## Section4.3Diagonalization, similarity, and powers of a matrix

The first example we considered in this chapter was the matrix $$A=\left[\begin{array}{rr} 1 \amp 2 \\ 2 \amp 1 \\ \end{array}\right] \text{,}$$ which has eigenvectors $$\vvec_1=\twovec{1}{1}$$ and $$\vvec_2 = \twovec{-1}{1}$$ and associated eigenvalues $$\lambda_1=3$$ and $$\lambda_2=-1\text{.}$$ In Subsection 4.1.2, we described how $$A$$ is, in some sense, equivalent to the diagonal matrix $$D = \left[\begin{array}{rr} 3 \amp 0 \\ 0 \amp -1\\ \end{array}\right] \text{.}$$

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 $$\vvec_1$$ direction and flipping them in the $$\vvec_2$$ direction. That is, the geometric effect of $$A$$ is the same as that of $$D$$ when viewed in a basis of eigenvectors of $$A\text{.}$$

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 Activity4.3.1.

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 $$A = \begin{bmatrix} \avec_1 \amp \avec_2 \end{bmatrix}\text{,}$$ express the product $$A\twovec2{-3}$$ in terms of $$\avec_1$$ and $$\avec_2\text{.}$$

2. What is the product $$A\twovec40$$ in terms of $$\avec_1$$ and $$\avec_2\text{?}$$

3. Next, remember how matrix-matrix multiplication is defined. Suppose that we have matrices $$A$$ and $$B$$ and that $$B = \begin{bmatrix} \bvec_1 \amp \bvec_2 \end{bmatrix}\text{.}$$ How can we express the matrix product $$AB$$ in terms of the columns of $$B\text{?}$$

4. Suppose that $$A$$ is a matrix having eigenvectors $$\vvec_1$$ and $$\vvec_2$$ with associated eigenvalues $$\lambda_1 = 4$$ and $$\lambda_2 = -1\text{.}$$ Express the product $$A(2\vvec_1+3\vvec_2)$$ in terms of $$\vvec_1$$ and $$\vvec_2\text{.}$$

5. Suppose that $$A$$ is the matrix from the previous part and that $$P=\begin{bmatrix} \vvec_1 \amp \vvec_2 \end{bmatrix}\text{.}$$ What is the matrix product

\begin{equation*} AP = A\begin{bmatrix} \vvec_1 \amp \vvec_2 \end{bmatrix}? \end{equation*}

### Subsection4.3.1Diagonalization of matrices

When working with an $$n\times n$$ matrix $$A\text{,}$$ Subsection 4.1.2 demonstrated the value of having a basis of $$\real^n$$ consisting of eigenvectors of $$A\text{.}$$ 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 $$D\text{.}$$

#### Activity4.3.2.

Suppose that $$A$$ is a $$2\times2$$ matrix having eigenvectors $$\vvec_1$$ and $$\vvec_2$$ with associated eigenvalues $$\lambda_1=3$$ and $$\lambda_2 = -6\text{.}$$ Because the eigenvalues are real and distinct, we know by Proposition 4.2.9 that these eigenvectors form a basis of $$\real^2\text{.}$$

1. What are the products $$A\vvec_1$$ and $$A\vvec_2$$ in terms of $$\vvec_1$$ and $$\vvec_2\text{?}$$

2. If we form the matrix $$P = \begin{bmatrix} \vvec_1 \amp \vvec_2 \end{bmatrix} \text{,}$$ what is the product $$AP$$ in terms of $$\vvec_1$$ and $$\vvec_2\text{?}$$

3. Use the eigenvalues to form the diagonal matrix $$D = \begin{bmatrix} 3 \amp 0 \\ 0 \amp -6 \end{bmatrix}$$ and determine the product $$PD$$ in terms of $$\vvec_1$$ and $$\vvec_2\text{.}$$

4. The results from the previous two parts of this activity demonstrate that $$AP=PD\text{.}$$ Using the fact that the eigenvectors $$\vvec_1$$ and $$\vvec_2$$ form a basis of $$\real^2\text{,}$$ explain why $$P$$ is invertible and that we must have $$A=PDP^{-1}\text{.}$$

