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

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

Now that we have developed some algebraic techniques for finding eigenvalues and eigenvectors, we will explore this observation more deeply. In particular, we will make precise the sense in which $A$ and $D$ are equivalent by using the coordinate system defined by the basis of eigenvectors $\vvec_1$ and $\vvec_2\text{.}$

##### Preview Activity4.3.1

Let's recall how a vector in $\real^2$ can be represented in a coordinate system defined by a basis $\bcal=\{\vvec_1, \vvec_2\}\text{.}$

1. Suppose that we consider the basis $\bcal$ defined by

\begin{equation*} \vvec_1 = \twovec{1}{1},\qquad \vvec_2 = \twovec{-1}{0} \text{.} \end{equation*}

Find the vector $\xvec$ whose representation in the coordinate system defined by $\bcal$ is $\coords{\xvec}{\bcal} = \twovec{-3}{2}\text{.}$

2. Consider the vector $\xvec=\twovec{4}{5}$ and find its representation $\coords{\xvec}{\bcal}$ in the coordinate system defined by $\bcal\text{.}$

3. How do we use the matrix $C_{\bcal} = \left[\begin{array}{rr} \vvec_1 \amp \vvec_2 \end{array}\right]$ to convert a vector's representation $\coords{\xvec}{\bcal}$ in the coordinate system defined by $\bcal$ into its standard representation $\xvec\text{?}$ How do we use this matrix to convert $\xvec$ into $\coords{\xvec}{\bcal}\text{?}$

4. Suppose that we have a matrix $A$ whose eigenvectors are $\vvec_1$ and $\vvec_2$ and associated eigenvalues are $\lambda_1=4$ and $\lambda_2 = 2\text{.}$ Express the vector $A(-3\vvec_1 +5\vvec_2)$ as a linear combination of $\vvec_1$ and $\vvec_2\text{.}$

5. If $\coords{\xvec}{\bcal} = \twovec{-3}{5}\text{,}$ find $\coords{A\xvec}{\bcal}\text{.}$

# Subsection4.3.1Diagonalization of matrices

If $A$ is an $n\times n$ matrix and there is a basis of $\real^n$ consisting of eigenvectors of $A\text{,}$ we will now see how $A$ is equivalent to a diagonal matrix $D\text{.}$

As we have investigated eigenvalues and eigenvectors of matrices in this chapter, we have frequently asked whether we can find a basis of eigenvectors, as in Question 4.1.7. In fact, Proposition 4.2.3 tells us that if $A$ is an $n\times n$ matrix having distinct and real eigenvalues, then there is a basis for $\real^n$ consisting of eigenvectors of $A\text{.}$ There are, in addition, other conditions on $A$ that guarantee such a basis, as we will see in subsequent chapters, but for now, suffice it to say that for many matrices, we can find a basis of eigenvectors.

Remember also that we have seen how to use a basis $\bcal=\{\vvec_1,\vvec_2,\ldots,\vvec_n\}$ of $\real^n$ to construct a coordinate system for $\real^n\text{.}$ In particular, $\coords{\xvec}{\bcal} = \fourvec{c_1}{c_2}{\vdots}{c_n}$ if $\xvec = c_1\vvec_1 + c_2\vvec_2 + \ldots + c_n\vvec_n\text{.}$ We also used matrix multiplication to express this fact: if $C_{\bcal} = \left[\begin{array}{rrrr} \vvec_1 \amp \vvec_2 \amp \ldots \amp \vvec_n \end{array}\right]\text{,}$ then

\begin{equation*} \xvec = C_{\bcal}\coords{\xvec}{\bcal}, \qquad \coords{\xvec}{\bcal} = C_{\bcal}^{-1}\xvec \text{.} \end{equation*}
##### Activity4.3.2

Once again, we will consider the matrices

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

The matrix $A$ has eigenvectors $\vvec_1=\twovec{1}{1}$ and $\vvec_2=\twovec{-1}{1}$ and eigenvalues $\lambda_1=3$ and $\lambda_2=-1\text{.}$ We will consider the basis of $\real^2$ consisting of eigenvectors $\bcal= \{\vvec_1, \vvec_2\}\text{.}$

1. If $\xvec= 2\vvec_1 - 3\vvec_2\text{,}$ write $A\xvec$ as a linear combination of $\vvec_1$ and $\vvec_2\text{.}$

2. If $\coords{\xvec}{\bcal}=\twovec{2}{-3}\text{,}$ find $\coords{A\xvec}{\bcal}\text{,}$ the representation of $A\xvec$ in the coordinate system defined by $\bcal\text{.}$

