Diagonalization, similarity, and powers of a matrix

Section 4.3 Diagonalization, similarity, and powers of a matrix

The first example we considered in this chapter was the matrix

A = [\begin{array}{rr} 1 & 2 \\ 2 & 1 \end{array}],

which has eigenvectors

v_{1} = [\begin{array}{r} 1 \\ 1 \end{array}]

and

v_{2} = [\begin{array}{r} - 1 \\ 1 \end{array}]

and associated eigenvalues

λ_{1} = 3

and

λ_{2} = - 1 .

In Subsection 4.1.2, we described how

A

is, in some sense, equivalent to the diagonal matrix

D = [\begin{array}{rr} 3 & 0 \\ 0 & - 1 \end{array}] .

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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

v_{1}

direction and flipping them in the

v_{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 .

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Figure 4.3.1. The matrix $A$ has the same geometric effect as the diagonal matrix $D$ when viewed in the basis of eigenvectors.
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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.

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Preview Activity 4.3.1.

In this preview activity, we will review some familiar properties about matrix multiplication that appear in this section.

Remember that matrix-vector multiplication constructs linear combinations of the columns of the matrix. For instance, if $A = [\begin{matrix} a_{1} & a_{2} \end{matrix}],$ express the product $A [\begin{array}{r} 2 \\ - 3 \end{array}]$ in terms of $a_{1}$ and $a_{2} .$
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What is the product $A [\begin{array}{r} 4 \\ 0 \end{array}]$ in terms of $a_{1}$ and $a_{2} ?$
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Next, remember how matrix-matrix multiplication is defined. Suppose that we have matrices $A$ and $B$ and that $B = [\begin{matrix} b_{1} & b_{2} \end{matrix}] .$ How can we express the matrix product $A B$ in terms of the columns of $B ?$
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Suppose that $A$ is a matrix having eigenvectors $v_{1}$ and $v_{2}$ with associated eigenvalues $λ_{1} = 4$ and $λ_{2} = - 1 .$ Express the product $A (2 v_{1} + 3 v_{2})$ in terms of $v_{1}$ and $v_{2} .$
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Suppose that $A$ is the matrix from the previous part and that $P = [\begin{matrix} v_{1} & v_{2} \end{matrix}] .$ What is the matrix product

$A P = A [\begin{matrix} v_{1} & v_{2} \end{matrix}] ?$

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Subsection 4.3.1 Diagonalization of matrices

When working with an

n \times n

matrix

A,

Subsection 4.1.2 demonstrated the value of having a basis of

R^{n}

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 .

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Activity 4.3.2.

Suppose that

A

is a

2 \times 2

matrix having eigenvectors

v_{1}

and

v_{2}

with associated eigenvalues

λ_{1} = 3

and

λ_{2} = - 6 .

Because the eigenvalues are real and distinct, we know by Proposition 4.2.9 that these eigenvectors form a basis of

R^{2} .

What are the products $A v_{1}$ and $A v_{2}$ in terms of $v_{1}$ and $v_{2} ?$
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If we form the matrix $P = [\begin{matrix} v_{1} & v_{2} \end{matrix}],$ what is the product $A P$ in terms of $v_{1}$ and $v_{2} ?$
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Use the eigenvalues to form the diagonal matrix $D = [\begin{matrix} 3 & 0 \\ 0 & - 6 \end{matrix}]$ and determine the product $P D$ in terms of $v_{1}$ and $v_{2} .$
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The results from the previous two parts of this activity demonstrate that $A P = P D .$ Using the fact that the eigenvectors $v_{1}$ and $v_{2}$ form a basis of $R^{2},$ explain why $P$ is invertible and that we must have $A = P D P^{- 1} .$
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Suppose that $A = [\begin{matrix} - 3 & 6 \\ 3 & 0 \end{matrix}] .$ Verify that $v_{1} = [\begin{array}{r} 1 \\ 1 \end{array}]$ and $v_{2} = [\begin{array}{r} 2 \\ - 1 \end{array}]$ are eigenvectors of $A$ with eigenvalues $λ_{1} = 3$ and $λ_{2} = - 6 .$
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Use the Sage cell below to define the matrices $P$ and $D$ and then verify that $A = P D P^{- 1} .$
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# enter the matrices P and D below
P =
D =
P*D*P.inverse()
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More generally, suppose that we have an

