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Section 4.3 Diagonalization, similarity, and powers of a matrix

The first example we considered in this chapter was the matrix A=[1221], which has eigenvectors v1=[11] and v2=[11] and associated eigenvalues λ1=3 and λ2=1. In Subsection 4.1.2, we described how A is, in some sense, equivalent to the diagonal matrix D=[3001].
This equivalence is summarized by Figure 4.3.1. The diagonal matrix D has the geometric effect of stretching vectors horizontally by a factor of 3 and flipping vectors vertically. The matrix A has the geometric effect of stretching vectors by a factor of 3 in the v1 direction and flipping them in the v2 direction. That is, the geometric effect of A is the same as that of D when viewed in a basis of eigenvectors of A.
Figure 4.3.1. The matrix A has the same geometric effect as the diagonal matrix D when viewed in the basis of eigenvectors.
Our goal in this section is to express this geometric observation in algebraic terms. In doing so, we will make precise the sense in which A and D are equivalent.

Preview Activity 4.3.1.

In this preview activity, we will review some familiar properties about matrix multiplication that appear in this section.
  1. Remember that matrix-vector multiplication constructs linear combinations of the columns of the matrix. For instance, if A=[a1a2], express the product A[23] in terms of a1 and a2.
  2. What is the product A[40] in terms of a1 and a2?
  3. Next, remember how matrix-matrix multiplication is defined. Suppose that we have matrices A and B and that B=[b1b2]. How can we express the matrix product AB in terms of the columns of B?
  4. Suppose that A is a matrix having eigenvectors v1 and v2 with associated eigenvalues λ1=4 and λ2=1. Express the product A(2v1+3v2) in terms of v1 and v2.
  5. Suppose that A is the matrix from the previous part and that P=[v1v2]. What is the matrix product
    AP=A[v1v2]?

Subsection 4.3.1 Diagonalization of matrices

When working with an n×n matrix A, Subsection 4.1.2 demonstrated the value of having a basis of Rn consisting of eigenvectors of A. In fact, Proposition 4.2.9 tells us that if the eigenvalues of A are real and distinct, then there is a such a basis. As we’ll see later, there are other conditions on A that guarantee a basis of eigenvectors. For now, suffice it to say that we can find a basis of eigenvectors for many matrices. With this assumption, we will see how the matrix A is equivalent to a diagonal matrix D.

Activity 4.3.2.

Suppose that A is a 2×2 matrix having eigenvectors v1 and v2 with associated eigenvalues λ1=3 and λ2=6. Because the eigenvalues are real and distinct, we know by Proposition 4.2.9 that these eigenvectors form a basis of R2.
  1. What are the products Av1 and Av2 in terms of v1 and v2?
  2. If we form the matrix P=[v1v2], what is the product AP in terms of v1 and v2?
  3. Use the eigenvalues to form the diagonal matrix D=[3006] and determine the product PD in terms of v1 and v2.
  4. The results from the previous two parts of this activity demonstrate that AP=PD. Using the fact that the eigenvectors v1 and v2 form a basis of R2, explain why P is invertible and that we must have A=PDP1.
  5. Suppose that A=[3630]. Verify that v1=[11] and v2=[21] are eigenvectors of A with eigenvalues λ1=3 and λ2=6.
  6. Use the Sage cell below to define the matrices P and D and then verify that A=PDP1.
More generally, suppose that we have an n×n matrix A and that there is a basis of Rn consisting of eigenvectors v1,v2,,vn of A with associated eigenvalues λ1,λ2,,λn. If we use the eigenvectors to form the matrix
P=[v1v2vn]
and the eigenvalues to form the diagonal matrix
D=[λ1000λ20000λn]
and apply the same reasoning demonstrated in the activity, we find that AP=PD and hence
A=PDP1.
We have now seen the following proposition.

Example 4.3.3.

We have seen that A=[1221] has eigenvectors v1=[11] and v2=[11] with associated eigenvalues λ1=3 and λ2=1. Forming the matrices
P=[v1v2]=[1111],   D=[3001],
we see that A=PDP1.
This is the sense in which we mean that A is equivalent to a diagonal matrix D. The expression A=PDP1 says that A, expressed in the basis defined by the columns of P, has the same geometric effect as D, expressed in the standard basis e1,e2,,en.

