## Compute the Jordan canonical form of a given matrix

Compute the Jordan canonical form of the matrix $A = \begin{bmatrix} 9 & 4 & 5 \\ -4 & 0 & -3 \\ -6 & -4 & -2 \end{bmatrix}$. Find an explicit matrix $P$ such that $P^{-1}AP$ is in Jordan canonical form. Is $A$ diagonalizable? if not, why not?

We will follow the algorithm given on page 496 of D&F.

First we determine a sequence of ERCOs which transforms $xI-A$ into Smith Normal Form. Evidently, the following sequence works.

1. $-\frac{1}{4}C_2 \rightarrow C_2$
2. $(9-x)C_2 + C_1 \rightarrow C_1$
3. $5C_2 + C_3 \rightarrow C_3$
4. $\frac{1}{4}xR_1 + R_2 \rightarrow R_2$
5. $R_1 + R_3 \rightarrow R_3$
6. $C_1 \leftrightarrow C_2$
7. $\frac{1}{5}xC_3 + C_2 \rightarrow C_2$
8. $-\frac{33}{25}C_3 + C_2 \rightarrow C_2$
9. $25C_2 \rightarrow C_2$
10. $\frac{1}{4}(5x-12)C_2 + C_3 \rightarrow C_3$
11. $(-5x^2 + 23x - 24)R_2 + R_3 \rightarrow R_3$
12. $\frac{4}{25}C_3 \rightarrow C_3$

The resulting matrix is $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & (x-2)^2(x-3) \end{bmatrix}$. We used three row operations, whose corresponding column operations are as follows.

1. $C_1 - \frac{1}{4}AC_2 \rightarrow C_1$
2. $C_1 - C_3 \rightarrow C_1$
3. $C_2 + (5A^2 - 23A + 24I)C_3 \rightarrow C_2$

The resulting matrix is $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$. Multiplying the only nonzero column successively by $(A-2I)(A-3I)$, $A-3I$, and $(A-2I)^2$ to construct the columns of a matrix $P$, we have $P = \begin{bmatrix} -2 & 5 & 3 \\ 1 & -3 & -2 \\ 2 & -5 & -2 \end{bmatrix}$. Evidently, $P^{-1} = \begin{bmatrix} -4 & -5 & -1 \\ -2 & -2 & -1 \\ 1 & 0 & 1 \end{bmatrix}$, and $P^{-1}AP = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ is in Jordan canonical form.