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.

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