Definition 4.3.4.

We say that the matrix A is diagonalizable if there is a diagonal matrix D and invertible matrix P such that
A=PDP1.

Example 4.3.5.

We will try to find a diagonalization of A=[5634] whose characteristic equation is
det(AλI)=(5λ)(4λ)+18=(2λ)(1λ)=0.
This shows that the eigenvalues of A are λ1=2 and λ2=1.
By constructing Nul(A(2)I), we find a basis for E2 consisting of the vector v1=[21]. Similarly, a basis for E1 consists of the vector v2=[11]. This shows that we can construct a basis {v1,v2} of R2 consisting of eigenvectors of A.
We now form the matrices
D=[2001],P=[v1v2]=[2111]
and verify that
PDP1=[2111][2001][1112]=[5634]=A.
There are, in fact, many ways to diagonalize A. For instance, we could change the order of the eigenvalues and eigenvectors and write
D=[1002],P=[v2v1]=[1211].
If we choose a different basis for the eigenspaces, we will also find a different matrix P that diagonalizes A. The point is that there are many ways in which A can be written in the form A=PDP1.

Example 4.3.6.

We will try to find a diagonalization of A=[0414].
Once again, we find the eigenvalues by solving the characteristic equation:
det(AλI)=λ(4λ)+4=(2λ)2=0.
In this case, there is a single eigenvalue λ=2.
We find a basis for the eigenspace E2 by describing Nul(A2I):
A2I=[2412][1200].
This shows that the eigenspace E2 is one-dimensional with v1=[21] forming a basis.
In this case, there is not a basis of R2 consisting of eigenvectors of A, which tells us that A is not diagonalizable.
In fact, if we only know that A=PDP1, we can say that the columns of P are eigenvectors of A and that the diagonal entries of D are the associated eigenvalues.

Example 4.3.8.

Suppose we know that A=PDP1 where
D=[2002],P=[v2v1]=[1112].
The columns of P form eigenvectors of A so that v1=[11] is an eigenvector of A with eigenvalue λ1=2 and v2=[12] is an eigenvector with eigenvalue λ2=2.
We can verify this by computing
A=PDP1=[6486]
and checking that Av1=[11]=2v1 and Av2=[12]=2v2.

Activity 4.3.3.

  1. Find a diagonalization of A, if one exists, when
    A=[3265].
  2. Can the diagonal matrix
    A=[2005]
    be diagonalized? If so, explain how to find the matrices P and D.
  3. Find a diagonalization of A, if one exists, when
    A=[200130203].
  4. Find a diagonalization of A, if one exists, when
    A=[200130213].
  5. Suppose that A=PDP1 where
    D=[3001],P=[v2v1]=[2211].
    1. Explain why A is invertible.
    2. Find a diagonalization of A1.
    3. Find a diagonalization of A3.

Subsection 4.3.2 Powers of a diagonalizable matrix

In several earlier examples, we have been interested in computing powers of a given matrix. For instance, in Activity 4.1.3, we had the matrix A=[0.80.60.20.4] and an initial vector x0=[10000], and we wanted to compute
x1=Ax0x2=Ax1=A2x0x3=Ax2=A3x0.
In particular, we wanted to find xk=Akx0 and determine what happens as k becomes very large. If a matrix A is diagonalizable, writing A=PDP1 can help us understand powers of A more easily.

Activity 4.3.4.

  1. Let’s begin with the diagonal matrix
    D=[2001].
    Find the powers D2, D3, and D4. What is Dk for a general value of k?
  2. Suppose that A is a matrix with eigenvector v and associated eigenvalue λ; that is, Av=λv. By considering A2v, explain why v is also an eigenvector of A with eigenvalue λ2.
  3. Suppose that A=PDP1 where
    D=[2001].
    Remembering that the columns of P are eigenvectors of A, explain why A2 is diagonalizable and find a diagonalization in terms of P and D.
  4. Give another explanation of the diagonalizability of A2 by writing
    A2=(PDP1)(PDP1)=PD(P1P)DP1.
  5. In the same way, find a diagonalization of A3, A4, and Ak.
  6. Suppose that A is a diagonalizable 2×2 matrix with eigenvalues λ1=0.5 and λ2=0.1. What happens to Ak as k becomes very large?
If A is diagonalizable, the activity demonstrates that any power of A is as well.