5. Suppose that $$A=\begin{bmatrix} -3 \amp 6 \\ 3 \amp 0 \\ \end{bmatrix}\text{.}$$ Verify that $$\vvec_1=\twovec11$$ and $$\vvec_2=\twovec2{-1}$$ are eigenvectors of $$A$$ with eigenvalues $$\lambda_1 = 3$$ and $$\lambda_2=-6\text{.}$$

6. Use the Sage cell below to define the matrices $$P$$ and $$D$$ and then verify that $$A=PDP^{-1}\text{.}$$

# enter the matrices P and D below
P =
D =
P*D*P.inverse()


More generally, suppose that we have an $$n\times n$$ matrix $$A$$ and that there is a basis of $$\real^n$$ consisting of eigenvectors $$\vvec_1,\vvec_2,\ldots,\vvec_n$$ of $$A$$ with associated eigenvalues $$\lambda_1,\lambda_2,\ldots,\lambda_n\text{.}$$ If we use the eigenvectors to form the matrix

\begin{equation*} P = \begin{bmatrix} \vvec_1 \amp \vvec_2 \amp \cdots \amp \vvec_n \end{bmatrix} \end{equation*}

and the eigenvalues to form the diagonal matrix

\begin{equation*} D = \left[\begin{array}{cccc} \lambda_1 \amp 0 \amp \ldots \amp 0 \\ 0 \amp \lambda_2 \amp \ldots \amp 0 \\ \vdots \amp \vdots \amp \ddots \amp 0 \\ 0 \amp 0 \amp \ldots \amp \lambda_n \\ \end{array}\right] \end{equation*}

and apply the same reasoning demonstrated in the activity, we find that $$AP = PD$$ and hence

\begin{equation*} A=PDP^{-1}. \end{equation*}

We have now seen the following proposition.

#### Example4.3.3.

We have seen that $$A = \begin{bmatrix} 1 \amp 2 \\ 2 \amp 1 \\ \end{bmatrix}$$ has eigenvectors $$\vvec_1 = \twovec11$$ and $$\vvec_2=\twovec{-1}1$$ with associated eigenvalues $$\lambda_1 = 3$$ and $$\lambda_2 = -1\text{.}$$ Forming the matrices

\begin{equation*} P = \begin{bmatrix} \vvec_1 \amp \vvec_2 \end{bmatrix} = \begin{bmatrix} 1 \amp -1 \\ 1 \amp 1 \\ \end{bmatrix},~~~ D = \begin{bmatrix} 3 \amp 0 \\ 0 \amp -1 \\ \end{bmatrix}, \end{equation*}

we see that $$A=PDP^{-1}\text{.}$$

This is the sense in which we mean that $$A$$ is equivalent to a diagonal matrix $$D\text{.}$$ The expression $$A=PDP^{-1}$$ says that $$A\text{,}$$ expressed in the basis defined by the columns of $$P\text{,}$$ has the same geometric effect as $$D\text{,}$$ expressed in the standard basis $$\evec_1, \evec_2,\ldots,\evec_n\text{.}$$

#### Definition4.3.4.

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

\begin{equation*} A = PDP^{-1}. \end{equation*}

#### Example4.3.5.

We will try to find a diagonalization of $$A = \left[\begin{array}{rr} -5 \amp 6 \\ -3 \amp 4 \\ \end{array}\right]$$ whose characteristic equation is

\begin{equation*} \det(A-\lambda I) = (-5-\lambda)(4-\lambda)+18 = (-2-\lambda)(1-\lambda) = 0\text{.} \end{equation*}

This shows that the eigenvalues of $$A$$ are $$\lambda_1 = -2$$ and $$\lambda_2 = 1\text{.}$$

By constructing $$\nul(A-(-2)I)\text{,}$$ we find a basis for $$E_{-2}$$ consisting of the vector $$\vvec_1 = \twovec{2}{1}\text{.}$$ Similarly, a basis for $$E_1$$ consists of the vector $$\vvec_2 = \twovec{1}{1}\text{.}$$ This shows that we can construct a basis $$\{\vvec_1,\vvec_2\}$$ of $$\real^2$$ consisting of eigenvectors of $$A\text{.}$$