n \times n

matrix

A

and that there is a basis of

R^{n}

consisting of eigenvectors

v_{1}, v_{2}, \dots, v_{n}

A

with associated eigenvalues

λ_{1}, λ_{2}, \dots, λ_{n} .

If we use the eigenvectors to form the matrix

P = [\begin{matrix} v_{1} & v_{2} & \dots & v_{n} \end{matrix}]

and the eigenvalues to form the diagonal matrix

D = [\begin{array}{cccc} λ_{1} & 0 & \dots & 0 \\ 0 & λ_{2} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & 0 \\ 0 & 0 & \dots & λ_{n} \end{array}]

and apply the same reasoning demonstrated in the activity, we find that

A P = P D

and hence

A = P D P^{- 1} .

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We have now seen the following proposition.

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Proposition 4.3.2.

A

is an

n \times n

matrix and there is a basis

{v_{1}, v_{2}, \dots, v_{n}}

R^{n}

consisting of eigenvectors of

A

having associated eigenvalues

λ_{1}, λ_{2}, \dots, λ_{n},

then we can write

A = P D P^{- 1}

where

D

is the diagonal matrix whose diagonal entries are the eigenvalues of

A

D = [\begin{array}{cccc} λ_{1} & 0 & \dots & 0 \\ 0 & λ_{2} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & 0 \\ 0 & 0 & \dots & λ_{n} \end{array}]

and the matrix

P = [\begin{array}{cccc} v_{1} & v_{2} & \dots & v_{n} \end{array}] .

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Example 4.3.3.

We have seen that

A = [\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix}]

has eigenvectors

v_{1} = [\begin{array}{r} 1 \\ 1 \end{array}]

and

v_{2} = [\begin{array}{r} - 1 \\ 1 \end{array}]

with associated eigenvalues

λ_{1} = 3

and

λ_{2} = - 1 .

Forming the matrices

P = [\begin{matrix} v_{1} & v_{2} \end{matrix}] = [\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}], D = [\begin{matrix} 3 & 0 \\ 0 & - 1 \end{matrix}],

we see that

A = P D P^{- 1} .

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This is the sense in which we mean that

A

is equivalent to a diagonal matrix

D .

The expression

A = P D P^{- 1}

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

e_{1}, e_{2}, \dots, e_{n} .

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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 = P D P^{- 1} .

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Example 4.3.5.

We will try to find a diagonalization of

A = [\begin{array}{rr} - 5 & 6 \\ - 3 & 4 \end{array}]

whose characteristic equation is

det (A - λ I) = (- 5 - λ) (4 - λ) + 18 = (- 2 - λ) (1 - λ) = 0 .

This shows that the eigenvalues of

A

are

λ_{1} = - 2

and

λ_{2} = 1 .

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By constructing

Nul (A - (- 2) I),

we find a basis for

E_{- 2}

consisting of the vector

v_{1} = [\begin{array}{r} 2 \\ 1 \end{array}] .

Similarly, a basis for

E_{1}

consists of the vector

v_{2} = [\begin{array}{r} 1 \\ 1 \end{array}] .

This shows that we can construct a basis

{v_{1}, v_{2}}

R^{2}

consisting of eigenvectors of

A .

