ads The Diagonalization Process

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Section 12.4 The Diagonalization Process

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Subsection 12.4.1 Eigenvalues and Eigenvectors

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We now have the background to understand the main ideas behind the diagonalization process.

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Definition 12.4.1. Eigenvalue, Eigenvector.

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Let

A

be an

n \times n

matrix over

R .

λ

is an eigenvalue of

A

if for some nonzero column vector

x \in R^{n}

we have

A x = λ x .

x

is called an eigenvector corresponding to the eigenvalue

λ .

Example 12.4.2. Examples of eigenvalues and eigenvectors.

Find the eigenvalues and corresponding eigenvectors of the matrix

A = (\begin{array}{cc} 2 & 1 \\ 2 & 3 \end{array}) .

We want to find nonzero vectors

X = (\begin{matrix} x_{1} \\ x_{2} \end{matrix})

and real numbers

λ

such that

\begin{array}{r} (12.4.1) & \begin{aligned} A X = λ X & \Leftrightarrow (\begin{array}{cc} 2 & 1 \\ 2 & 3 \end{array}) (\begin{array}{c} x_{1} \\ x_{2} \end{array}) = λ (\begin{array}{c} x_{1} \\ x_{2} \end{array}) \\ \Leftrightarrow (\begin{array}{cc} 2 & 1 \\ 2 & 3 \end{array}) (\begin{array}{c} x_{1} \\ x_{2} \end{array}) - λ (\begin{array}{c} x_{1} \\ x_{2} \end{array}) = (\begin{array}{c} 0 \\ 0 \end{array}) \\ \Leftrightarrow (\begin{array}{cc} 2 & 1 \\ 2 & 3 \end{array}) (\begin{array}{c} x_{1} \\ x_{2} \end{array}) - λ (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}) (\begin{array}{c} x_{1} \\ x_{2} \end{array}) = (\begin{array}{c} 0 \\ 0 \end{array}) \\ \Leftrightarrow ((\begin{array}{cc} 2 & 1 \\ 2 & 3 \end{array}) - λ (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array})) (\begin{array}{c} x_{1} \\ x_{2} \end{array}) = (\begin{array}{c} 0 \\ 0 \end{array}) \\ \Leftrightarrow (\begin{array}{cc} 2 - λ & 1 \\ 2 & 3 - λ \end{array}) (\begin{array}{c} x_{1} \\ x_{2} \end{array}) = (\begin{array}{c} 0 \\ 0 \end{array}) \end{aligned} \end{array}

The last matrix equation will have nonzero solutions if and only if

det (\begin{array}{cc} 2 - λ & 1 \\ 2 & 3 - λ \end{array}) = 0

(2 - λ) (3 - λ) - 2 = 0,

which simplifies to

λ^{2} - 5 λ + 4 = 0 .

Therefore, the solutions to this quadratic equation,

λ_{1} = 1

and

λ_{2} = 4,

are the eigenvalues of

A .

We now have to find eigenvectors associated with each eigenvalue.

Case 1. For

λ_{1} = 1,

(12.4.1) becomes:

(\begin{array}{cc} 2 - 1 & 1 \\ 2 & 3 - 1 \end{array}) (\begin{matrix} x_{1} \\ x_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) (\begin{array}{cc} 1 & 1 \\ 2 & 2 \end{array}) (\begin{matrix} x_{1} \\ x_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})

which reduces to the single equation,

x_{1} + x_{2} = 0 .

From this,

x_{1} = - x_{2} .

This means the solution set of this equation is (in column notation)

E_{1} = {(\begin{matrix} - c \\ c \end{matrix}) | c \in R}

So any column vector of the form

(\begin{matrix} - c \\ c \end{matrix})

where

c

is any nonzero real number is an eigenvector associated with

λ_{1} = 1 .

The reader should verify that, for example,

(\begin{array}{cc} 2 & 1 \\ 2 & 3 \end{array}) (\begin{matrix} 1 \\ - 1 \end{matrix}) = 1 (\begin{matrix} 1 \\ - 1 \end{matrix})

so that

(\begin{matrix} 1 \\ - 1 \end{matrix})

is an eigenvector associated with eigenvalue 1.