3. If $\coords{\xvec}{\bcal}=\twovec{c_1}{c_2}\text{,}$ find $\coords{A\xvec}{\bcal}\text{,}$ the representation of $A\xvec$ in the coordinate system defined by $\bcal\text{.}$

4. Explain why $\coords{A\xvec}{\bcal} = D\coords{\xvec}{\bcal}\text{.}$

5. Explain why $C_{\bcal}^{-1}A\xvec = DC_{\bcal}^{-1}\xvec$ for all vectors $\xvec$ and hence

\begin{equation*} C_{\bcal}^{-1}A = DC_{\bcal}^{-1} \text{.} \end{equation*}
6. Explain why $A = C_{\bcal}DC_{\bcal}^{-1}$ and verify this relationship by computing $C_{\bcal}DC_{\bcal}^{-1}$ in the Sage cell below.

The key to understanding the equivalence of a matrix $A$ and a diagonal matrix $D$ is through the coordinate system defined by a basis consisting of eigenvectors of $A\text{.}$ We will assume that $A$ is an $n\times n$ matrix and that there is a basis $\bcal=\{\vvec_1,\vvec_2,\ldots,\vvec_n\}$ consisting of eigenvectors of $A$ with associated eigenvalues $\lambda_1, \lambda_2,\ldots,\lambda_n\text{.}$

We know that if

\begin{equation*} \xvec = c_1\vvec_1 + c_2\vvec_2 + \ldots + c_n\vvec_n \text{,} \end{equation*}

then

\begin{equation*} A\xvec = \lambda_1c_1\vvec_1 + \lambda_2c_2\vvec_2 + \ldots + \lambda_nc_n\vvec_n \text{.} \end{equation*}

This fact is conveniently expressed using the coordinate system defined by $\bcal\text{;}$ in particular,

\begin{equation*} \coords{\xvec}{\bcal} = \fourvec{c_1}{c_2}{\vdots}{c_n},\qquad \coords{A\xvec}{\bcal} = \fourvec{\lambda_1c_1}{\lambda_2c_2}{\vdots}{\lambda_nc_n} \text{.} \end{equation*}

Forming 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] \text{,} \end{equation*}

we see that

\begin{equation*} \coords{A\xvec}{\bcal} = D\coords{\xvec}{\bcal} \text{.} \end{equation*}

We now use the fact that the matrix $C_{\bcal} = \left[\begin{array}{ccc} \vvec_1 \amp \vvec_2 \amp \ldots \amp \vvec_n \end{array}\right]$ performs the change of coordinates; that is, $\coords{A\xvec}{\bcal} = C_{\bcal}^{-1}A\xvec$ and $\coords{\xvec}{\bcal} = C_{\bcal}^{-1}\xvec\text{.}$ This says that

\begin{equation*} C_{\bcal}^{-1}A\xvec = DC_{\bcal}^{-1}\xvec \text{,} \end{equation*}

for all vectors $\xvec\text{,}$ which means that $C_{\bcal}^{-1}A = DC_{\bcal}^{-1}$ or

\begin{equation*} A = C_{\bcal}^{-1}DC_{\bcal}^{-1} \text{.} \end{equation*}

So that the form of this expression stands out more clearly, it is customary to denote the matrix $C_{\bcal}$ as $P$ so that we have $P = C_{\bcal}\text{.}$ In that case, we write

\begin{equation*} A = PDP^{-1} \text{.} \end{equation*}
##### Definition4.3.2

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} \text{.} \end{equation*}

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 $\evec_1, \evec_2,\ldots,\evec_n\text{.}$

We have now seen the following proposition.

In fact, if we only know that $A = PDP^{-1}$ where $P = \left[\begin{array}{cccc} \vvec_1 \amp \vvec_2 \amp \ldots \vvec_n \end{array}\right]\text{,}$ we can say that the vectors $\vvec_j$ are eigenvectors of $A$ and that the associated eigenvalue of the $j^{th}$ diagonal entry of $D\text{.}$

##### Example4.3.4

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

First, we find the eigenvalues of $A$ by solving the characteristic equation

\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 $\bcal=\{\vvec_1,\vvec_2\}$ 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, of course, 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 $A=PDP^{-1}\text{.}$

##### Example4.3.5

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.

##### Example4.3.6

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] \text{.} \end{equation*}

In this case, we know that the columns of $P$ form eigenvectors of $A\text{.}$ For instance, $\vvec_1 = \twovec{1}{1}$ is an eigenvector of $A$ with eigenvalue $\lambda_1 = 2\text{.}$ Also, $\vvec_2 = \twovec{1}{2}$ 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] \text{.} \end{equation*}

Then, we can compute 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 are given the matrix $A = \left[\begin{array}{rr} 0.8 \amp 0.6 \\ 0.2 \amp 0.4 \\ \end{array}\right]$ and an initial vector $\xvec_0=\twovec{1000}{0}\text{,}$ 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*}

More generally, we would like 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 $A$ 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 of it.

