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Section 26.4 Examples

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Subsection 26.4.1 The algebraic pattern of similarity

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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

A=[11027−1051],B=[−112−232−181−511],P=[10202−3110].

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. In other words, we should have

Ap1=−p1+2p2+p3,Ap2=12p1−p2−5p3,Ap3=−23p1+8p2+11p3,

where p1,p2,p3 represent the columns of P, and the coefficients in each of these linear combinations are taken from the corresponding column of B.

We have

p1=[101],p2=[021],p3=[2−30],

and we can compute

Ap1=[111],Ap2=[21311],Ap3=[−1−17−15].

On the other hand, we have

−p1+2p2+p3=−[101]+2[021]+[2−30]=[111],12p1−p2−5p3=12[101]−[021]−5[2−30]=[21311],−23p1+8p2+11p3=−23[101]+8[021]+11[2−30]=[−1−17−15],

each of which agree with the corresponding Apj calculation above.

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Remark 26.4.2.

In the example above, we are really just computing and comparing AP and PB, 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 Rn (or Cn, as appropriate) to bear on the analysis of various specific similarity patterns that we will explore in the chapters to come.

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Subsection 26.4.2 Computing P−1AP by row reduction

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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 (✠✠) in Subsubsection 22.3.5.3.) The same modified pattern can be used to compute the product of P−1 and AP:

[PAP]row→reduce[IP−1(AP)].
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Example 26.4.3.

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

First, compute

AP=[12−1113−17111−15].

Now augment P with AP and reduce:

[PAP]=[10212−102−3113−17110111−15]XXR3−R1⟶[10212−102−3113−1701−209−14]XR2↔R3⟶[10212−101−209−1402−3113−17]XXR3−2R2⟶[10212−101−209−140011−511]R1−2R3R2+2R3X⟶[100−112−230102−180011−511].

Comparing with the matrix B from Example 26.4.1, we see that P−1AP=B as expected.