Case 2. For

λ_{2} = 4

(12.4.1) becomes:

(\begin{array}{cc} 2 - 4 & 1 \\ 2 & 3 - 4 \end{array}) (\begin{matrix} x_{1} \\ x_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) (\begin{array}{cc} - 2 & 1 \\ 2 & - 1 \end{array}) (\begin{matrix} x_{1} \\ x_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})

which reduces to the single equation

- 2 x_{1} + x_{2} = 0,

so that

x_{2} = 2 x_{1} .

The solution set of the equation is

E_{2} = {(\begin{matrix} c \\ 2 c \end{matrix}) | c \in R}

Therefore, all eigenvectors of

A

associated with the eigenvalue

λ_{2} = 4

are of the form

(\begin{matrix} c \\ 2 c \end{matrix}),

where

c

can be any nonzero number.

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The following theorems summarize the most important aspects of the previous example.

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Theorem 12.4.3. Characterization of Eigenvalues of a Square Matrix.

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Let

A

be any

n \times n

matrix over

R .

Then

λ \in R

is an eigenvalue of

A

if and only if

det (A - λ I) = 0 .

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The equation

det (A - λ I) = 0

is called the characteristic equation, and the left side of this equation is called the characteristic polynomial of

A .

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Theorem 12.4.4. Linear Independence of Eigenvectors.

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Nonzero eigenvectors corresponding to distinct eigenvalues are linearly independent.

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The solution space of

(A - λ I) x = 0

is called the eigenspace of

A

corresponding to

λ .

This terminology is justified by Exercise 2 of this section.

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Subsection 12.4.2 Diagonalization

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We now consider the main aim of this section. Given an

n \times n

(square) matrix

A,

we would like to transform

A

into a diagonal matrix

D,

perform our tasks with the simpler matrix

D,

and then describe the results in terms of the given matrix

A .

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Definition 12.4.5. Diagonalizable Matrix.

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n \times n

matrix

A

is called diagonalizable if there exists an invertible

n \times n

matrix

P

such that

P^{- 1} A P

is a diagonal matrix

D .

The matrix

P

is said to diagonalize the matrix

A .

Example 12.4.6. Diagonalization of a Matrix.

We will now diagonalize the matrix

A

of Example 12.4.2. We form the matrix

P

as follows: Let

P^{(1)}

be the first column of

P .

Choose for

P^{(1)}

any eigenvector from

E_{1} .

We may as well choose a simple vector in

E_{1}

P^{(1)} = (\begin{matrix} 1 \\ - 1 \end{matrix})

is our candidate.

Similarly, let

P^{(2)}

be the second column of

P,

and choose for

P^{(2)}

any eigenvector from

E_{2} .

The vector

P^{(2)} = (\begin{matrix} 1 \\ 2 \end{matrix})

is a reasonable choice, thus

P = (\begin{array}{cc} 1 & 1 \\ - 1 & 2 \end{array}) and P^{- 1} = \frac{1}{3} (\begin{array}{cc} 2 & - 1 \\ 1 & 1 \end{array}) = (\begin{array}{cc} \frac{2}{3} & - \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} \end{array})

so that

P^{- 1} A P = \frac{1}{3} (\begin{array}{cc} 2 & - 1 \\ 1 & 1 \end{array}) (\begin{array}{cc} 2 & 1 \\ 2 & 3 \end{array}) (\begin{array}{cc} 1 & 1 \\ - 1 & 2 \end{array}) = (\begin{array}{cc} 1 & 0 \\ 0 & 4 \end{array})

Notice that the elements on the main diagonal of

D

are the eigenvalues of

A,

where

D_{i i}

is the eigenvalue corresponding to the eigenvector

P^{(i)}

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Note 12.4.7.

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The first step in the diagonalization process is the determination of the eigenvalues. The ordering of the eigenvalues is purely arbitrary. If we designate $λ_{1} = 4$ and $λ_{2} = 1,$ the columns of $P$ would be interchanged and $D$ would be $(\begin{array}{cc} 4 & 0 \\ 0 & 1 \end{array})$ (see Exercise 3b of this section). Nonetheless, the final outcome of the application to which we are applying the diagonalization process would be the same.
If $A$ is an $n \times n$ matrix with distinct eigenvalues, then $P$ is also an $n \times n$ matrix whose columns $P^{(1)},$ $P^{(2)}, \dots,$ $P^{(n)}$ are $n$ linearly independent vectors.

Example 12.4.8. Diagonalization of a 3 by 3 matrix.

Diagonalize the matrix

A = (\begin{array}{ccc} 1 & 12 & - 18 \\ 0 & - 11 & 18 \\ 0 & - 6 & 10 \end{array}) .