We now form the matrices

\begin{equation*} D = \left[\begin{array}{rr} -2 \amp 0 \\ 0 \amp 1 \\ \end{array}\right],\qquad P = \left[\begin{array}{cc} \vvec_1 \amp \vvec_2 \end{array}\right] = \left[\begin{array}{rr} 2 \amp 1 \\ 1 \amp 1 \\ \end{array}\right] \end{equation*}

and verify that

\begin{equation*} PDP^{-1} = \left[\begin{array}{rr} 2 \amp 1 \\ 1 \amp 1 \\ \end{array}\right] \left[\begin{array}{rr} -2 \amp 0 \\ 0 \amp 1 \\ \end{array}\right] \left[\begin{array}{rr} 1 \amp -1 \\ -1 \amp 2 \\ \end{array}\right] = \left[\begin{array}{rr} -5 \amp 6 \\ -3 \amp 4 \\ \end{array}\right] = A\text{.} \end{equation*}

There are, in fact, many ways to diagonalize $$A\text{.}$$ For instance, we could change the order of the eigenvalues and eigenvectors and write

\begin{equation*} D = \left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp -2 \\ \end{array}\right],\qquad P = \left[\begin{array}{cc} \vvec_2 \amp \vvec_1 \end{array}\right] = \left[\begin{array}{rr} 1 \amp 2 \\ 1 \amp 1 \\ \end{array}\right]\text{.} \end{equation*}

If we choose a different basis for the eigenspaces, we will also find a different matrix $$P$$ that diagonalizes $$A\text{.}$$ The point is that there are many ways in which $$A$$ can be written in the form $$A=PDP^{-1}\text{.}$$

#### Example4.3.6.

We will try to find a diagonalization of $$A = \left[\begin{array}{rr} 0 \amp 4 \\ -1 \amp 4 \\ \end{array}\right] \text{.}$$

Once again, we find the eigenvalues by solving the characteristic equation:

\begin{equation*} \det(A-\lambda I) = -\lambda(4-\lambda) + 4 = (2-\lambda)^2 = 0\text{.} \end{equation*}

In this case, there is a single eigenvalue $$\lambda=2\text{.}$$

We find a basis for the eigenspace $$E_2$$ by describing $$\nul(A-2I)\text{:}$$

\begin{equation*} A-2I = \left[\begin{array}{rr} -2 \amp 4 \\ -1 \amp 2 \\ \end{array}\right] \sim \left[\begin{array}{rr} 1 \amp -2 \\ 0 \amp 0 \\ \end{array}\right]\text{.} \end{equation*}

This shows that the eigenspace $$E_2$$ is one-dimensional with $$\vvec_1=\twovec{2}{1}$$ forming a basis.

In this case, there is not a basis of $$\real^2$$ consisting of eigenvectors of $$A\text{,}$$ which tells us that $$A$$ is not diagonalizable.

In fact, if we only know that $$A = PDP^{-1}\text{,}$$ we can say that the columns of $$P$$ are eigenvectors of $$A$$ and that the diagonal entries of $$D$$ are the associated eigenvalues.

#### Example4.3.8.

Suppose we know that $$A=PDP^{-1}$$ where

\begin{equation*} D = \left[\begin{array}{rr} 2 \amp 0 \\ 0 \amp -2 \\ \end{array}\right],\qquad P = \left[\begin{array}{cc} \vvec_2 \amp \vvec_1 \end{array}\right] = \left[\begin{array}{rr} 1 \amp 1 \\ 1 \amp 2 \\ \end{array}\right]. \end{equation*}

The columns of $$P$$ form eigenvectors of $$A$$ so that $$\vvec_1 = \twovec{1}{1}$$ is an eigenvector of $$A$$ with eigenvalue $$\lambda_1 = 2$$ and $$\vvec_2 = \twovec{1}{2}$$ is an eigenvector with eigenvalue $$\lambda_2=-2\text{.}$$

We can verify this by computing

\begin{equation*} A = PDP^{-1} = \left[\begin{array}{rr} 6 \amp -4 \\ 8 \amp -6 \\ \end{array}\right] \end{equation*}

and checking that $$A\vvec_1 = \twovec{1}{1}=2\vvec_1$$ and $$A\vvec_2 = \twovec{1}{2} = -2\vvec_2\text{.}$$

