Chapter 6 Eigenvalues and Eigenvectors - 6.1 Eigenvalues...

1 Introduction

2 EigenvectorsAlgebraically

3 FindingEigenvalues

4 FindingEigenspaces

5 Putting itTogether

Chapter 6 Eigenvalues and Eigenvectors6.1 Eigenvalues and Eigenvectors

1 Introduction

What are eigenvalues and eigenvalues?

DefinitionGiven an n × n matrix A, a nonzero vector u in Rn is calledan eigenvector of A if there exists a scalar λ such thatAu = λu. The scalar λ is called an eigenvalue of A.

Note: λ is allowed to be 0.

1 Introduction

What are eigenvalues and eigenvalues?

DefinitionGiven an n × n matrix A, a nonzero vector u in Rn is calledan eigenvector of A if there exists a scalar λ such thatAu = λu. The scalar λ is called an eigenvalue of A.

Note: λ is allowed to be 0.

1 Introduction

Why are eigenvalues and eigenvectors important?Tacoma Narrows Bridge, 1940

Does anyone know what the natural frequency of an objectlike a bridge is?

DefinitionThe natural frequency is the frequency at which a systemnaturally vibrates once it has been set into motion. That is,it is motion a structure takes on in response to wind or beingwalked on.

The reason this bridge collapsed is because the naturalfrequency of the bridge was too close to the naturalfrequency of the wind. When frequencies match, thecompound provided too strong a force for the bridge.

Mathematically, the natural frequency can be characterizedby the eigenvalue of smallest magnitude.

1 Introduction

Why are eigenvalues and eigenvectors important?

Linear transformations: Eigenvectors make understandinglinear transformations easy. They are the “axes” (directions)along which a linear transformation acts simply bystretching/compressing and/or reflecting.

Eigenvalues give you the factors by which this compressionoccurs.

The more directions you have along which you understandthe behavior of a linear transformation, the easier it is tounderstand the linear transformation; so you want to have asmany linearly independent eigenvectors as possibleassociated to a single linear transformation.

1 Introduction

Eigenvalues tell us a whole lot about a matrix:

1. If A is singular then λ = 0

2. If A is symmetric, then all eigenvalues are real

3. If A has rank n then the eigenvectors form a basis for Rn

4. The eigenvectors of AAT form a basis for col(A)

5. The eigenvectors of ATA form a basis for row(A)

1 Introduction

How these work

Example

Verify that u is an eigenvector for the matrix A anddetermine the associated eigenvalue.

[3 11 3

Therefore, we have shown u is an eigenvector of A andλ = 4 is the associated eigenvalue.

1 Introduction

How these work

Example

[3 11 3

1 Introduction

How these work

Example

[3 11 3

1 Introduction

How these work

Example

[3 11 3

1 Introduction

Visualization

1 Introduction

Visualization

1 Introduction

Another Example

Example

10 12 −68 11 −4

34 44 −19

10 12 −68 11 −4

34 44 −19

1 Introduction

Another Example

Example

10 12 −68 11 −4

34 44 −19

10 12 −68 11 −4

34 44 −19

1 Introduction

Another Example

Example

10 12 −68 11 −4

34 44 −19

10 12 −68 11 −4

34 44 −19

1 Introduction

Another Example

Example

10 12 −68 11 −4

34 44 −19

10 12 −68 11 −4

34 44 −19

1 Introduction

How we find eigenvalues

Example

Find the characteristic polynomial and eigenvalues for A.

[3 11 3

The characteristic polynomial is found by findingdet(A− λIn), so in this case we need det(A− λI2),