First, we find the eigenvalues of

A .

\begin{aligned} det (A - λ I) & = det (\begin{array}{ccc} 1 - λ & 12 & - 18 \\ 0 & - λ - 11 & 18 \\ 0 & - 6 & 10 - λ \end{array}) \\ = (1 - λ) det (\begin{array}{cc} - λ - 11 & 18 \\ - 6 & 10 - λ \end{array}) \\ = (1 - λ) ((- λ - 11) (10 - λ) + 108) = (1 - λ) (λ^{2} + λ - 2) \end{aligned}

Hence, the equation

det (A - λ I)

becomes

(1 - λ) (λ^{2} + λ - 2) = - (λ - 1)^{2} (λ + 2)

Therefore, our eigenvalues for

A

are

λ_{1} = - 2

and

λ_{2} = 1 .

We note that we do not have three distinct eigenvalues, but we proceed as in the previous example.

Case 1. For

λ_{1} = - 2

the equation

(A - λ I) x = 0

becomes

(\begin{array}{ccc} 3 & 12 & - 18 \\ 0 & - 9 & 18 \\ 0 & - 6 & 12 \end{array}) (\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})

We can row reduce the matrix of coefficients to

(\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & - 2 \\ 0 & 0 & 0 \end{array}) .

The matrix equation is then equivalent to the equations

x_{1} = - 2 x_{3} and x_{2} = 2 x_{3} .

Therefore, the solution set, or eigenspace, corresponding to

λ_{1} = - 2

consists of vectors of the form

(\begin{matrix} - 2 x_{3} \\ 2 x_{3} \\ x_{3} \end{matrix}) = x_{3} (\begin{matrix} - 2 \\ 2 \\ 1 \end{matrix})

Therefore

(\begin{matrix} - 2 \\ 2 \\ 1 \end{matrix})

is an eigenvector corresponding to the eigenvalue

λ_{1} = - 2,

and can be used for our first column of

P :

P = (\begin{array}{ccc} - 2 & ? & ? \\ 2 & ? & ? \\ 1 & ? & ? \end{array})

Before we continue we make the observation:

E_{1}

is a subspace of

R^{3}

with basis

{P^{(1)}}

and

\dim E_{1} = 1 .

Case 2. If

λ_{2} = 1,

then the equation

(A - λ I) x = 0

becomes

(\begin{array}{ccc} 0 & 12 & - 18 \\ 0 & - 12 & 18 \\ 0 & - 6 & 9 \end{array}) (\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})

Without the aid of any computer technology, it should be clear that all three equations that correspond to this matrix equation are equivalent to

2 x_{2} - 3 x_{3} = 0,

x_{2} = \frac{3}{2} x_{3} .

Notice that

x_{1}

can take on any value, so any vector of the form

(\begin{matrix} x_{1} \\ \frac{3}{2} x_{3} \\ x_{3} \end{matrix}) = x_{1} (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + x_{3} (\begin{matrix} 0 \\ \frac{3}{2} \\ 1 \end{matrix})

will solve the matrix equation.

We note that the solution set contains two independent variables,

x_{1}

and

x_{3} .

Further, note that we cannot express the eigenspace

E_{2}

as a linear combination of a single vector as in Case 1. However, it can be written as

E_{2} = {x_{1} (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + x_{3} (\begin{matrix} 0 \\ \frac{3}{2} \\ 1 \end{matrix}) ∣ x_{1}, x_{3} \in R} .

We can replace any vector in a basis is with a nonzero multiple of that vector. Simply for aesthetic reasons, we will multiply the second vector that generates

E_{2}

by 2. Therefore, the eigenspace

E_{2}

is a subspace of

R^{3}

with basis

{(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 3 \\ 2 \end{matrix})}

and so

\dim E_{2} = 2 .

What this means with respect to the diagonalization process is that

λ_{2} = 1

gives us both Column 2 and Column 3 the diagonalizing matrix. The order is not important so we have

P = (\begin{array}{ccc} - 2 & 1 & 0 \\ 2 & 0 & 3 \\ 1 & 0 & 2 \end{array})

The reader can verify (see Exercise 5 of this section) that

P^{- 1} = (\begin{array}{ccc} 0 & 2 & - 3 \\ 1 & 4 & - 6 \\ 0 & - 1 & 2 \end{array})

and

P^{- 1} A P = (\begin{array}{ccc} - 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array})

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In doing Example 12.4.8, the given

3 \times 3

matrix

A

produced only two, not three, distinct eigenvalues, yet we were still able to diagonalize

A .