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?

We begin by noting that the eigenvectors of a matrix $A$ are also eigenvectors of the powers of $A\text{.}$ For instance, if $A\vvec = \lambda\vvec\text{,}$ then

\begin{equation*} A^2\vvec = A(A\vvec) = A(\lambda\vvec) = \lambda A\vvec = \lambda^2\vvec \text{.} \end{equation*}

In this way, we see that $\vvec$ is an eigenvector of $A^2$ with eigenvalue $\lambda^2\text{.}$ Furthermore, for any $k\text{,}$ $\vvec$ is an eigenvector of $A^k$ with eigenvalue $\lambda^k\text{.}$

Now if $A$ is diagonalizable, we can write $A=PDP^{-1}$ where the columns of $P$ are eigenvectors of $A$ and the diagonal entries of $D$ are the eigenvalues. If $D = \left[\begin{array}{rr} \lambda_1 \amp 0 \\ 0 \amp \lambda_2 \\ \end{array}\right] \text{,}$ then

\begin{equation*} A^2 = P\left[\begin{array}{rr} \lambda_1^2 \amp 0 \\ 0 \amp \lambda_2^2 \\ \end{array}\right] P^{-1} = PD^2P^{-1} \text{.} \end{equation*}

We have the same matrix $P$ in this expression since the eigenvectors of $A^2$ are also the eigenvectors of $A\text{.}$

Another way to see this is to note that

\begin{equation*} \begin{aligned} A^2 \amp {}={} (PDP^{-1})(PDP^{-1}) \\ \amp {}={} PD(P^{-1}P)DP^{-1} \\ \amp {}={} PDIDP^{-1} \\ \amp {}={} PDDP^{-1} \\ \amp {}={} PD^2P^{-1}\text{.} \end{aligned} \end{equation*}

Similarly, any power of $A$ is diagaonalizable; in particular, $A^k = PD^kP^{-1}\text{.}$

In the next section, we will see some important uses of our ability to deal with powers in this way. Until then, consider the case where $D = \left[\begin{array}{rr} 0.5 \amp 0 \\ 0 \amp 0.1 \\ \end{array}\right]$ so that $D^k = \left[\begin{array}{rr} 0.5^k \amp 0 \\ 0 \amp 0.1^k \\ \end{array}\right] \text{.}$ As $k$ becomes very large, the diagonal entries become increasingly close to zero. This means that $D^k$ becomes increasingly close to the zero matrix $\left[\begin{array}{rr} 0 \amp 0 \\ 0 \amp 0 \\ \end{array}\right]$ as does $A^k = PD^kP^{-1}\text{.}$ In other words, no matter what vector $\xvec_0$ we begin with, the vectors $A^k\xvec_0$ becomes increasingly close to $\zerovec\text{.}$

# 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 $D\text{.}$ If we write $A=PDP^{-1}\text{,}$ we see that multiplying a vector by $A$ in the coordinates defined by the columns of $P$ is the same as multiplying by $D$ in standard coordinates. Under this change of coordinates, $A$ and $D$ have the same effect on vectors.

More generally, if we have two matrices $A$ and $B$ such that $A=PBP^{-1}\text{,}$ we may regard multiplication by $A$ and $B$ as having the same effect on vectors under the change of coordinates defined by the columns of $P\text{.}$ That is, if $\bcal$ is the basis formed by the columns of $P\text{,}$ then $\coords{A\xvec}{\bcal} = B\coords{\xvec}{\bcal}\text{.}$ This leads to the following definition.

##### Definition4.3.7

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. We have, however, seen several examples of a matrix $A$ that is not diagonalizable. In this case, it is natural to ask if there is some simpler matrix to which $A$ is similar.

##### Example4.3.8

Let's consider the matrix $A = \left[\begin{array}{rr} -2 \amp 2 \\ -5 \amp 4 \\ \end{array}\right]$ whose characteristic equation is

\begin{equation*} \det(A-\lambda I) = (-2-\lambda)(4-\lambda)+10 = 2 - 2\lambda + \lambda^2 = 0 \text{.} \end{equation*}

Applying the quadratic formula to find the eigenvalues, we obtain

\begin{equation*} \lambda = \frac{2\pm\sqrt{(-2)^2-4\cdot1\cdot2}}{2}=1\pm i \text{.} \end{equation*}

Here we see that the matrix $A$ has two complex eigenvalues and is therefore not diagonalizable.

