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Section 26.4 Examples

Subsection 26.4.1 The algebraic pattern of similarity

Example 26.4.1.

Let's verify the algebraic pattern of similarity identified in Discovery 26.3 and Subsection 26.3.2 in a specific example.

Consider matrices

\begin{align*} A \amp = \left[\begin{array}{rrr} 1 \amp 1 \amp 0 \\ 2 \amp 7 \amp -1 \\ 0 \amp 5 \amp 1 \end{array}\right] \text{,} \amp B \amp = \left[\begin{array}{rrr} -1 \amp 12 \amp -23 \\ 2 \amp -1 \amp 8 \\ 1 \amp -5 \amp 11 \end{array}\right] \text{,} \amp P \amp = \left[\begin{array}{rrr} 1 \amp 0 \amp 2 \\ 0 \amp 2 \amp -3 \\ 1 \amp 1 \amp 0 \end{array}\right] \text{.} \end{align*}

Let's verify that \(A\) and \(B\) are similar and that \(P\) is a transition matrix that realizes that similarity relationship using the aforementioned algebraic pattern of similarity.

Pattern 26.3.1 says that each column of \(B\) should tell us how to decompose the result of computing \(A\) times the corresponding column of \(P\) as a linear combination of all the columns of \(P\text{.}\) In other words, we should have

\begin{equation*} \begin{array}{rcrcrcr} A \uvec{p}_1 \amp = \amp - \uvec{p}_1 \amp + \amp 2 \uvec{p}_2 \amp + \amp \uvec{p}_3 \text{,} \\ A \uvec{p}_2 \amp = \amp 12 \uvec{p}_1 \amp - \amp \uvec{p}_2 \amp - \amp 5 \uvec{p}_3 \text{,} \\ A \uvec{p}_3 \amp = \amp -23 \uvec{p}_1 \amp + \amp 8 \uvec{p}_2 \amp + \amp 11 \uvec{p}_3 \text{,} \end{array} \end{equation*}

where \(\uvec{p}_1, \uvec{p}_2, \uvec{p}_3\) represent the columns of \(P\text{,}\) and the coefficients in each of these linear combinations are taken from the corresponding column of \(B\text{.}\)

We have

\begin{align*} \uvec{p}_1 \amp = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \text{,} \amp \uvec{p}_2 \amp = \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} \text{,} \amp \uvec{p}_3 \amp = \left[\begin{array}{r} 2 \\ -3 \\ 0 \end{array}\right] \text{,} \end{align*}

and we can compute

\begin{align*} A \uvec{p}_1 \amp = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \text{,} \amp A \uvec{p}_2 \amp = \begin{bmatrix} 2 \\ 13 \\ 11 \end{bmatrix} \text{,} \amp A \uvec{p}_3 \amp = \left[\begin{array}{r} -1 \\ -17 \\ -15 \end{array}\right] \text{.} \end{align*}

On the other hand, we have

\begin{align*} - \uvec{p}_1 + 2 \uvec{p}_2 + \uvec{p}_3 = - \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + 2 \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} + \left[\begin{array}{r} 2 \\ -3 \\ 0 \end{array}\right] \amp = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \text{,}\\ 12 \uvec{p}_1 - \uvec{p}_2 - 5 \uvec{p}_3 = 12 \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} - \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} - 5 \left[\begin{array}{r} 2 \\ -3 \\ 0 \end{array}\right] \amp = \begin{bmatrix} 2 \\ 13 \\ 11 \end{bmatrix} \text{,}\\ -23 \uvec{p}_1 + 8 \uvec{p}_2 + 11 \uvec{p}_3 = -23 \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + 8 \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} + 11 \left[\begin{array}{r} 2 \\ -3 \\ 0 \end{array}\right] \amp = \left[\begin{array}{r} -1 \\ -17 \\ -15 \end{array}\right] \text{,} \end{align*}

each of which agree with the corresponding \(A \uvec{p}_j\) calculation above.

Remark 26.4.2.

