Show explicitly that two matrices are similar

Let B = \begin{bmatrix} 0 & -4 & 85 \\ 1 & 4 & -30 \\ 0 & 0 & 3 \end{bmatrix} and let C = \begin{bmatrix} 2 & 2 & 1 \\ 0 & 2 & -1 \\ 0 & 0 & 3 \end{bmatrix}. Find matrices P and Q such that P^{-1}BP and Q^{-1}CQ are in Jordan Canonical Form. Give an explicit matrix R such that R^{-1}BR = C.


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

We begin by finding a sequence of ERCOs which transforms the matrix xI-B into Smith Normal Form. Evidently the following sequence works.

  1. xR_2 + R_1 \rightarrow R_1
  2. -C_1 \rightarrow C_1
  3. (4-x)C_1 + C_2 \rightarrow C_2
  4. -30C_1 + C_3 \rightarrow C_3
  5. -30R_3 + R_1 \rightarrow R_1
  6. -\frac{1}{5}(x-2)^2C_3 + C_2 \rightarrow C_2
  7. \frac{1}{5}R_1 \rightarrow R_1
  8. (3-x)R_1 + R_3 \rightarrow R_3
  9. C_1 \leftrightarrow C_3
  10. C_2 \leftrightarrow C_3
  11. -5C_3 \rightarrow C_3

The result is \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & (x-2)^2(x+3) \end{bmatrix}.

In particular, the elementary divisors of B are (x-2)^2 and x-3, so that the Jordan canonical form of B is \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}.

We used four row operations, whose corresponding column operations are as follows.

  1. C_2 - BC_1 \rightarrow C_2
  2. C_3 + 30C_1 \rightarrow C_3
  3. C_1 + (B-3I)C_3 \rightarrow C_1
  4. 5C_1 \rightarrow C_1

We now perform this sequence of column operations (in order) on the identity matrix. The result is \begin{bmatrix} 0 & 0 & 30 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}. Multiplying the only nonzero column by each of (B-2I)(B-3I), B-3I, and B-2I to construct a matrix P, we have P = \begin{bmatrix} 10 & -5 & 35 \\ -5 & 0 & -5 \\ 0 & 0 & 1 \end{bmatrix}. Using Gauss-Jordan elimination on [P|I], we can compute P^{-1} = \frac{1}{5}\begin{bmatrix} 0 & -1 & -5 \\ -1 & -2 & 25 \\ 0 & 0 & 5 \end{bmatrix}. We can easily verify that P^{-1}BP is in Jordan Canonical Form. (WolframAlpha agrees.)

Next we consider C. Evidently xI-C is put into Smith normal form by the following sequence of ERCOs.

  1. R_2 + R_1 \rightarrow R_1
  2. -C_1 + C_2 \rightarrow C_2
  3. -\frac{1}{2}C_2 \rightarrow C_2
  4. (2-x)C_2 + C_1 \rightarrow C_1
  5. \frac{1}{2}(x-2)R_1 + R_2 \rightarrow R_2
  6. -\frac{1}{2}(x-2)^2C_3 + C_1 \rightarrow C_1
  7. (3-x)R_2 + R_3 \rightarrow R_3
  8. C_1 \leftrightarrow C_2
  9. C_2 \leftrightarrow C_3
  10. -2C_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}, so that the elementary divisors of C are (x-2)^2 and x-3. At this point, we know that B and C are similar since they have the same elementary divisors.

We used three elementary row operations, whose corresponding column operations are as follows.

  1. C_1 - C_2 \rightarrow C_2
  2. C_1 - \frac{1}{2}(C-2I)C_2 \rightarrow C_1
  3. C_2 + (C-3I)C_3 \rightarrow C_2

Performing these operations (in order) on the identity matrix yields \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}. Multiplying the only nonzero column of this matrix by (C-2I)^2, (C-2I)(C-3I), and C-3I to construct a matrix Q, we have Q = \begin{bmatrix} -2 & 1 & -1 \\ 0 & -1 & -1 \\ 0 & 0 & 1 \end{bmatrix}. Evidently Q^{-1} = \frac{1}{2} \begin{bmatrix} -1 & -1 & -2 \\ 0 & -2 & -2 \\ 0 & 0 & 2 \end{bmatrix}, and Q^{-1}CP is in Jordan Canonical Form. (WolframAlpha agrees.)

In particular, we have P^{-1}BP = Q^{-1}CQ, so that (PQ^{-1})^{-1}B(PQ^{-1}) = C.

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