det(A− λI2) =

∣∣∣∣ 3 − λ 11 3 − λ

∣∣∣∣= (3 − λ)2 − 1

= 9 − 6λ+ λ2 − 1

= λ2 − 6λ+ 8

1 Introduction

Example

[3 11 3

det(A− λI2) =

∣∣∣∣ 3 − λ 11 3 − λ

∣∣∣∣= (3 − λ)2 − 1

= 9 − 6λ+ λ2 − 1

= λ2 − 6λ+ 8

1 Introduction

Example

[3 11 3

det(A− λI2) =

∣∣∣∣ 3 − λ 11 3 − λ

∣∣∣∣

= (3 − λ)2 − 1

= 9 − 6λ+ λ2 − 1

= λ2 − 6λ+ 8

1 Introduction

Example

[3 11 3

det(A− λI2) =

∣∣∣∣ 3 − λ 11 3 − λ

∣∣∣∣= (3 − λ)2 − 1

= 9 − 6λ+ λ2 − 1

= λ2 − 6λ+ 8

1 Introduction

Example

[3 11 3

det(A− λI2) =

∣∣∣∣ 3 − λ 11 3 − λ

∣∣∣∣= (3 − λ)2 − 1

= 9 − 6λ+ λ2 − 1

= λ2 − 6λ+ 8

1 Introduction

Example

[3 11 3

det(A− λI2) =

∣∣∣∣ 3 − λ 11 3 − λ

∣∣∣∣= (3 − λ)2 − 1

= 9 − 6λ+ λ2 − 1

= λ2 − 6λ+ 8

1 Introduction

So, our characteristic polynomial is λ2 − 6λ+ 8.

The eigenvalues are the values of λ such thatdet(A− λIn) = 0.

So, here we have

λ2 − 6λ+ 8 = 0

(λ− 2)(λ− 4) = 0

λ = 2, 4

So, the eigenvalues associated with

[3 11 3

]are λ = 2 and

λ = 4.

1 Introduction

So, here we have

λ2 − 6λ+ 8 = 0

(λ− 2)(λ− 4) = 0

λ = 2, 4

[3 11 3

]are λ = 2 and

λ = 4.

1 Introduction

So, here we have

λ2 − 6λ+ 8 = 0

(λ− 2)(λ− 4) = 0

λ = 2, 4

[3 11 3

]are λ = 2 and

λ = 4.

1 Introduction

So, here we have

λ2 − 6λ+ 8 = 0

(λ− 2)(λ− 4) = 0

λ = 2, 4

[3 11 3

]are λ = 2 and

λ = 4.

1 Introduction

So, here we have

λ2 − 6λ+ 8 = 0

(λ− 2)(λ− 4) = 0

λ = 2, 4

[3 11 3

]are λ = 2 and

λ = 4.

1 Introduction

So, here we have

λ2 − 6λ+ 8 = 0

(λ− 2)(λ− 4) = 0

λ = 2, 4

[3 11 3

]are λ = 2 and

λ = 4.

1 Introduction

Another Example

Example

[−7 10−5 8

det(A− λI2) =

∣∣∣∣ −7 − λ 10-5 8 − λ

∣∣∣∣= (−7 − λ)(8 − λ) + 50

= −56 − λ+ λ2 + 50

= λ2 − λ− 6

1 Introduction

Another Example

Example

[−7 10−5 8

det(A− λI2) =

∣∣∣∣ −7 − λ 10-5 8 − λ

∣∣∣∣

= (−7 − λ)(8 − λ) + 50

= −56 − λ+ λ2 + 50

= λ2 − λ− 6

1 Introduction

Another Example

Example

[−7 10−5 8

det(A− λI2) =

∣∣∣∣ −7 − λ 10-5 8 − λ

∣∣∣∣= (−7 − λ)(8 − λ) + 50

= −56 − λ+ λ2 + 50

= λ2 − λ− 6

1 Introduction

Another Example

Example

[−7 10−5 8

det(A− λI2) =

∣∣∣∣ −7 − λ 10-5 8 − λ

∣∣∣∣= (−7 − λ)(8 − λ) + 50

= −56 − λ+ λ2 + 50

= λ2 − λ− 6

1 Introduction

Another Example

Example

[−7 10−5 8

det(A− λI2) =

∣∣∣∣ −7 − λ 10-5 8 − λ

∣∣∣∣= (−7 − λ)(8 − λ) + 50

= −56 − λ+ λ2 + 50

= λ2 − λ− 6

1 Introduction

Another Example

λ2 − λ− 6 = 0

(λ− 3)(λ+ 2) = 0

λ = 3,−2

Therefore, the eigenvalues associated with

[−7 10−5 8

λ = 3 and λ = −2.

1 Introduction

Another Example

λ2 − λ− 6 = 0

(λ− 3)(λ+ 2) = 0

λ = 3,−2

[−7 10−5 8

λ = 3 and λ = −2.