#### Activity4.3.3.

1. Find a diagonalization of $$A\text{,}$$ if one exists, when

\begin{equation*} A = \left[\begin{array}{rr} 3 \amp -2 \\ 6 \amp -5 \\ \end{array}\right]\text{.} \end{equation*}
2. Can the diagonal matrix

\begin{equation*} A = \left[\begin{array}{rr} 2 \amp 0 \\ 0 \amp -5 \\ \end{array}\right] \end{equation*}

be diagonalized? If so, explain how to find the matrices $$P$$ and $$D\text{.}$$

3. Find a diagonalization of $$A\text{,}$$ if one exists, when

\begin{equation*} A = \left[\begin{array}{rrr} -2 \amp 0 \amp 0 \\ 1 \amp -3\amp 0 \\ 2 \amp 0 \amp -3 \\ \end{array}\right]\text{.} \end{equation*}


4. Find a diagonalization of $$A\text{,}$$ if one exists, when

\begin{equation*} A = \left[\begin{array}{rrr} -2 \amp 0 \amp 0 \\ 1 \amp -3\amp 0 \\ 2 \amp 1 \amp -3 \\ \end{array}\right]\text{.} \end{equation*}


5. Suppose that $$A=PDP^{-1}$$ where

\begin{equation*} D = \left[\begin{array}{rr} 3 \amp 0 \\ 0 \amp -1 \\ \end{array}\right],\qquad P = \left[\begin{array}{cc} \vvec_2 \amp \vvec_1 \end{array}\right] = \left[\begin{array}{rr} 2 \amp 2 \\ 1 \amp -1 \\ \end{array}\right]\text{.} \end{equation*}
1. Explain why $$A$$ is invertible.

2. Find a diagonalization of $$A^{-1}\text{.}$$

3. Find a diagonalization of $$A^3\text{.}$$

### Subsection4.3.2Powers 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 $$A = \left[\begin{array}{rr} 0.8 \amp 0.6 \\ 0.2 \amp 0.4 \\ \end{array}\right]$$ and an initial vector $$\xvec_0=\ctwovec{1000}{0}\text{,}$$ and we wanted to compute

\begin{equation*} \begin{aligned} \xvec_1 \amp {}={} A\xvec_0 \\ \xvec_2 \amp {}={} A\xvec_1 = A^2\xvec_0 \\ \xvec_3 \amp {}={} A\xvec_2 = A^3\xvec_0\text{.} \\ \end{aligned} \end{equation*}

In particular, we wanted to find $$\xvec_k=A^k\xvec_0$$ and determine what happens as $$k$$ becomes very large. If a matrix $$A$$ is diagonalizable, writing $$A=PDP^{-1}$$ can help us understand powers of $$A$$ more easily.

#### Activity4.3.4.

1. Let's begin with the diagonal matrix

\begin{equation*} D = \left[\begin{array}{rr} 2 \amp 0 \\ 0 \amp -1 \\ \end{array}\right]\text{.} \end{equation*}

Find the powers $$D^2\text{,}$$ $$D^3\text{,}$$ and $$D^4\text{.}$$ What is $$D^k$$ for a general value of $$k\text{?}$$

2. Suppose that $$A$$ is a matrix with eigenvector $$\vvec$$ and associated eigenvalue $$\lambda\text{;}$$ that is, $$A\vvec = \lambda\vvec\text{.}$$ By considering $$A^2\vvec\text{,}$$ explain why $$\vvec$$ is also an eigenvector of $$A$$ with eigenvalue $$\lambda^2\text{.}$$

3. Suppose that $$A= PDP^{-1}$$ where

\begin{equation*} D = \left[\begin{array}{rr} 2 \amp 0 \\ 0 \amp -1 \\ \end{array}\right]\text{.} \end{equation*}

Remembering that the columns of $$P$$ are eigenvectors of $$A\text{,}$$ explain why $$A^2$$ is diagonalizable and find a diagonalization in terms of $$P$$ and $$D\text{.}$$

4. Give another explanation of the diagonalizability of $$A^2$$ by writing

\begin{equation*} A^2 = (PDP^{-1})(PDP^{-1}) = PD(P^{-1}P)DP^{-1}\text{.} \end{equation*}
5. In the same way, find a diagonalization of $$A^3\text{,}$$ $$A^4\text{,}$$ and $$A^k\text{.}$$

6. Suppose that $$A$$ is a diagonalizable $$2\times2$$ matrix with eigenvalues $$\lambda_1 = 0.5$$ and $$\lambda_2=0.1\text{.}$$ What happens to $$A^k$$ as $$k$$ becomes very large?