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We now form the matrices

D = [\begin{array}{rr} - 2 & 0 \\ 0 & 1 \end{array}], P = [\begin{array}{cc} v_{1} & v_{2} \end{array}] = [\begin{array}{rr} 2 & 1 \\ 1 & 1 \end{array}]

and verify that

P D P^{- 1} = [\begin{array}{rr} 2 & 1 \\ 1 & 1 \end{array}] [\begin{array}{rr} - 2 & 0 \\ 0 & 1 \end{array}] [\begin{array}{rr} 1 & - 1 \\ - 1 & 2 \end{array}] = [\begin{array}{rr} - 5 & 6 \\ - 3 & 4 \end{array}] = A .

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There are, in fact, many ways to diagonalize

A .

For instance, we could change the order of the eigenvalues and eigenvectors and write

D = [\begin{array}{rr} 1 & 0 \\ 0 & - 2 \end{array}], P = [\begin{array}{cc} v_{2} & v_{1} \end{array}] = [\begin{array}{rr} 1 & 2 \\ 1 & 1 \end{array}] .

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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 = P D P^{- 1} .

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Example 4.3.6.

We will try to find a diagonalization of

A = [\begin{array}{rr} 0 & 4 \\ - 1 & 4 \end{array}] .

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

det (A - λ I) = - λ (4 - λ) + 4 = (2 - λ)^{2} = 0 .

In this case, there is a single eigenvalue

λ = 2 .

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We find a basis for the eigenspace

E_{2}

by describing

Nul (A - 2 I) :

A - 2 I = [\begin{array}{rr} - 2 & 4 \\ - 1 & 2 \end{array}] \sim [\begin{array}{rr} 1 & - 2 \\ 0 & 0 \end{array}] .

This shows that the eigenspace

E_{2}

is one-dimensional with

v_{1} = [\begin{array}{r} 2 \\ 1 \end{array}]

forming a basis.

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In this case, there is not a basis of

R^{2}

consisting of eigenvectors of

A,

which tells us that

A

is not diagonalizable.

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In fact, if we only know that

A = P D P^{- 1},

we can say that the columns of

P

are eigenvectors of

A

and that the diagonal entries of

D

are the associated eigenvalues.

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Proposition 4.3.7.

n \times n

matrix

A

is diagonalizable if and only if there is a basis of

R^{n}

consisting of eigenvectors of

A .

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Example 4.3.8.

Suppose we know that

A = P D P^{- 1}

where

D = [\begin{array}{rr} 2 & 0 \\ 0 & - 2 \end{array}], P = [\begin{array}{cc} v_{2} & v_{1} \end{array}] = [\begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array}] .

The columns of

P

form eigenvectors of

A

so that

v_{1} = [\begin{array}{r} 1 \\ 1 \end{array}]

is an eigenvector of

A

with eigenvalue

λ_{1} = 2

and

v_{2} = [\begin{array}{r} 1 \\ 2 \end{array}]

is an eigenvector with eigenvalue

λ_{2} = - 2 .

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We can verify this by computing

A = P D P^{- 1} = [\begin{array}{rr} 6 & - 4 \\ 8 & - 6 \end{array}]

and checking that

A v_{1} = [\begin{array}{r} 1 \\ 1 \end{array}] = 2 v_{1}

and

A v_{2} = [\begin{array}{r} 1 \\ 2 \end{array}] = - 2 v_{2} .

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Activity 4.3.3.

Find a diagonalization of $A,$ if one exists, when

$A = [\begin{array}{rr} 3 & - 2 \\ 6 & - 5 \end{array}] .$

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Can the diagonal matrix

$A = [\begin{array}{rr} 2 & 0 \\ 0 & - 5 \end{array}]$

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

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Find a diagonalization of $A,$ if one exists, when

$A = [\begin{array}{rrr} - 2 & 0 & 0 \\ 1 & - 3 & 0 \\ 2 & 0 & - 3 \end{array}] .$

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Find a diagonalization of $A,$ if one exists, when

$A = [\begin{array}{rrr} - 2 & 0 & 0 \\ 1 & - 3 & 0 \\ 2 & 1 & - 3 \end{array}] .$

