## Compute the Jordan Canonical Form for a given matrix

Let $A = \begin{bmatrix} -8 & -10 & -1 \\ 7 & 9 & 1 \\ 3 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} -3 & 2 & -4 \\ 4 & -1 & 4 \\ 4 & -2 & 5 \end{bmatrix}$. Show that these matrices have the same characteristic polynomial, but that one is diagonalizable and the other not. Compute the Jordan canonical form for each.

Let $P = \begin{bmatrix} -1 & -1 & 0 \\ 4 & 4 & 1 \\ -5 & -6 & -1 \end{bmatrix}$ and $Q = \begin{bmatrix} -2 & 1 & -2 \\ 2 & -1 & 3 \\ 1 & 0 & 1 \end{bmatrix}$. Evidently we have $P^{-1}AP = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ and $Q^{-1}BQ = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$, which are both in Jordan canonical form.

Recall that characteristic polynomials are invariant under conjugation; in particular, evidently both $A$ and $B$ have characteristic polynomial $(x+1)(x-1)^2$. By Corollary 24 in D&F, $A$ is not diagonalizable.