If $$A$$ is diagonalizable, the activity demonstrates that any power of $$A$$ is as well.

#### Example4.3.10.

Let's revisit Activity 4.1.3 where we had the matrix $$A = \begin{bmatrix} 0.8 \amp 0.6 \\ 0.2 \amp 0.4 \\ \end{bmatrix}$$ and the initial vector $$\xvec_0 = \ctwovec{1000}0\text{.}$$ We were interested in understanding the sequence of vectors $$\xvec_{k+1} = A\xvec_k\text{,}$$ which means that $$\xvec_k = A^k\xvec_0\text{.}$$

We can verify that $$\vvec_1 = \twovec31$$ and $$\vvec_2 = \twovec{-1}1$$ are eigenvectors of $$A$$ having associated eigenvalues $$\lambda_1=1$$ and $$\lambda_2 = 0.2\text{.}$$ This means that $$A = PDP^{-1}$$ where

\begin{equation*} P = \begin{bmatrix} 3 \amp -1 \\ 1 \amp 1 \\ \end{bmatrix},~~~ D = \begin{bmatrix} 1 \amp 0 \\ 0 \amp 0.2 \\ \end{bmatrix}. \end{equation*}

Therefore, the powers of $$A$$ have the form $$A^k = PD^kP^{-1}\text{.}$$

Notice that $$D^k = \begin{bmatrix} 1^k \amp 0 \\ 0 \amp 0.2^k \\ \end{bmatrix} = \begin{bmatrix} 1 \amp 0 \\ 0 \amp 0.2^k \end{bmatrix} \text{.}$$ As $$k$$ increases, $$0.2^k$$ becomes closer and closer to zero. This means that for very large powers $$k\text{,}$$ we have

\begin{equation*} D^k \approx \begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \\ \end{bmatrix} \end{equation*}

and therefore

\begin{equation*} A^k = PD^kP^{-1} \approx P\begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \\ \end{bmatrix}P^{-1} = \begin{bmatrix} \frac 34 \amp \frac 34 \\ \frac 14 \amp \frac 14 \end{bmatrix}. \end{equation*}

Beginning with the vector $$\xvec_0 = \ctwovec{1000}{0}\text{,}$$ we find that $$\xvec_k = A^k\xvec_0\approx \twovec{750}{250}$$ when $$k$$ is very large.

### Subsection4.3.3Similarity and complex eigenvalues

We have been interested in diagonalizing a matrix $$A$$ because doing so relates a matrix $$A$$ to a simpler diagonal matrix $$D\text{.}$$ In particular, the effect of multiplying a vector by $$A=PDP^{-1}\text{,}$$ viewed in the basis defined by the columns of $$P\text{,}$$ 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 $$\real^n$$ 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.

#### Definition4.3.11.

We say that $$A$$ is similar to $$B$$ if there is an invertible matrix $$P$$ such that $$A = PBP^{-1}\text{.}$$

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 $$A=PCP^{-1}$$ has the same effect, when viewed in the basis defined by the columns of $$P\text{,}$$ as $$C\text{,}$$ when viewed in the standard basis.

To begin, suppose that $$A$$ is a $$2\times2$$ matrix having a complex eigenvalue $$\lambda = a+bi\text{.}$$ It turns out that $$A$$ is similar to $$C=\begin{bmatrix} a \amp -b \\ b \amp a \\ \end{bmatrix} \text{.}$$

The next activity shows that $$C$$ has a simple geometric effect on $$\real^2\text{.}$$ First, however, we will use polar coordinates to rewrite $$C\text{.}$$ As shown in the figure, the point $$(a,b)$$ defines $$r\text{,}$$ the distance from the origin, and $$\theta\text{,}$$ the angle formed with the positive horizontal axis. We then have

\begin{equation*} \begin{aligned} a \amp {}={} r\cos\theta \\ b \amp {}={} r\sin\theta\text{.} \\ \end{aligned} \end{equation*}

