Lesson14: Eigenvalues And Eigenvectors

Lesson 14Eigenvalues and Eigenvectors

Math 20

October 22, 2007

Announcements

I Midterm almost done

I Problem Set 5 is on the WS. Due October 24

I OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323)

I Prob. Sess.: Sundays 6–7 (SC B-10), Tuesdays 1–2 (SC 116)

Math 20 - October 22, 2007.GWBMonday, Oct 22, 2007

Page1of16

Geometric effect of a diagonal linear transformation

Example

Let A =

(2 00 −1

). Draw the effect of the linear transformation

which is multiplication by A.

x

y

e1 Ae1

e2

Ae2

v

Av


Example

Let A =

(2 00 −1



x

y

e1 Ae1

e2

Ae2

v

Av


Example

Let A =

(2 00 −1



x

y

e1

Ae1

e2

Ae2

v

Av


Page2of16


Example

Let A =

(2 00 −1



x

y

e1 Ae1

e2

Ae2

v

Av


Example

Let A =

(2 00 −1



x

y

e1 Ae1

e2

Ae2

v

Av


Example

Let A =

(2 00 −1



x

y

e1 Ae1

e2

Ae2

v

Av


Page4of16


Example

Let A =

(2 00 −1



x

y

e1 Ae1

e2

Ae2

v

Av


Page4of16


Page5of16


Example

Let A =

(2 00 −1



x

y

e1 Ae1

e2

Ae2

v

Av


Example

Let A =

(2 00 −1



x

y

e1 Ae1

e2

Ae2

v

Av


Example

Let A =

(2 00 −1



x

y

e1 Ae1

e2

Ae2

v

Av


Example

Let A =

(2 00 −1



x

y

e1 Ae1

e2

Ae2

v

Av

Example

Draw the effect of the linear transformation which is multiplicationby A2.

x

y

Example


x

y

Example


x

y

Geometric effect of a non-diagonal linear transformation

Example

Let B =

(1/2 3/23/2 1/2


which is multiplication by B.

x

y

e1

Be1e2

Be2

v

Bv


Example

Let B =

(1/2 3/23/2 1/2



x

y

e1

Be1e2

Be2

v

Bv


Example

Let B =

(1/2 3/23/2 1/2



x

y

e1

Be1

e2

Be2

v

Bv


Example

Let B =

(1/2 3/23/2 1/2



x

y

e1

Be1e2

Be2

v

Bv


Example

Let B =

(1/2 3/23/2 1/2



x

y

e1

Be1e2

Be2

v

Bv


Example

Let B =

(1/2 3/23/2 1/2



x

y

e1

Be1e2

Be2

v

Bv


Example

Let B =

(1/2 3/23/2 1/2



x

y

e1

Be1e2

Be2

v

Bv


Page6of16


Example

Let B =

(1/2 3/23/2 1/2



x

y

e1

Be1e2

Be2

v

Bv


Example

Let B =

(1/2 3/23/2 1/2



x

y

e1

Be1e2

Be2

v

Bv


Example

Let B =

(1/2 3/23/2 1/2



x

y

e1

Be1e2

Be2

v

Bv


Example

Let B =

(1/2 3/23/2 1/2



x

y

v1

Bv1

v2

Bv2

2e2

2Be2


Page8of16


Example

Let B =

(1/2 3/23/2 1/2



x

y

v1

Bv1

v2

Bv2

2e2

2Be2


Example

Let B =

(1/2 3/23/2 1/2



x

y

v1

Bv1

v2

Bv2

2e2

2Be2


Example

Let B =

(1/2 3/23/2 1/2



x

y

v1

Bv1

v2

Bv2

2e2

2Be2


Page9of16


Example

Let B =

(1/2 3/23/2 1/2



x

y

v1

Bv1

v2

Bv2

2e2

2Be2


Example

Let B =

(1/2 3/23/2 1/2



x

y

v1

Bv1

v2

Bv2

2e2

2Be2


Example

Let B =

(1/2 3/23/2 1/2



x

y

v1

Bv1

v2

Bv2

2e2

2Be2


Example

Let B =

(1/2 3/23/2 1/2



x

y

v1

Bv1

v2

Bv2

2e2

2Be2


Example

Let B =

(1/2 3/23/2 1/2



x

y

v1

Bv1

v2

Bv2

2e2

2Be2


Example

Let B =

(1/2 3/23/2 1/2



x

y

v1

Bv1

v2

Bv2

2e2

2Be2


Example

Let B =

(1/2 3/23/2 1/2



x

y

v1

Bv1

v2

Bv2

2e2

2Be2


Example

Let B =

(1/2 3/23/2 1/2



x

y

v1

Bv1

v2

Bv2

2e2

2Be2

The big concept

DefinitionLet A be an n × n matrix. The number λ is called an eigenvalueof A if there exists a nonzero vector x ∈ Rn such that

Ax = λx. (1)

Every nonzero vector satisfying (1) is called an eigenvector of Aassociated with the eigenvalue λ.


