n-1}$ not equal $0$

Question

If J is the Jordan matrix, with $0$ on the diagonal, how do I show that $J^{n-1}$ is not equal to zero, given that (by Cayley-Hamilton) $J^{n}=0$?

Answer 1

Think about the generalized eigenvectors given by the block. I recently read Artin and it gives a nice explanation.

See that if we have the block given by (for ease 4x4 example here):

$J=\begin{bmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix}$

The eigenvector, and then generalized eigenvectors are given by the eigenvector $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ \end{bmatrix}$ (as it maps to zero in the Jordan form w/ diagonal $0=\lambda$). The generalized eigenvectors are given below. Note that the generalized eigenvector of degree $4$, that is $v=\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix}$ will never be mapped to zero under the transformation $k-1$ times, that is $J^{k-1}v\neq 0$.

$\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix}\rightarrow \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix}\rightarrow \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ \end{bmatrix}\rightarrow \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ \end{bmatrix}$

I hope this helps!