Notice that the Pythagorean theorem says that $$r=\sqrt{a^2+b^2}\text{.}$$

#### Activity4.3.5.

We begin by rewriting $$C$$ in terms of $$r$$ and $$\theta$$ and noting that

\begin{equation*} C = \left[\begin{array}{rr} a \amp -b \\ b \amp a \\ \end{array}\right] = \left[\begin{array}{rr} r\cos\theta \amp -r\sin\theta \\ r\sin\theta \amp r\cos\theta \\ \end{array}\right] = \left[\begin{array}{rr} r \amp 0 \\ 0 \amp r \\ \end{array}\right] \left[\begin{array}{rr} \cos\theta \amp -\sin\theta \\ \sin\theta \amp \cos\theta \\ \end{array}\right]. \end{equation*}
1. Explain why $$C$$ has the geometric effect of rotating vectors by $$\theta$$ and scaling them by a factor of $$r\text{.}$$

2. Let's now consider the matrix

\begin{equation*} A = \left[\begin{array}{rr} -2 \amp 2 \\ -5 \amp 4 \\ \end{array}\right] \end{equation*}

whose eigenvalues are $$\lambda_1 = 1+i$$ and $$\lambda_2 = 1-i\text{.}$$ We will choose to focus on one of the eigenvalues $$\lambda_1 = a+bi= 1+i.$$

Form the matrix $$C$$ using these values of $$a$$ and $$b\text{.}$$ Then rewrite the point $$(a,b)$$ in polar coordinates by identifying the values of $$r$$ and $$\theta\text{.}$$ Explain the geometric effect of multiplying vectors by $$C\text{.}$$

3. Suppose that $$P=\left[\begin{array}{rr} 1 \amp 1 \\ 2 \amp 1 \\ \end{array}\right] \text{.}$$ Verify that $$A = PCP^{-1}\text{.}$$

C =
P =
P*C*P.inverse()

4. Explain why $$A^k = PC^kP^{-1}\text{.}$$

5. We formed the matrix $$C$$ by choosing the eigenvalue $$\lambda_1=1+i\text{.}$$ Suppose we had instead chosen $$\lambda_2 = 1-i\text{.}$$ Form the matrix $$C'$$ and use polar coordinates to describe the geometric effect of $$C\text{.}$$

6. Using the matrix $$P' = \left[\begin{array}{rr} 1 \amp -1 \\ 2 \amp -1 \\ \end{array}\right] \text{,}$$ show that $$A = P'C'P'^{-1}\text{.}$$

If the $$2\times2$$ matrix $$A$$ has a complex eigenvalue $$\lambda = a + bi\text{,}$$ it turns out that $$A$$ is always similar to the matrix $$C = \left[\begin{array}{rr} a \amp -b \\ b \amp a \\ \end{array}\right],$$ 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 $$A=PCP^{-1}$$ that we'll see in Exercise 4.3.5.8. For now, we note that $$A$$ has the same geometric effect as $$C\text{,}$$ when viewed in the basis provided by the columns of $$P\text{.}$$ We will put this fact to use in the next section to understand certain dynamical systems.

### Subsection4.3.4Summary

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 $$A = PDP^{-1}$$ 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 $$n\times n$$ matrix $$A$$ is diagonalizable if and only if there is a basis of $$\real^n$$ consisting of eigenvectors of $$A\text{.}$$

• We said that $$A$$ and $$B$$ are similar if there is an invertible matrix $$P$$ such that $$A=PBP^{-1}\text{.}$$ In this case, $$A^k = PB^kP^{-1}\text{.}$$

• If $$A$$ is a $$2\times2$$ matrix with complex eigenvalue $$\lambda = a+bi\text{,}$$ then $$A$$ is similar to $$C = \left[\begin{array}{rr} a \amp -b \\ b \amp a \\ \end{array} \right] \text{.}$$ Writing the point $$(a,b)$$ in polar coordinates $$r$$ and $$\theta\text{,}$$ we see that $$C$$ rotates vectors through an angle $$\theta$$ and scales them by a factor of $$r=\sqrt{a^2+b^2}\text{.}$$

### Exercises4.3.5Exercises

#### 1.

Determine whether the following matrices are diagonalizable. If so, find matrices $$D$$ and $$P$$ such that $$A=PDP^{-1}\text{.}$$

1. $$A = \left[\begin{array}{rr} -2 \amp -2 \\ -2 \amp 1 \\ \end{array}\right] \text{.}$$

2. $$A = \left[\begin{array}{rr} -1 \amp 1 \\ -1 \amp -3 \\ \end{array}\right] \text{.}$$