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Suppose that $A = P D P^{- 1}$ where

$D = [\begin{array}{rr} 3 & 0 \\ 0 & - 1 \end{array}], P = [\begin{array}{cc} v_{2} & v_{1} \end{array}] = [\begin{array}{rr} 2 & 2 \\ 1 & - 1 \end{array}] .$
1. Explain why $A$ is invertible.
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2. Find a diagonalization of $A^{- 1} .$
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3. Find a diagonalization of $A^{3} .$
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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 = [\begin{array}{rr} 0.8 & 0.6 \\ 0.2 & 0.4 \end{array}]

and an initial vector

x_{0} = [\begin{matrix} 1000 \\ 0 \end{matrix}],

and we wanted to compute

\begin{aligned} x_{1} & = A x_{0} \\ x_{2} & = A x_{1} = A^{2} x_{0} \\ x_{3} & = A x_{2} = A^{3} x_{0} . \end{aligned}

In particular, we wanted to find

x_{k} = A^{k} x_{0}

and determine what happens as

k

becomes very large. If a matrix

A

is diagonalizable, writing

A = P D P^{- 1}

can help us understand powers of

A

more easily.

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Activity 4.3.4.

Let’s begin with the diagonal matrix

$D = [\begin{array}{rr} 2 & 0 \\ 0 & - 1 \end{array}] .$

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

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Suppose that $A$ is a matrix with eigenvector $v$ and associated eigenvalue $λ;$ that is, $A v = λ v .$ By considering $A^{2} v,$ explain why $v$ is also an eigenvector of $A$ with eigenvalue $λ^{2} .$
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Suppose that $A = P D P^{- 1}$ where

$D = [\begin{array}{rr} 2 & 0 \\ 0 & - 1 \end{array}] .$

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

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Give another explanation of the diagonalizability of $A^{2}$ by writing

$A^{2} = (P D P^{- 1}) (P D P^{- 1}) = P D (P^{- 1} P) D P^{- 1} .$

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In the same way, find a diagonalization of $A^{3},$ $A^{4},$ and $A^{k} .$
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Suppose that $A$ is a diagonalizable $2 \times 2$ matrix with eigenvalues $λ_{1} = 0.5$ and $λ_{2} = 0.1 .$ What happens to $A^{k}$ as $k$ becomes very large?
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A

is diagonalizable, the activity demonstrates that any power of

A

is as well.

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Proposition 4.3.9.

A = P D P^{- 1},

then

A^{k} = P D^{k} P^{- 1} .

When

A

is invertible, we also have

A^{- 1} = P D^{- 1} P^{- 1} .

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Example 4.3.10.

Let’s revisit Activity 4.1.3 where we had the matrix

A = [\begin{matrix} 0.8 & 0.6 \\ 0.2 & 0.4 \end{matrix}]

and the initial vector

x_{0} = [\begin{matrix} 1000 \\ 0 \end{matrix}] .

We were interested in understanding the sequence of vectors

x_{k + 1} = A x_{k},

which means that

x_{k} = A^{k} x_{0} .

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We can verify that

v_{1} = [\begin{array}{r} 3 \\ 1 \end{array}]

and

v_{2} = [\begin{array}{r} - 1 \\ 1 \end{array}]

are eigenvectors of

A

having associated eigenvalues

λ_{1} = 1

and

λ_{2} = 0.2 .

This means that

A = P D P^{- 1}

where

P = [\begin{matrix} 3 & - 1 \\ 1 & 1 \end{matrix}], D = [\begin{matrix} 1 & 0 \\ 0 & 0.2 \end{matrix}] .

Therefore, the powers of

A

have the form

A^{k} = P D^{k} P^{- 1} .

