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.

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