3. $$A = \left[\begin{array}{rr} 3 \amp -4 \\ 2 \amp -1 \\ \end{array}\right] \text{.}$$

4. $$A = \left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 2 \amp -2 \amp 0 \\ 0 \amp 1 \amp 4 \\ \end{array}\right] \text{.}$$

5. $$A = \left[\begin{array}{rrr} 1 \amp 2 \amp 2 \\ 2 \amp 1 \amp 2 \\ 2 \amp 2 \amp 1 \\ \end{array}\right] \text{.}$$

#### 2.

Determine whether the following matrices have complex eigenvalues. If so, find the matrix $$C$$ such that $$A = PCP^{-1}\text{.}$$

1. $$A = \left[\begin{array}{rr} -2 \amp -2 \\ -2 \amp 1 \\ \end{array}\right] \text{.}$$

2. $$A = \left[\begin{array}{rr} -1 \amp 1 \\ -1 \amp -3 \\ \end{array}\right] \text{.}$$

3. $$A = \left[\begin{array}{rr} 3 \amp -4 \\ 2 \amp -1 \\ \end{array}\right] \text{.}$$

#### 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 $$n\times n$$ matrix, then there is a basis of $$\real^n$$ consisting of eigenvectors of $$A\text{.}$$

4. If $$A$$ is diagonalizable, then $$A^{10}$$ 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 $$3\times3$$ matrix having eigenvalues $$\lambda = 2, 3, -4\text{,}$$ can you guarantee that $$A$$ is diagonalizable?

2. If $$A$$ is a $$2\times 2$$ matrix with a complex eigenvalue, can you guarantee that $$A$$ is diagonalizable?

3. If $$A$$ is similar to the matrix $$B = \left[\begin{array}{rrr} -5 \amp 0 \amp 0 \\ 0 \amp -5 \amp 0 \\ 0 \amp 0 \amp 3 \\ \end{array}\right] \text{,}$$ is $$A$$ diagonalizable?

4. What can you say about a matrix that is similar to the identity matrix?

5. If $$A$$ is a diagonalizable $$2\times2$$ matrix with a single eigenvalue $$\lambda = 4\text{,}$$ what is $$A\text{?}$$

#### 5.

Describe geometric effect that the following matrices have on $$\real^2\text{:}$$

1. $$\displaystyle A = \left[\begin{array}{rr} 2 \amp 0 \\ 0 \amp 2 \\ \end{array}\right]$$

2. $$\displaystyle A = \left[\begin{array}{rr} 4 \amp 2 \\ 0 \amp 4 \\ \end{array}\right]$$

3. $$\displaystyle A = \left[\begin{array}{rr} 3 \amp -6 \\ 6 \amp 3 \\ \end{array}\right]$$

4. $$\displaystyle A = \left[\begin{array}{rr} 4 \amp 0 \\ 0 \amp -2 \\ \end{array}\right]$$

5. $$\displaystyle A = \left[\begin{array}{rr} 1 \amp 3 \\ 3 \amp 1 \\ \end{array}\right]$$

#### 6.

We say that $$A$$ is similar to $$B$$ if there is a matrix $$P$$ such that $$A = PBP^{-1}\text{.}$$

1. If $$A$$ is similar to $$B\text{,}$$ explain why $$B$$ is similar to $$A\text{.}$$

2. If $$A$$ is similar to $$B$$ and $$B$$ is similar to $$C\text{,}$$ explain why $$A$$ is similar to $$C\text{.}$$

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 $$\det(A-\lambda I) = \det(B-\lambda I)\text{.}$$

5. If $$A$$ and $$B$$ are similar, explain why $$A$$ and $$B$$ have the same eigenvalues.

#### 7.

Suppose that $$A = PDP^{-1}$$ where

\begin{equation*} D = \left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp 0 \\ \end{array}\right],\qquad P = \left[\begin{array}{rr} 1 \amp -2 \\ 2 \amp 1 \\ \end{array}\right]\text{.} \end{equation*}
1. Explain the geometric effect that $$D$$ has on vectors in $$\real^2\text{.}$$