1 Introduction

Another Example

λ2 − λ− 6 = 0

(λ− 3)(λ+ 2) = 0

λ = 3,−2

[−7 10−5 8

λ = 3 and λ = −2.

1 Introduction

Another Example

λ2 − λ− 6 = 0

(λ− 3)(λ+ 2) = 0

λ = 3,−2

[−7 10−5 8

λ = 3 and λ = −2.

1 Introduction

One More

Example

1 0 −20 1 22 2 1

1 Introduction

One More

det(A− λI3) =

∣∣∣∣∣∣1 − λ 0 -2

0 1 − λ 22 2 1 − λ

∣∣∣∣∣∣

= (1 − λ)

∣∣∣∣ 1 − λ 22 1 − λ

∣∣∣∣− 2

∣∣∣∣ 0 1 − λ2 2

∣∣∣∣= (1 − λ)((1 − λ)2 − 4) − 2(0 − 2(1 − λ))

= (1 − λ)(λ2 − 2λ+ 3) + 4(1 − λ)

= −λ3 + 3λ2 − 3λ+ 1

1 Introduction

One More

det(A− λI3) =

∣∣∣∣∣∣1 − λ 0 -2

0 1 − λ 22 2 1 − λ

∣∣∣∣∣∣= (1 − λ)

∣∣∣∣ 1 − λ 22 1 − λ

∣∣∣∣− 2

∣∣∣∣ 0 1 − λ2 2

∣∣∣∣

= (1 − λ)((1 − λ)2 − 4) − 2(0 − 2(1 − λ))

= (1 − λ)(λ2 − 2λ+ 3) + 4(1 − λ)

= −λ3 + 3λ2 − 3λ+ 1

1 Introduction

One More

det(A− λI3) =

∣∣∣∣∣∣1 − λ 0 -2

0 1 − λ 22 2 1 − λ

∣∣∣∣∣∣= (1 − λ)

∣∣∣∣ 1 − λ 22 1 − λ

∣∣∣∣− 2

∣∣∣∣ 0 1 − λ2 2

∣∣∣∣= (1 − λ)((1 − λ)2 − 4) − 2(0 − 2(1 − λ))

= (1 − λ)(λ2 − 2λ+ 3) + 4(1 − λ)

= −λ3 + 3λ2 − 3λ+ 1

1 Introduction

One More

det(A− λI3) =

∣∣∣∣∣∣1 − λ 0 -2

0 1 − λ 22 2 1 − λ

∣∣∣∣∣∣= (1 − λ)

∣∣∣∣ 1 − λ 22 1 − λ

∣∣∣∣− 2

∣∣∣∣ 0 1 − λ2 2

∣∣∣∣= (1 − λ)((1 − λ)2 − 4) − 2(0 − 2(1 − λ))

= (1 − λ)(λ2 − 2λ+ 3) + 4(1 − λ)

= −λ3 + 3λ2 − 3λ+ 1

1 Introduction

One More

det(A− λI3) =

∣∣∣∣∣∣1 − λ 0 -2

0 1 − λ 22 2 1 − λ

∣∣∣∣∣∣= (1 − λ)

∣∣∣∣ 1 − λ 22 1 − λ

∣∣∣∣− 2

∣∣∣∣ 0 1 − λ2 2

∣∣∣∣= (1 − λ)((1 − λ)2 − 4) − 2(0 − 2(1 − λ))

= (1 − λ)(λ2 − 2λ+ 3) + 4(1 − λ)

= −λ3 + 3λ2 − 3λ+ 1

1 Introduction

One More

−λ3 + 3λ2 − 3λ+ 1 = 0

(1 − λ)3 = 0

λ = 1

So, the only eigenvalue associated with

1 −0 −20 1 22 2 1

λ = 1.

1 Introduction

One More

−λ3 + 3λ2 − 3λ+ 1 = 0

(1 − λ)3 = 0

λ = 1

1 −0 −20 1 22 2 1

λ = 1.

1 Introduction

One More

−λ3 + 3λ2 − 3λ+ 1 = 0

(1 − λ)3 = 0

λ = 1

1 −0 −20 1 22 2 1

λ = 1.

1 Introduction

One More

−λ3 + 3λ2 − 3λ+ 1 = 0

(1 − λ)3 = 0

λ = 1

1 −0 −20 1 22 2 1

λ = 1.