In the example above, we are really just computing and comparing \(AP\) and \(PB\text{,}\) since those two products are where we got our patten of similarity. Rather than a computational tool, the algebraic pattern of similarity identified in Discovery 26.3 and Subsection 26.3.2 will be a theoretical tool to bring the theory of the vector space \(\R^n\) (or \(\C^n\text{,}\) as appropriate) to bear on the analysis of various specific similarity patterns that we will explore in the chapters to come.

Subsection 26.4.2 Computing \(\inv{P} A P\) by row reduction

In Subsubsection 22.3.5.3 we modified the inverse-by-row-reduction computation pattern to show that multiplication by an inverse can be computed by row reduction. (See pattern (\(\maltese\maltese\)) in Subsubsection 22.3.5.3.) The same modified pattern can be used to compute the product of \(\inv{P}\) and \(AP\text{:}\)

\begin{equation*} \left[\begin{array}{c|c} P \amp AP \end{array}\right] \qquad \rowredarrow \qquad \left[\begin{array}{c|c} I \amp \inv{P}(AP) \end{array}\right]\text{.} \end{equation*}
Example 26.4.3.

Let's use the matrices of Example 26.4.1 to demonstrate.

First, compute

\begin{equation*} AP = \left[\begin{array}{rrr} 1 \amp 2 \amp -1 \\ 1 \amp 13 \amp -17 \\ 1 \amp 11 \amp -15 \end{array}\right]\text{.} \end{equation*}

Now augment \(P\) with \(AP\) and reduce:

\begin{align*} \left[\begin{array}{c|c} P \amp AP \end{array}\right] = \amp \left[\begin{array}{rrr|rrr} 1 \amp 0 \amp 2 \amp 1 \amp 2 \amp -1 \\ 0 \amp 2 \amp -3 \amp 1 \amp 13 \amp -17 \\ 1 \amp 1 \amp 0 \amp 1 \amp 11 \amp -15 \end{array}\right] \begin{matrix} \phantom{X} \\ \phantom{X} \\ R_3 - R_1 \end{matrix}\\ \longrightarrow \amp \left[\begin{array}{rrr|rrr} 1 \amp 0 \amp 2 \amp 1 \amp 2 \amp -1 \\ 0 \amp 2 \amp -3 \amp 1 \amp 13 \amp -17 \\ 0 \amp 1 \amp -2 \amp 0 \amp 9 \amp -14 \end{array}\right] \begin{matrix} \phantom{X} \\ R_2 \leftrightarrow R_3 \end{matrix}\\ \longrightarrow \amp \left[\begin{array}{rrr|rrr} 1 \amp 0 \amp 2 \amp 1 \amp 2 \amp -1 \\ 0 \amp 1 \amp -2 \amp 0 \amp 9 \amp -14 \\ 0 \amp 2 \amp -3 \amp 1 \amp 13 \amp -17 \end{array}\right] \begin{matrix} \phantom{X} \\ \phantom{X} \\ R_3 - 2R_2 \end{matrix}\\ \longrightarrow \amp \left[\begin{array}{rrr|rrr} 1 \amp 0 \amp 2 \amp 1 \amp 2 \amp -1 \\ 0 \amp 1 \amp -2 \amp 0 \amp 9 \amp -14 \\ 0 \amp 0 \amp 1 \amp 1 \amp -5 \amp 11 \end{array}\right] \begin{matrix} R_1 - 2R_3 \\ R_2 + 2R_3 \\ \phantom{X} \end{matrix}\\ \longrightarrow \amp \left[\begin{array}{rrr|rrr} 1 \amp 0 \amp 0 \amp -1 \amp 12 \amp -23 \\ 0 \amp 1 \amp 0 \amp 2 \amp -1 \amp 8 \\ 0 \amp 0 \amp 1 \amp 1 \amp -5 \amp 11 \end{array}\right]\text{.} \end{align*}

Comparing with the matrix \(B\) from Example 26.4.1, we see that \(\inv{P} A P = B\) as expected.