2. Explain the geometric effect that $$A$$ has on vectors in $$\real^2\text{.}$$

3. What can you say about $$A^2$$ and other powers of $$A\text{?}$$

4. Is $$A$$ invertible?

#### 8.

When $$A$$ is a $$2\times2$$ matrix with a complex eigenvalue $$\lambda = a+bi\text{,}$$ we have said that there is a matrix $$P$$ such that $$A=PCP^{-1}$$ where $$C=\left[\begin{array}{rr} a \amp -b \\ b \amp a \\ \end{array}\right] \text{.}$$ In this exercise, we will learn how to find the matrix $$P\text{.}$$ As an example, we will consider the matrix $$A = \left[\begin{array}{rr} 2 \amp 2 \\ -1 \amp 4 \\ \end{array}\right] \text{.}$$

1. Show that the eigenvalues of $$A$$ are complex.

2. Choose one of the complex eigenvalues $$\lambda=a+bi$$ and construct the usual matrix $$C\text{.}$$

3. Using the same eigenvalue, we will find an eigenvector $$\vvec$$ where the entries of $$\vvec$$ are complex numbers. As always, we will describe $$\nul(A-\lambda I)$$ by constructing the matrix $$A-\lambda I$$ 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 $$\vvec\text{.}$$ Write $$\vvec = \vvec_1 - i \vvec_2$$ to identify vectors $$\vvec_1$$ and $$\vvec_2$$ having real entries.

5. Construct the matrix $$P = \left[\begin{array}{rr} \vvec_1 \amp \vvec_2 \end{array}\right]$$ and verify that $$A=PCP^{-1}\text{.}$$

#### 9.

For each of the following matrices, sketch the vector $$\xvec = \twovec{1}{0}$$ and powers $$A^k\xvec$$ for $$k=1,2,3,4\text{.}$$

1. $$A = \left[\begin{array}{rr} 0 \amp -1.4 \\ 1.4 \amp 0 \\ \end{array}\right] \text{.}$$

2. $$A = \left[\begin{array}{rr} 0 \amp -0.8 \\ 0.8 \amp 0 \\ \end{array}\right] \text{.}$$

3. $$A = \left[\begin{array}{rr} 0 \amp -1 \\ 1 \amp 0 \\ \end{array}\right] \text{.}$$

4. Consider a matrix of the form $$C=\left[\begin{array}{rr} a \amp -b \\ b \amp a \\ \end{array}\right]$$ with $$r=\sqrt{a^2+b^2}\text{.}$$ What happens when $$k$$ becomes very large when

1. $$r \lt 1\text{.}$$

2. $$r = 1\text{.}$$

3. $$r \gt 1\text{.}$$

#### 10.

For each of the following matrices and vectors, sketch the vector $$\xvec$$ along with $$A^k\xvec$$ for $$k=1,2,3,4\text{.}$$

1. \begin{equation*} \begin{aligned} A \amp {}={} \left[\begin{array}{rr} 1.4 \amp 0 \\ 0 \amp 0.7 \\ \end{array}\right] \\ \\ \xvec \amp {}={} \twovec{1}{2}\text{.} \end{aligned}\text{.} \end{equation*}
2. \begin{equation*} \begin{aligned} A \amp {}={} \left[\begin{array}{rr} 0.6 \amp 0 \\ 0 \amp 0.9 \\ \end{array}\right] \\ \\ \xvec \amp {}={} \twovec{4}{3}\text{.} \end{aligned} \end{equation*}
3. \begin{equation*} \begin{aligned} A \amp {}={} \left[\begin{array}{rr} 1.2 \amp 0 \\ 0 \amp 1.4 \\ \end{array}\right] \\ \\ \xvec\amp{}={}\twovec{2}{1}\text{.} \end{aligned} \end{equation*}
4. \begin{equation*} \begin{aligned} A \amp {}={} \left[\begin{array}{rr} 0.95 \amp 0.25 \\ 0.25 \amp 0.95 \\ \end{array}\right] \\ \\ \xvec\amp{}={}\twovec{3}{0}\text{.} \end{aligned} \end{equation*}

Find the eigenvalues and eigenvectors of $$A$$ to create your sketch.

5. If $$A$$ is a $$2\times2$$ matrix with eigenvalues $$\lambda_1=0.7$$ and $$\lambda_2=0.5$$ and $$\xvec$$ is any vector, what happens to $$A^k\xvec$$ when $$k$$ becomes very large?