1 Introduction

Finding the eigenvectors

DefinitionIf λ is an eigenvalue for an n× n matrix A, the eigenspace ofλ is the solution set to the homogeneous linear system(A− λIn)x = 0, that is to say, the subspace ker(A− λIn).

So, to find the vectors we use the matrix A− λIn and rowreduce to find the form of the infinite number of solutions.The vector can be represented as a basis for the eigenspaceassociated with the given eigenvalue.

1 Introduction

Finding the eigenvectors

DefinitionIf λ is an eigenvalue for an n× n matrix A, the eigenspace ofλ is the solution set to the homogeneous linear system(A− λIn)x = 0, that is to say, the subspace ker(A− λIn).

So, to find the vectors we use the matrix A− λIn and rowreduce to find the form of the infinite number of solutions.The vector can be represented as a basis for the eigenspaceassociated with the given eigenvalue.

1 Introduction

Finding the bases for the eigenspaces

Example

Find a basis for the eigenspace of A associated with eacheigenvalue.

[3 11 3

From earlier, we found that the eigenvalues were λ = 2 andλ = 4.

1 Introduction

Finding the bases for the eigenspaces

Example

Find a basis for the eigenspace of A associated with eacheigenvalue.

[3 11 3

From earlier, we found that the eigenvalues were λ = 2 andλ = 4.

1 Introduction

Finding the bases for the eigenspacesλ = 2: Remember, we are solving the homogeneous system...

A− 2I2 =

[3 − λ 1

1 3 − λ

[1 11 1

1 10 0

Solving this system, we get x = s

[1−1

]So, a basis for the eigenspace associated with λ = 2 is

{[1−1

1 Introduction

A− 2I2 =

[3 − λ 1

1 3 − λ

[1 11 1

1 10 0

[1−1

{[1−1

1 Introduction

A− 2I2 =

[3 − λ 1

1 3 − λ

[1 11 1

1 10 0

[1−1

{[1−1

1 Introduction

A− 2I2 =

[3 − λ 1

1 3 − λ

[1 11 1

1 10 0

[1−1

So, a basis for the eigenspace associated with λ = 2 is

{[1−1

1 Introduction

A− 2I2 =

[3 − λ 1

1 3 − λ

[1 11 1

1 10 0

[1−1

{[1−1

1 Introduction

Find the bases for the eigenspaces

λ = 4:

A− 4I2 =

[3 − λ 1

1 3 − λ

[−1 11 −1

1 −10 0

1 Introduction

λ = 4:

A− 4I2 =

[3 − λ 1

1 3 − λ

[−1 11 −1

1 −10 0

1 Introduction

λ = 4:

A− 4I2 =

[3 − λ 1

1 3 − λ

[−1 11 −1

1 −10 0

1 Introduction

λ = 4:

A− 4I2 =

[3 − λ 1

1 3 − λ

[−1 11 −1

1 −10 0

So, a basis for the eigenspace associated with λ = 4 is

1 Introduction

λ = 4:

A− 4I2 =

[3 − λ 1

1 3 − λ

[−1 11 −1

1 −10 0

1 Introduction

The Big Theorem - Version 8

TheoremLet A = {a1, . . . , an} be a set of n vectors in Rn, letA = [a1 a2 . . . an] and let T : Rn → Rn be given by T (x) = Ax.Then the following are equivalent:

(a) A spans Rn

(b) A is linearly independent

(c) Ax = b has a unique solution ∀b ∈ Rn

(d) T is onto

(e) T is 1-1

(f) A is invertible

(g) ker(T ) = {0}(h) A is a basis for Rn

(i) col(A) = Rn

(j) row(A) = Rn

(k) rank(A) = n

(l) det(A) 6= 0

(m) λ = 0 is not an eigenvalue of A

1 Introduction

(a) A spans Rn

(d) T is onto

(e) T is 1-1

(f) A is invertible

(i) col(A) = Rn

(j) row(A) = Rn

(k) rank(A) = n

(l) det(A) 6= 0

1 Introduction

(a) A spans Rn

(d) T is onto

(e) T is 1-1

(f) A is invertible

(i) col(A) = Rn

(j) row(A) = Rn

(k) rank(A) = n

(l) det(A) 6= 0

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