Example 4.3.10.

Let’s revisit Activity 4.1.3 where we had the matrix A=[0.80.60.20.4] and the initial vector x0=[10000]. We were interested in understanding the sequence of vectors xk+1=Axk, which means that xk=Akx0.
We can verify that v1=[31] and v2=[11] are eigenvectors of A having associated eigenvalues λ1=1 and λ2=0.2. This means that A=PDP1 where
P=[3111],   D=[1000.2].
Therefore, the powers of A have the form Ak=PDkP1.
Notice that Dk=[1k000.2k]=[1000.2k]. As k increases, 0.2k becomes closer and closer to zero. This means that for very large powers k, we have
Dk[1000]
and therefore
Ak=PDkP1P[1000]P1=[34341414].
Beginning with the vector x0=[10000], we find that xk=Akx0[750250] when k is very large.

Subsection 4.3.3 Similarity and complex eigenvalues

We have been interested in diagonalizing a matrix A because doing so relates a matrix A to a simpler diagonal matrix D. In particular, the effect of multiplying a vector by A=PDP1, viewed in the basis defined by the columns of P, is the same as the effect of multiplying by D in the standard basis.
While many matrices are diagonalizable, there are some that are not. For example, if a matrix has complex eigenvalues, it is not possible to find a basis of Rn consisting of eigenvectors, which means that the matrix is not diagonalizable. In this case, however, we can still relate the matrix to a simpler form that explains the geometric effect this matrix has on vectors.

Definition 4.3.11.

We say that A is similar to B if there is an invertible matrix P such that A=PBP1.
Notice that a matrix is diagonalizable if and only if it is similar to a diagonal matrix. In case a matrix A has complex eigenvalues, we will find a simpler matrix C that is similar to A and note that A=PCP1 has the same effect, when viewed in the basis defined by the columns of P, as C, when viewed in the standard basis.
To begin, suppose that A is a 2×2 matrix having a complex eigenvalue λ=a+bi. It turns out that A is similar to C=[abba].
The next activity shows that C has a simple geometric effect on R2. First, however, we will use polar coordinates to rewrite C. As shown in the figure, the point (a,b) defines r, the distance from the origin, and θ, the angle formed with the positive horizontal axis. We then have
a=rcosθb=rsinθ.
Notice that the Pythagorean theorem says that r=a2+b2.

Activity 4.3.5.

We begin by rewriting C in terms of r and θ and noting that
C=[abba]=[rcosθrsinθrsinθrcosθ]=[r00r][cosθsinθsinθcosθ].
  1. Explain why C has the geometric effect of rotating vectors by θ and scaling them by a factor of r.
  2. Let’s now consider the matrix
    A=[2254]
    whose eigenvalues are λ1=1+i and λ2=1i. We will choose to focus on one of the eigenvalues λ1=a+bi=1+i.
    Form the matrix C using these values of a and b. Then rewrite the point (a,b) in polar coordinates by identifying the values of r and θ. Explain the geometric effect of multiplying vectors by C.
  3. Suppose that P=[1121]. Verify that A=PCP1.
  4. Explain why Ak=PCkP1.
  5. We formed the matrix C by choosing the eigenvalue λ1=1+i. Suppose we had instead chosen λ2=1i. Form the matrix C and use polar coordinates to describe the geometric effect of C.
  6. Using the matrix P=[1121], show that A=PCP1.
If the 2×2 matrix A has a complex eigenvalue λ=a+bi, it turns out that A is always similar to the matrix C=[abba], whose geometric effect on vectors can be described in terms of a rotation and a scaling. There is, in fact, a method for finding the matrix P so that A=PCP1 that we’ll see in Exercise 4.3.5.8. For now, we note that A has the same geometric effect as C, when viewed in the basis provided by the columns of P. We will put this fact to use in the next section to understand certain dynamical systems.