The reason we were able to do so was because we were able to find three linearly independent eigenvectors. Again, the main idea is to produce a matrix

P

that does the diagonalizing. If

A

is an

n \times n

matrix,

P

will be an

n \times n

matrix, and its

n

columns must be linearly independent eigenvectors. The main question in the study of diagonalizability is “When can it be done?” This is summarized in the following theorem.

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Theorem 12.4.9. A condition for diagonalizability.

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Let

A

be an

n \times n

matrix. Then

A

is diagonalizable if and only if

A

has

n

linearly independent eigenvectors.

Proof.

Outline of a proof: (

⟸

) Assume that

A

has linearly independent eigenvectors,

P^{(1)}, P^{(2)}, \dots, P^{(n)},

with corresponding eigenvalues

λ_{1},

λ_{2}, \dots

λ_{n} .

We want to prove that

A

is diagonalizable. Column

i

of the

n \times n

matrix

A P

A P^{(i)}

(see Exercise 7 of this section). Then, since the

P^{(i)}

is an eigenvector of

A

associated with the eigenvalue

λ_{i}

we have

A P^{(i)} = λ_{i} P^{(i)}

for

i = 1, 2, \dots, n .

But this means that

A P = P D,

where

D

is the diagonal matrix with diagonal entries

λ_{1},

λ_{2}, \dots,

λ_{n} .

If we multiply both sides of the equation by

P^{- 1}

we get the desired

P^{- 1} A P = D .

(

⟹

) The proof in this direction involves a concept that is not covered in this text (rank of a matrix); so we refer the interested reader to virtually any linear algebra text for a proof.

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We now give an example of a matrix that is not diagonalizable.

Example 12.4.10. A Matrix that is Not Diagonalizable.

Let us attempt to diagonalize the matrix

A = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 1 & - 1 & 4 \end{array})

First, we determine the eigenvalues.

\begin{aligned} det (A - λ I) & = det (\begin{array}{ccc} 1 - λ & 0 & 0 \\ 0 & 2 - λ & 1 \\ 1 & - 1 & 4 - λ \end{array}) \\ = (1 - λ) det (\begin{array}{cc} 2 - λ & 1 \\ - 1 & 4 - λ \end{array}) \\ = (1 - λ) ((2 - λ) (4 - λ) + 1) \\ = (1 - λ) (λ^{2} - 6 λ + 9) \\ = (1 - λ) (λ - 3)^{2} \end{aligned}

Therefore there are two eigenvalues,

λ_{1} = 1

and

λ_{2} = 3 .

Since

λ_{1}

is an eigenvalue of degree one, it will have an eigenspace of dimension 1. Since

λ_{2}

is a double root of the characteristic equation, the dimension of its eigenspace must be 2 in order to be able to diagonalize.

Case 1. For

λ_{1} = 1,

the equation

(A - λ I) x = 0

becomes

(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & - 1 & 3 \end{array}) (\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})

Row reduction of this system reveals one free variable and eigenspace

(\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} - 4 x_{3} \\ - x_{3} \\ x_{3} \end{matrix}) = x_{3} (\begin{matrix} - 4 \\ - 1 \\ 1 \end{matrix})

Hence,

{(\begin{matrix} - 4 \\ - 1 \\ 1 \end{matrix})}

is a basis for the eigenspace of

λ_{1} = 1 .

Case 2. For

λ_{2} = 3,

the equation

(A - λ I) x = 0

becomes

(\begin{array}{ccc} - 2 & 0 & 0 \\ 0 & - 1 & 1 \\ 1 & - 1 & 1 \end{array}) (\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})

Once again there is only one free variable in the row reduction and so the dimension of the eigenspace will be one:

(\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} 0 \\ x_{3} \\ x_{3} \end{matrix}) = x_{3} (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix})

Hence,

{(\begin{matrix} 0 \\ 1 \\ 1 \end{matrix})}

is a basis for the eigenspace of

λ_{2} = 3 .

This means that

λ_{2} = 3

produces only one column for

P .

Since we began with only two eigenvalues, we had hoped that

λ_{2} = 3

would produce a vector space of dimension two, or, in matrix terms, two linearly independent columns for

P .