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Notice that

D^{k} = [\begin{matrix} 1^{k} & 0 \\ 0 & {0.2}^{k} \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & {0.2}^{k} \end{matrix}] .

k

increases,

{0.2}^{k}

becomes closer and closer to zero. This means that for very large powers

k,

we have

D^{k} \approx [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}]

and therefore

A^{k} = P D^{k} P^{- 1} \approx P [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}] P^{- 1} = [\begin{matrix} \frac{3}{4} & \frac{3}{4} \\ \frac{1}{4} & \frac{1}{4} \end{matrix}] .

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Beginning with the vector

x_{0} = [\begin{matrix} 1000 \\ 0 \end{matrix}],

we find that

x_{k} = A^{k} x_{0} \approx [\begin{array}{r} 750 \\ 250 \end{array}]

when

k

is very large.

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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 = P D P^{- 1},

viewed in the basis defined by the columns of

P,

is the same as the effect of multiplying by

D

in the standard basis.

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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

R^{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.

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Definition 4.3.11.

We say that

A

is similar to

B

if there is an invertible matrix

P

such that

A = P B P^{- 1} .

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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 = P C P^{- 1}

has the same effect, when viewed in the basis defined by the columns of

P,

C,

when viewed in the standard basis.

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To begin, suppose that

A

is a

2 \times 2

matrix having a complex eigenvalue

λ = a + b i .

It turns out that

A

is similar to

C = [\begin{matrix} a & - b \\ b & a \end{matrix}] .

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The next activity shows that

C

has a simple geometric effect on

R^{2} .

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

\begin{aligned} a & = r \cos θ \\ b & = r \sin θ . \end{aligned}

Notice that the Pythagorean theorem says that

r = \sqrt{a^{2} + b^{2}} .

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Activity 4.3.5.

We begin by rewriting

C

in terms of

r

and

θ

and noting that

C = [\begin{array}{rr} a & - b \\ b & a \end{array}] = [\begin{array}{rr} r \cos θ & - r \sin θ \\ r \sin θ & r \cos θ \end{array}] = [\begin{array}{rr} r & 0 \\ 0 & r \end{array}] [\begin{array}{rr} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array}] .

Explain why $C$ has the geometric effect of rotating vectors by $θ$ and scaling them by a factor of $r .$
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Let’s now consider the matrix

$A = [\begin{array}{rr} - 2 & 2 \\ - 5 & 4 \end{array}]$

whose eigenvalues are $λ_{1} = 1 + i$ and $λ_{2} = 1 - i .$ We will choose to focus on one of the eigenvalues $λ_{1} = a + b i = 1 + i .$

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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 $C .$
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Suppose that $P = [\begin{array}{rr} 1 & 1 \\ 2 & 1 \end{array}] .$ Verify that $A = P C P^{- 1} .$
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C =
P =
P*C*P.inverse()
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Explain why $A^{k} = P C^{k} P^{- 1} .$
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We formed the matrix $C$ by choosing the eigenvalue $λ_{1} = 1 + i .$ Suppose we had instead chosen $λ_{2} = 1 - i .$ Form the matrix $C^{'}$ and use polar coordinates to describe the geometric effect of $C .$
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Using the matrix $P^{'} = [\begin{array}{rr} 1 & - 1 \\ 2 & - 1 \end{array}],$ show that $A = P^{'} C^{'} P^{' - 1} .$
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If the

2 \times 2

matrix

A

has a complex eigenvalue

λ = a + b i,

it turns out that

A

is always similar to the matrix

C = [\begin{array}{rr} a & - b \\ b & a \end{array}],

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 = P C P^{- 1}

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.

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Proposition 4.3.12.

A

is a

2 \times 2

matrix with a complex eigenvalue

λ = a + b i,

then

A

is similar to

C = [\begin{matrix} a & - b \\ b & a \end{matrix}];

that is, there is a matrix

P

such that

A = P C P^{- 1} .