Subsection 4.3.4 Summary

Our goal in this section has been to use the eigenvalues and eigenvectors of a matrix A to relate A to a simpler matrix.
  • We said that A is diagonalizable if we can write A=PDP1 where D is a diagonal matrix. The columns of P consist of eigenvectors of A and the diagonal entries of D are the associated eigenvalues.
  • An n×n matrix A is diagonalizable if and only if there is a basis of Rn consisting of eigenvectors of A.
  • We said that A and B are similar if there is an invertible matrix P such that A=PBP1. In this case, Ak=PBkP1.
  • If A is a 2×2 matrix with complex eigenvalue λ=a+bi, then A is similar to C=[abba]. Writing the point (a,b) in polar coordinates r and θ, we see that C rotates vectors through an angle θ and scales them by a factor of r=a2+b2.

Exercises 4.3.5 Exercises

1.

Determine whether the following matrices are diagonalizable. If so, find matrices D and P such that A=PDP1.
  1. A=[2221].
  2. A=[1113].
  3. A=[3421].
  4. A=[100220014].
  5. A=[122212221].

2.

Determine whether the following matrices have complex eigenvalues. If so, find the matrix C such that A=PCP1.
  1. A=[2221].
  2. A=[1113].
  3. A=[3421].

3.

Determine whether the following statements are true or false and provide a justification for your response.
  1. If A is invertible, then A is diagonalizable.
  2. If A and B are similar and A is invertible, then B is also invertible.
  3. If A is a diagonalizable n×n matrix, then there is a basis of Rn consisting of eigenvectors of A.
  4. If A is diagonalizable, then A10 is also diagonalizable.
  5. If A is diagonalizable, then A is invertible.

4.

Provide a justification for your response to the following questions.
  1. If A is a 3×3 matrix having eigenvalues λ=2,3,4, can you guarantee that A is diagonalizable?
  2. If A is a 2×2 matrix with a complex eigenvalue, can you guarantee that A is diagonalizable?
  3. If A is similar to the matrix B=[500050003], is A diagonalizable?
  4. What can you say about a matrix that is similar to the identity matrix?
  5. If A is a diagonalizable 2×2 matrix with a single eigenvalue λ=4, what is A?

6.

We say that A is similar to B if there is a matrix P such that A=PBP1.
  1. If A is similar to B, explain why B is similar to A.
  2. If A is similar to B and B is similar to C, explain why A is similar to C.
  3. If A is similar to B and B is diagonalizable, explain why A is diagonalizable.
  4. If A and B are similar, explain why A and B have the same characteristic polynomial; that is, explain why det(AλI)=det(BλI).
  5. If A and B are similar, explain why A and B have the same eigenvalues.

7.

Suppose that A=PDP1 where
D=[1000],P=[1221].
  1. Explain the geometric effect that D has on vectors in R2.
  2. Explain the geometric effect that A has on vectors in R2.
  3. What can you say about A2 and other powers of A?
  4. Is A invertible?

8.

When A is a 2×2 matrix with a complex eigenvalue λ=a+bi, we have said that there is a matrix P such that A=PCP1 where C=[abba]. In this exercise, we will learn how to find the matrix P. As an example, we will consider the matrix A=[2214].
  1. Show that the eigenvalues of A are complex.
  2. Choose one of the complex eigenvalues λ=a+bi and construct the usual matrix C.
  3. Using the same eigenvalue, we will find an eigenvector v where the entries of v are complex numbers. As always, we will describe Nul(AλI) by constructing the matrix AλI and finding its reduced row echelon form. In doing so, we will necessarily need to use complex arithmetic.
  4. We have now found a complex eigenvector v. Write v=v1iv2 to identify vectors v1 and v2 having real entries.
  5. Construct the matrix P=[v1v2] and verify that A=PCP1.

9.

For each of the following matrices, sketch the vector x=[10] and powers Akx for k=1,2,3,4.
  1. A=[01.41.40].
  2. A=[00.80.80].
  3. A=[0110].
  4. Consider a matrix of the form C=[abba] with r=a2+b2. What happens when k becomes very large when
    1. r<1.
    2. r=1.
    3. r>1.

10.

For each of the following matrices and vectors, sketch the vector x along with Akx for k=1,2,3,4.
  1. A=[1.4000.7]x=[12]..
  2. A=[0.6000.9]x=[43].
  3. A=[1.2001.4]x=[21].
  4. A=[0.950.250.250.95]x=[30].
    Find the eigenvalues and eigenvectors of A to create your sketch.
  5. If A is a 2×2 matrix with eigenvalues λ1=0.7 and λ2=0.5 and x is any vector, what happens to Akx when k becomes very large?
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