Since

A

does not have three linearly independent eigenvectors

A

cannot be diagonalized.

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Subsection 12.4.3 SageMath Note - Diagonalization

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We demonstrate how diagonalization can be done in Sage. We start by defining the matrix to be diagonalized, and also declare

D

and

P

to be variables.

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var (' D, P')
A = Matrix (QQ, [[4, 1, 0], [1, 5, 1], [0, 1, 4]]);A

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We have been working with “right eigenvectors” since the

x

A x = λ x

is a column vector. It’s not so common but still desirable in some situations to consider “left eigenvectors,” so SageMath allows either one. The right_eigenmatrix method returns a pair of matrices. The diagonal matrix,

D,

with eigenvalues and the diagonalizing matrix,

P,

which is made up of columns that are eigenvectors corresponding to the eigenvectors of

D .

xxxxxxxxxx
 
(D,P)=A.right_eigenmatrix();(D,P)

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We should note here that

P

is not unique because even if an eigenspace has dimension one, any nonzero vector in that space will serve as an eigenvector. For that reason, the

P

generated by Sage isn’t necessarily the same as the one computed by any other computer algebra system such as Mathematica. Here we verify the result for our Sage calculation. Recall that an asterisk is used for matrix multiplication in Sage.

xxxxxxxxxx
 
P.inverse()*A*P

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Here is a second matrix to diagonalize.

xxxxxxxxxx
 
A2=Matrix(QQ,[[8,1,0],[1,5,1],[0,1,7]]);A2

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Here we’ve already specified that the underlying system is the rational numbers. Since the eigenvalues are not rational, Sage will revert to approximate number by default. We’ll just pull out the matrix of eigenvectors this time and display rounded entries.

xxxxxxxxxx
 
P=A2.right_eigenmatrix()[1]
P.numerical_approx(digits=3)
print('------------------')
D=(P.inverse()*A2*P);D.numerical_approx(digits=3)

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Finally, we examine how Sage reacts to the matrix from Example 12.4.10 that couldn’t be diagonalized. Notice that the last column is a zero column, indicating the absence of one needed eigenvector.

xxxxxxxxxx
 
A3=Matrix(QQ,[[1, 0, 0],[0,2,1],[1,-1,4]])
(D,P)=A3.right_eigenmatrix();(D,P)

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Exercises 12.4.4 Exercises

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1.

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List three different eigenvectors of $A = (\begin{array}{cc} 2 & 1 \\ 2 & 3 \end{array}),$ the matrix of Example 12.4.2, associated with each of the two eigenvalues 1 and 4. Verify your results.
Choose one of the three eigenvectors corresponding to 1 and one of the three eigenvectors corresponding to 4, and show that the two chosen vectors are linearly independent.

Answer.

Any nonzero multiple of $(\begin{matrix} 1 \\ - 1 \end{matrix})$ is an eigenvector associated with $λ = 1 .$
Any nonzero multiple of $(\begin{matrix} 1 \\ 2 \end{matrix})$ is an eigenvector associated with $λ = 4 .$
Let $x_{1} = (\begin{matrix} a \\ - a \end{matrix})$ and $x_{2} = (\begin{matrix} b \\ 2 b \end{matrix})$ . You can verify that $c_{1} x_{1} + c_{2} x_{2} = (\begin{matrix} 0 \\ 0 \end{matrix})$ if and only if $c_{1} = c_{2} = 0.$ Therefore, ${x_{1}, x_{2}}$ is linearly independent.

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2.

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Verify that $E_{1}$ and $E_{2}$ in Example 12.4.2 are vector spaces over $R .$ Since they are also subsets of $R^{2},$ they are called subvector-spaces, or subspaces for short, of $R^{2} .$ Since these are subspaces consisting of eigenvectors, they are called eigenspaces.
Use the definition of dimension in the previous section to find $\dim E_{1}$ and $\dim E_{2}$ . Note that $\dim E_{1} + \dim E_{2} = \dim R^{2} .$ This is not a coincidence.

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3.

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Verify that $P^{- 1} A P$ is indeed equal to $(\begin{array}{cc} 1 & 0 \\ 0 & 4 \end{array}),$ as indicated in Example 12.4.6.
Choose $P^{(1)} = (\begin{matrix} 1 \\ 2 \end{matrix})$ and $P^{(2)} = (\begin{matrix} 1 \\ - 1 \end{matrix})$ and verify that the new value of $P$ satisfies $P^{- 1} A P = (\begin{array}{cc} 4 & 0 \\ 0 & 1 \end{array}) .$
Take two different (from the previous part) linearly independent eigenvectors of the matrix $A$ of Example 12.4.6 and verify that $P^{- 1} A P$ is a diagonal matrix.