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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 $A = P D P^{- 1}$ 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.
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An $n \times n$ matrix $A$ is diagonalizable if and only if there is a basis of $R^{n}$ consisting of eigenvectors of $A .$
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We said that $A$ and $B$ are similar if there is an invertible matrix $P$ such that $A = P B P^{- 1} .$ In this case, $A^{k} = P B^{k} P^{- 1} .$
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If $A$ is a $2 \times 2$ matrix with complex eigenvalue $λ = a + b i,$ then $A$ is similar to $C = [\begin{array}{rr} a & - b \\ b & a \end{array}] .$ 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 $r = \sqrt{a^{2} + b^{2}} .$
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Exercises 4.3.5 Exercises

1.

Determine whether the following matrices are diagonalizable. If so, find matrices

D

and

P

such that

A = P D P^{- 1} .

$A = [\begin{array}{rr} - 2 & - 2 \\ - 2 & 1 \end{array}] .$
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$A = [\begin{array}{rr} - 1 & 1 \\ - 1 & - 3 \end{array}] .$
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$A = [\begin{array}{rr} 3 & - 4 \\ 2 & - 1 \end{array}] .$
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$A = [\begin{array}{rrr} 1 & 0 & 0 \\ 2 & - 2 & 0 \\ 0 & 1 & 4 \end{array}] .$
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$A = [\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}] .$
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2.

Determine whether the following matrices have complex eigenvalues. If so, find the matrix

C

such that

A = P C P^{- 1} .

$A = [\begin{array}{rr} - 2 & - 2 \\ - 2 & 1 \end{array}] .$
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$A = [\begin{array}{rr} - 1 & 1 \\ - 1 & - 3 \end{array}] .$
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$A = [\begin{array}{rr} 3 & - 4 \\ 2 & - 1 \end{array}] .$
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3.

Determine whether the following statements are true or false and provide a justification for your response.

If $A$ is invertible, then $A$ is diagonalizable.
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If $A$ and $B$ are similar and $A$ is invertible, then $B$ is also invertible.
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If $A$ is a diagonalizable $n \times n$ matrix, then there is a basis of $R^{n}$ consisting of eigenvectors of $A .$
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If $A$ is diagonalizable, then $A^{10}$ is also diagonalizable.
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If $A$ is diagonalizable, then $A$ is invertible.
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4.

Provide a justification for your response to the following questions.

If $A$ is a $3 \times 3$ matrix having eigenvalues $λ = 2, 3, - 4,$ can you guarantee that $A$ is diagonalizable?
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If $A$ is a $2 \times 2$ matrix with a complex eigenvalue, can you guarantee that $A$ is diagonalizable?
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If $A$ is similar to the matrix $B = [\begin{array}{rrr} - 5 & 0 & 0 \\ 0 & - 5 & 0 \\ 0 & 0 & 3 \end{array}],$ is $A$ diagonalizable?
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What can you say about a matrix that is similar to the identity matrix?
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If $A$ is a diagonalizable $2 \times 2$ matrix with a single eigenvalue $λ = 4,$ what is $A ?$
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5.

Describe geometric effect that the following matrices have on

R^{2} :

$A = [\begin{array}{rr} 2 & 0 \\ 0 & 2 \end{array}]$
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$A = [\begin{array}{rr} 4 & 2 \\ 0 & 4 \end{array}]$
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$A = [\begin{array}{rr} 3 & - 6 \\ 6 & 3 \end{array}]$
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$A = [\begin{array}{rr} 4 & 0 \\ 0 & - 2 \end{array}]$
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$A = [\begin{array}{rr} 1 & 3 \\ 3 & 1 \end{array}]$
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6.

We say that

A

is similar to

B

if there is a matrix

P

such that

A = P B P^{- 1} .

If $A$ is similar to $B,$ explain why $B$ is similar to $A .$
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If $A$ is similar to $B$ and $B$ is similar to $C,$ explain why $A$ is similar to $C .$
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If $A$ is similar to $B$ and $B$ is diagonalizable, explain why $A$ is diagonalizable.
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If $A$ and $B$ are similar, explain why $A$ and $B$ have the same characteristic polynomial; that is, explain why $det (A - λ I) = det (B - λ I) .$
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If $A$ and $B$ are similar, explain why $A$ and $B$ have the same eigenvalues.
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7.

