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Section5.3Similarity

Objectives
  1. Learn to interpret similar matrices geoemetrically.
  2. Understand the relationship between the eigenvalues, eigenvectors, and characteristic polynomials of similar matrices.
  3. Recipe: compute Ax in terms of B , C for A = CBC − 1 .
  4. Picture: the geometry of similar matrices.
  5. Vocabulary word: similarity.

Some matrices are easy to understand. For instance, a diagonal matrix

D = K 2001 / 2 L

just scales the coordinates of a vector: D A xy B = A 2 xy / 2 B . The purpose of most of the rest of this chapter is to understand complicated-looking matrices by analyzing to what extent they “behave like” simple matrices. For instance, the matrix

A = 1 10 K 116914 L

has eigenvalues 2 and 1 / 2, with corresponding eigenvectors v 1 = A 2 / 31 B and v 2 = A − 11 B . Notice that

D ( xe 1 + ye 2 )= xDe 1 + yDe 2 = 2 xe 1 − 1 2 ye 2 A ( xv 1 + yv 2 )= xAv 1 + yAv 2 = 2 xv 1 − 1 2 yv 2 .

Using v 1 , v 2 instead of the usual coordinates makes A “behave” like a diagonal matrix.

Figure1The matrices A and D behave similarly. Click “multiply” to multiply the colored points by D on the left and A on the right. (We will see in Section 5.4 why the points follow hyperbolic paths.)

The other case of particular importance will be matrices that “behave” like a rotation matrix: indeed, this will be crucial for understanding Section 5.5 geometrically. See this important note.

In this section, we study in detail the situation when two matrices behave similarly with respect to different coordinate systems. In Section 5.4 and Section 5.5, we will show how to use eigenvalues and eigenvectors to find a simpler matrix that behaves like a given matrix.

Subsection5.3.1Similar Matrices

We begin with the algebraic definition of similarity.

Definition

Two n × n matrices A and B are similar if there exists an invertible n × n matrix C such that A = CBC − 1 .

Example
Example

As in the above example, one can show that I n is the only matrix that is similar to I n , and likewise for any scalar multiple of I n .

Similarity is unrelated to row equivalence. Any invertible matrix is row equivalent to I n , but I n is the only matrix similar to I n . For instance,

K 2102 L and K 1001 L

are row equivalent but not similar.

As suggested by its name, similarity is what is called an equivalence relation. This means that it satisfies the following properties.

Proof
Example

We conclude with an observation about similarity and powers of matrices.

Proof

First note that

A 2 = AA =( CBC − 1 )( CBC − 1 )= CB ( C − 1 C ) BC − 1 = CBI n BC − 1 = CB 2 C − 1 .

Next we have

A 3 = A 2 A =( CB 2 C − 1 )( CBC − 1 )= CB 2 ( C − 1 C ) BC − 1 = CB 3 C − 1 .

The pattern is clear.

Example

Subsection5.3.2Geometry of Similar Matrices

Similarity is a very interesting construction when viewed geometrically. We will see that, roughly, similar matrices do the same thing in different coordinate systems. The reader might want to review B -coordinates and nonstandard coordinate grids in Section 2.8 before reading this subsection.

By conditions 4 and 5 of the invertible matrix theorem in Section 5.1, an n × n matrix C is invertible if and only if its columns v 1 , v 2 ,..., v n form a basis for R n . This means we can speak of the B -coordinates of a vector in R n , where B is the basis of columns of C . Recall that

[ x ] B = EIIG c 1 c 2 ... c n FJJH means x = c 1 v 1 + c 2 v 2 + ··· + c n v n = C ||| v 1 v 2 ··· v n ||| DEIIG c 1 c 2 ... c n FJJH .

Since C is the matrix with columns v 1 , v 2 ,..., v n , this says that x = C [ x ] B . Multiplying both sides by C − 1 gives [ x ] B = C − 1 x . To summarize:

Let C be an invertible n × n matrix with columns v 1 , v 2 ,..., v n , and let B = { v 1 , v 2 ,..., v n } , a basis for R n . Then for any x in R n , we have

C [ x ] B = x and C − 1 x =[ x ] B .