In case a matrix $A$ has complex eigenvalues, we will find a simpler matrix $C$ that is similar to $A\text{.}$ In particular, if $A$ has an eigenvalue $\lambda = a+bi\text{,}$ then $A$ is similar to $C=\left[\begin{array}{rr} a \amp -b \\ b \amp a \\ \end{array}\right] \text{.}$

The next activity shows that $C$ has a simple geometric effect on $\real^2\text{.}$ First, however, we will rewrite $C$ in polar coordinates, as shown in the figure. We form the point $(a,b)\text{,}$ which 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 $r=\sqrt{a^2+b^2}\text{.}$

##### Activity4.3.5

1. We will rewrite $C$ in terms of $r$ and $\theta\text{.}$ Explain why

\begin{equation*} \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] \text{.} \end{equation*}
2. Explain why $C$ has the geometric effect of rotating vectors by $\theta$ and stretching them by a factor of $r\text{.}$

3. Let's now consider the matrix $A$ from Example 8:

\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 of $C\text{.}$

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

5. Explain why $A^kk = PC^kP^{-1}\text{.}$

6. 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 $C\text{.}$

7. 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 $\lambda = a + bi\text{,}$ this activity demonstrates the fact that $A$ is similar to the matrix $C = \left[\begin{array}{rr} a \amp -b \\ b \amp a \\ \end{array}\right] \text{.}$ When we consider the matrix $A = \left[\begin{array}{rr} -2 \amp 2 \\ -5 \amp 4 \\ \end{array}\right] \text{,}$ we find the complex eigenvalue $\lambda=1+i\text{,}$ which leads to the matrix

\begin{equation*} C = \left[\begin{array}{rr} 1 \amp -1 \\ 1 \amp 1 \\ \end{array}\right] = \left[\begin{array}{rr} \sqrt{2} \amp 0 \\ 0 \amp \sqrt{2} \\ \end{array}\right] \left[\begin{array}{rr} \cos(45^\circ) \amp -\sin(45^\circ) \\ \sin(45^\circ) \amp \cos(45^\circ) \\ \end{array}\right] \text{.} \end{equation*}

The matrix has the geometric effect of rotating vectors by $45^\circ$ and stretching them by a factor of $\sqrt{2}\text{,}$ as shown in the figure.

As we saw in the activity, our original matrix $A$ is similar to $C\text{.}$ That is, we saw that there is a matrix $P$ such that $A=PCP^{-1}\text{.}$ This means that, when expressed in the coordinates defined by the columns of $P\text{,}$ multiplying a vector by $A$ is equivalent to multiplying by $C\text{;}$ that is, if $\bcal$ is the basis formed by the columns of $A\text{,}$ then $\coords{A\xvec}{\bcal} = C\coords{\xvec}{\bcal}\text{.}$

Had we chosen the other eigenvalue $\lambda_2 = 1-i\text{,}$ we would have formed the matrix

\begin{equation*} C' = \left[\begin{array}{rr} 1 \amp 1 \\ -1 \amp 1 \\ \end{array}\right] = \left[\begin{array}{rr} \sqrt{2} \amp 0 \\ 0 \amp \sqrt{2} \\ \end{array}\right] \left[\begin{array}{rr} \cos(-45^\circ) \amp -\sin(-45^\circ) \\ \sin(-45^\circ) \amp \cos(-45^\circ) \\ \end{array}\right] \text{.} \end{equation*}

In other words, this matrix $C'$ rotates vectors by $-45^\circ$ and stretches them by a factor of $\sqrt{2}\text{.}$ The original matrix $A$ is also similar to $C'\text{.}$

Depending on which eigenvalue we choose, we find a matrix $C$ that performs either a counterclockwise or a clockwise rotation. For our future uses, we will not be concerned about the direction of the rotation. Instead, we will focus on $r\text{,}$ the stretching factor.

# Subsection4.3.4Summary

The ideas in this section demonstrate how the eigenvalues and eigenvectors of a matrix $A$ can provide us with a new coordinate system in which multiplying by $A$ reduces to a simpler operation.

• 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 $D$ and 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 stretches them by a factor of $r=\sqrt{a^2+b^2}\text{.}$

# Subsection4.3.5Exercises

##### 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 matrices are 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. $A = \left[\begin{array}{rr} 2 \amp 0 \\ 0 \amp 2 \\ \end{array}\right]$

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

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

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

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

##### 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. $r \tt 1\text{.}$

##### 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} 1.2 \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} 0.4 \amp 0 \\ 0 \amp 0.7 \\ \end{array}\right] \\ \\ \xvec \amp {}={} \twovec{3}{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{1}{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{1}{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?