Suppose that

A = P D P^{- 1}

where

D = [\begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array}], P = [\begin{array}{rr} 1 & - 2 \\ 2 & 1 \end{array}] .

Explain the geometric effect that $D$ has on vectors in $R^{2} .$
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Explain the geometric effect that $A$ has on vectors in $R^{2} .$
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What can you say about $A^{2}$ and other powers of $A ?$
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Is $A$ invertible?
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8.

When

A

is a

2 \times 2

matrix with a complex eigenvalue

λ = a + b i,

we have said that there is a matrix

P

such that

A = P C P^{- 1}

where

C = [\begin{array}{rr} a & - b \\ b & a \end{array}] .

In this exercise, we will learn how to find the matrix

P .

As an example, we will consider the matrix

A = [\begin{array}{rr} 2 & 2 \\ - 1 & 4 \end{array}] .

Show that the eigenvalues of $A$ are complex.
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Choose one of the complex eigenvalues $λ = a + b i$ and construct the usual matrix $C .$
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Using the same eigenvalue, we will find an eigenvector $v$ where the entries of $v$ are complex numbers. As always, we will describe $Nul (A - λ I)$ by constructing the matrix $A - λ I$ and finding its reduced row echelon form. In doing so, we will necessarily need to use complex arithmetic.
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We have now found a complex eigenvector $v .$ Write $v = v_{1} - i v_{2}$ to identify vectors $v_{1}$ and $v_{2}$ having real entries.
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Construct the matrix $P = [\begin{array}{rr} v_{1} & v_{2} \end{array}]$ and verify that $A = P C P^{- 1} .$
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9.

For each of the following matrices, sketch the vector

x = [\begin{array}{r} 1 \\ 0 \end{array}]

and powers

A^{k} x

for

k = 1, 2, 3, 4 .

$A = [\begin{array}{rr} 0 & - 1.4 \\ 1.4 & 0 \end{array}] .$
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$A = [\begin{array}{rr} 0 & - 0.8 \\ 0.8 & 0 \end{array}] .$
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$A = [\begin{array}{rr} 0 & - 1 \\ 1 & 0 \end{array}] .$
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Consider a matrix of the form $C = [\begin{array}{rr} a & - b \\ b & a \end{array}]$ with $r = \sqrt{a^{2} + b^{2}} .$ What happens when $k$ becomes very large when
1. $r < 1 .$
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2. $r = 1 .$
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3. $r > 1 .$
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10.

For each of the following matrices and vectors, sketch the vector

x

along with

A^{k} x

for

k = 1, 2, 3, 4 .

$\begin{aligned} A & = [\begin{array}{rr} 1.4 & 0 \\ 0 & 0.7 \end{array}] \\ x & = [\begin{array}{r} 1 \\ 2 \end{array}] . \end{aligned} .$

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$\begin{aligned} A & = [\begin{array}{rr} 0.6 & 0 \\ 0 & 0.9 \end{array}] \\ x & = [\begin{array}{r} 4 \\ 3 \end{array}] . \end{aligned}$

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$\begin{aligned} A & = [\begin{array}{rr} 1.2 & 0 \\ 0 & 1.4 \end{array}] \\ x & = [\begin{array}{r} 2 \\ 1 \end{array}] . \end{aligned}$

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$\begin{aligned} A & = [\begin{array}{rr} 0.95 & 0.25 \\ 0.25 & 0.95 \end{array}] \\ x & = [\begin{array}{r} 3 \\ 0 \end{array}] . \end{aligned}$

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

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If $A$ is a $2 \times 2$ matrix with eigenvalues $λ_{1} = 0.7$ and $λ_{2} = 0.5$ and $x$ is any vector, what happens to $A^{k} x$ when $k$ becomes very large?
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