This says that C changes from the B -coordinates to the usual coordinates, and C − 1 changes from the usual coordinates to the B -coordinates.

Suppose that A = CBC − 1 . The above observation gives us another way of computing Ax for a vector x in R n . Recall that CBC − 1 x = C ( B ( C − 1 x )) , so that multiplying CBC − 1 by x means first multiplying by C − 1 , then by B , then by C . See this example in Section 3.4.

Recipe: Computing Ax in terms of B

Suppose that A = CBC − 1 , where C is an invertible matrix with columns v 1 , v 2 ,..., v n . Let B = { v 1 , v 2 ,..., v n } , a basis for R n . Let x be a vector in R n . To compute Ax , one does the following:

  1. Multiply x by C − 1 , which changes to the B -coordinates: [ x ] B = C − 1 x .
  2. Multiply this by B : B [ x ] B = BC − 1 x .
  3. Interpreting this vector as a B -coordinate vector, we multiply it by C to change back to the usual coordinates: Ax = CBC − 1 x = CB [ x ] B .

B -coordinates [ x ] B B [ x ] B multiplyby C − 1 multiplyby C usualcoordinates x Ax

To summarize: if A = CBC − 1 , then A and B do the same thing, only in different coordinate systems.

The following example is the heart of this section.

Example

Consider the matrices

A = K 1 / 23 / 23 / 21 / 2 L B = K 200 − 1 L C = K 111 − 1 L .

One can verify that A = CBC − 1 : see this example in Section 5.4. Let v 1 = A 11 B and v 2 = A 1 − 1 B , the columns of C , and let B = { v 1 , v 2 } , a basis of R 2 .

The matrix B is diagonal: it scales the x -direction by a factor of 2 and the y -direction by a factor of − 1.

e 1 e 2 Be 1 Be 2 B

To compute Ax , first we multiply by C − 1 to find the B -coordinates of x , then we multiply by B , then we multiply by C again. For instance, let x = A 0 − 2 B .

  1. We see from the B -coordinate grid below that x = − v 1 + v 2 . Therefore, C − 1 x =[ x ] B = A − 11 B .
  2. Multiplying by B scales the coordinates: B [ x ] B = A − 2 − 1 B .
  3. Interpreting A − 2 − 1 B as a B -coordinate vector, we multiply by C to get
    Ax = C K − 2 − 1 L = − 2 v 1 − v 2 = K − 3 − 1 L .
    Of course, this vector lies at ( − 2, − 1 ) on the B -coordinate grid.

B -coordinates [ x ] B B [ x ] B multiplyby C − 1 scale x by2scale y by − 1 multiplyby C usualcoordinates x Ax

Now let x = 1 2 A 5 − 3 B .

  1. We see from the B -coordinate grid that x = 1 2 v 1 + 2 v 2 . Therefore, C − 1 x =[ x ] B = A 1 / 22 B .
  2. Multiplying by B scales the coordinates: B [ x ] B = A 1 − 2 B .
  3. Interpreting A 1 − 2 B as a B -coordinate vector, we multiply by C to get
    Ax = C K 1 − 2 L = v 1 − 2 v 2 = K − 13 L .
    This vector lies at ( 1, − 2 ) on the B -coordinate grid.

B -coordinates [ x ] B B [ x ] B multiplyby C − 1 scale x by2scale y by − 1 multiplyby C usualcoordinates x Ax

To summarize:

  • B scales the e 1 -direction by 2 and the e 2 -direction by − 1.
  • A scales the v 1 -direction by 2 and the v 2 -direction by − 1.

e 1 e 2 Be 1 Be 2 v 1 v 2 Av 1 Av 2 B A C − 1 C
Figure13The geometric relationship between the similar matrices A and B acting on R 2 . Click and drag the heads of x and [ x ] B . Study this picture until you can reliably predict where the other three vectors will be after moving one of them: this is the essence of the geometry of similar matrices.
Interactive: Another matrix similar to B
Example(A matrix similar to a rotation matrix)

To summarize and generalize the previous example:

A Matrix Similar to a Rotation Matrix

Let

B = K cos θ − sin θ sin θ cos θ L C = C || v 1 v 2 || D A = CBC − 1 ,

where C is assumed invertible. Then:

  • B rotates the plane by an angle of θ around the circle centered at the origin and passing through e 1 and e 2 , in the direction from e 1 to e 2 .
  • A rotates the plane by an angle of θ around the ellipse centered at the origin and passing through v 1 and v 2 , in the direction from v 1 to v 2 .

e 1 e 2 Be 1 Be 2 v 1 v 2 Av 1 Av 2 B A A C − 1 C
Interactive: Similar 3 × 3 matrices

Subsection5.3.3Eigenvalues of Similar Matrices

Since similar matrices behave in the same way with respect to different coordinate systems, we should expect their eigenvalues and eigenvectors to be closely related.

Proof

Suppose that A = CBC − 1 , where A , B , C are n × n matrices. We calculate

A − λ I n = CBC − 1 − λ CC − 1 = CBC − 1 − C λ C − 1 = CBC − 1 − C λ I n C − 1 = C ( B − λ I n ) C − 1 .

Therefore,

det ( A − λ I n )= det ( C ( B − λ I n ) C − 1 )= det ( C ) det ( B − λ I n ) det ( C ) − 1 = det ( B − λ I n ) .

Here we have used the multiplicativity property in Section 4.1 and its corollary in Section 4.1.

Since the eigenvalues of a matrix are the roots of its characteristic polynomial, we have shown:

Similar matrices have the same eigenvalues.

By this theorem in Section 5.2, similar matrices also have the same trace and determinant.

Note

The converse of the fact is false. Indeed, the matrices

K 1101 L and K 1001 L

both have characteristic polynomial f ( λ )=( λ − 1 ) 2 , but they are not similar, because the only matrix that is similar to I 2 is I 2 itself.

Given that similar matrices have the same eigenvalues, one might guess that they have the same eigenvectors as well. Upon reflection, this is not what one should expect: indeed, the eigenvectors should only match up after changing from one coordinate system to another. This is the content of the next fact, remembering that C and C − 1 change between the usual coordinates and the B -coordinates.

Proof

Suppose that v is an eigenvector of A with eigenvalue λ , so that Av = λ v . Then

B ( C − 1 v )= C − 1 ( CBC − 1 v )= C − 1 ( Av )= C − 1 λ v = λ ( C − 1 v ) ,

so that C − 1 v is an eigenvector of B with eigenvalue λ . Likewise if v is an eigenvector of B with eigenvalue λ , then Bv = λ v , and we have

A ( Cv )=( CBC − 1 ) Cv = CBv = C ( λ v )= λ ( Cv ) ,

so that Cv is an eigenvalue of A with eigenvalue λ .

If A = CBC − 1 , then C − 1 takes the λ -eigenspace of A to the λ -eigenspace of B , and C takes the λ -eigenspace of B to the λ -eigenspace of A .

Example

We continue with the above example: let

A = K 1 / 23 / 23 / 21 / 2 L B = K 200 − 1 L C = K 111 − 1 L ,

so A = CBC − 1 . Let v 1 = A 11 B and v 2 = A 1 − 1 B , the columns of C . Recall that:

  • B scales the e 1 -direction by 2 and the e 2 -direction by − 1.
  • A scales the v 1 -direction by 2 and the v 2 -direction by − 1.

This means that the x -axis is the 2 -eigenspace of B , and the y -axis is the − 1 -eigenspace of B ; likewise, the “v 1 -axis” is the 2 -eigenspace of A , and the “v 2 -axis” is the − 1 -eigenspace of A . This is consistent with the fact, as multiplication by C changes e 1 into Ce 1 = v 1 and e 2 into Ce 2 = v 2 .

2-eigenspace − 1-eigenspace 2-eigenspace − 1-eigenspace C
Figure27The eigenspaces of A are the lines through v 1 and v 2 . These are the images under C of the coordinate axes, which are the eigenspaces of B .
Interactive: Another matrix similar to B
Interactive: Similar 3 × 3 matrices