Answer.

Part c: You should obtain

(\begin{array}{cc} 4 & 0 \\ 0 & 1 \end{array})

(\begin{array}{cc} 1 & 0 \\ 0 & 4 \end{array}),

depending on how you order the eigenvalues.

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4.

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Let $A$ be the matrix in Example 12.4.8 and $P = (\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & 2 \end{array}) .$ Without doing any actual matrix multiplications, determine the value of $P^{- 1} A P$
If you choose the columns of $P$ in the reverse order, what is $P^{- 1} A P ?$

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5.

🔗

Diagonalize the following, if possible:

$(\begin{array}{cc} 1 & 2 \\ 3 & 2 \end{array})$
$(\begin{array}{cc} - 2 & 1 \\ - 7 & 6 \end{array})$
$(\begin{array}{cc} 3 & 0 \\ 0 & 4 \end{array})$
$(\begin{array}{ccc} 1 & - 1 & 4 \\ 3 & 2 & - 1 \\ 2 & 1 & - 1 \end{array})$
$(\begin{array}{ccc} 6 & 0 & 0 \\ 0 & 7 & - 4 \\ 9 & 1 & 3 \end{array})$
$(\begin{array}{ccc} 1 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 1 \end{array})$

Answer.

If $P = (\begin{array}{cc} 2 & 1 \\ 3 & - 1 \end{array}),$ then $P^{- 1} A P = (\begin{array}{cc} 4 & 0 \\ 0 & - 1 \end{array}) .$
If $P = (\begin{array}{cc} 1 & 1 \\ 7 & 1 \end{array}),$ then $P^{- 1} A P = (\begin{array}{cc} 5 & 0 \\ 0 & - 1 \end{array}) .$
If $P = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}),$ then $P^{- 1} A P = (\begin{array}{cc} 3 & 0 \\ 0 & 4 \end{array}) .$
If $P = (\begin{array}{ccc} 1 & - 1 & 1 \\ - 1 & 4 & 2 \\ - 1 & 1 & 1 \end{array}),$ then $P^{- 1} A P = (\begin{array}{ccc} - 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}) .$
$A$ is not diagonalizable. Five is a double root of the characteristic equation, but has an eigenspace with dimension only 1.
If $P = (\begin{array}{ccc} 1 & 1 & 1 \\ - 2 & 0 & 1 \\ 1 & - 1 & 1 \end{array}),$ then $P^{- 1} A P = (\begin{array}{ccc} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}) .$

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6.

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Diagonalize the following, if possible:

$(\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array})$
$(\begin{array}{cc} 2 & 1 \\ 4 & 2 \end{array})$
$(\begin{array}{cc} 2 & - 1 \\ 1 & 0 \end{array})$
$(\begin{array}{ccc} 1 & 3 & 6 \\ - 3 & - 5 & - 6 \\ 3 & 3 & 6 \end{array})$
$(\begin{array}{ccc} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{array})$
$(\begin{array}{ccc} 2 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 2 \end{array})$

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7.

🔗

Let

A

and

P

be as in Example 12.4.8. Show that the columns of the matrix

A P

can be found by computing

A P^{(1)},

A P^{(2)}, \dots,

A P^{(n)} .

Answer.

This is a direct application of the definition of matrix multiplication. Let

A_{(i)}

be the

i^{th}

row of

A,

and let

P^{(j)}

be the

j^{th}

column of

P .

Then the

j^{th}

column of the product

A P

(\begin{matrix} A_{(1)} P^{(j)} \\ A_{(2)} P^{(j)} \\ ⋮ \\ A_{(n)} P^{(j)} \end{matrix})

Hence,

(A P)^{(j)} = A (P^{(j)})

for

j = 1, 2, \dots, n .

Thus, each column of

A P

depends on

A

and the

j^{th}

column of

P .

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8.

🔗

Prove that if

P

is an

n \times n

matrix and

D

is a diagonal matrix with diagonal entries

d_{1},

d_{2}, \dots,

d_{n},

then

P D

is the matrix obtained from

P,

by multiplying column

i

P

d_{i},

i = 1, 2, \dots, n .

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