Page10of16

Finding the eigenvector(s) corresponding to an eigenvalue

Example (Worksheet Problem 1)

Let A =

(0 −2−3 1

). The number 3 is an eigenvalue for A. Find

an eigenvector corresponding to this eigenvalue.

SolutionWe seek x such that

Ax = 3x =⇒ (A− 3I)x = 0.

[A− 3I 0

]=

[−3 −2 0−3 −2 0

]

[1 2/3 00 0 0

]So

(−2/3

1

), or

(−23

), are possible eigenvectors.


Page11of16



Let A =

(0 −2−3 1




Ax = 3x =⇒ (A− 3I)x = 0.

[A− 3I 0

]=

[−3 −2 0−3 −2 0

]

[1 2/3 00 0 0

]So

(−2/3

1

), or

(−23




Let A =

(0 −2−3 1




Ax = 3x =⇒ (A− 3I)x = 0.

[A− 3I 0

]=

[−3 −2 0−3 −2 0

]

[1 2/3 00 0 0

]

So

(−2/3

1

), or

(−23



Page12of16



Let A =

(0 −2−3 1




Ax = 3x =⇒ (A− 3I)x = 0.

[A− 3I 0

]=

[−3 −2 0−3 −2 0

]

[1 2/3 00 0 0

]So

(−2/3

1

), or

(−23



Page13of16


The number −2 is also an eigenvalue for A. Find an eigenvector.

SolutionWe have

A + 2I =

(2 −2−3 3

)

(1 −10 0

).

So

(11

)is an eigenvector for the eigenvalue −2.


The number −2 is also an eigenvalue for A. Find an eigenvector.

SolutionWe have

A + 2I =

(2 −2−3 3

)

(1 −10 0

).

So

(11

)is an eigenvector for the eigenvalue −2.

Summary

To find the eigenvector(s) of a matrix corresponding to aneigenvalue λ, do Gaussian Elimination on A− λI.

Find the eigenvalues of a matrix


Determine the eigenvalues of

A =

(−4 −33 6

).

SolutionAx = λx has a nonzero solution if and only if (A− λI)x = 0 has anonzero solution, if and only if A− λI is not invertible. Whatmagic number determines whether a matrix is invertible? Thedeterminant! So to find the possible values λ for which this couldbe true, we need to find the determinant of A− λI.




A =

(−4 −33 6

).

SolutionAx = λx has a nonzero solution if and only if (A− λI)x = 0 has anonzero solution, if and only if A− λI is not invertible.

Whatmagic number determines whether a matrix is invertible? Thedeterminant! So to find the possible values λ for which this couldbe true, we need to find the determinant of A− λI.


Page14of16


Page15of16




A =

(−4 −33 6

).

SolutionAx = λx has a nonzero solution if and only if (A− λI)x = 0 has anonzero solution, if and only if A− λI is not invertible. Whatmagic number determines whether a matrix is invertible?

Thedeterminant! So to find the possible values λ for which this couldbe true, we need to find the determinant of A− λI.




A =

(−4 −33 6

).

SolutionAx = λx has a nonzero solution if and only if (A− λI)x = 0 has anonzero solution, if and only if A− λI is not invertible. Whatmagic number determines whether a matrix is invertible? Thedeterminant!

So to find the possible values λ for which this couldbe true, we need to find the determinant of A− λI.




A =

(−4 −33 6

).

SolutionAx = λx has a nonzero solution if and only if (A− λI)x = 0 has anonzero solution, if and only if A− λI is not invertible. Whatmagic number determines whether a matrix is invertible? Thedeterminant! So to find the possible values λ for which this couldbe true, we need to find the determinant of A− λI.

det(A− λI) =

∣∣∣∣−4− λ −33 6− λ

∣∣∣∣= (−4− λ)(6− λ)− (−3)(3)

= (−24− 2λ+ λ2) + 9 = λ2 − 2λ− 15

= (λ+ 3)(λ− 5)

So λ = −3 and λ = 5 are the eigenvalues for A.

We can find the eigenvectors now, based on what we did before.We have

A + 3I =

(−1 −33 9

)

(1 30 0

)So

(−31

)is an eigenvector for this eigenvalue. Also,

A− 5I =

(−9 −33 1

)

(1 1/30 0

)

So

(−1/3

1

)or

(−13

)would be an eigenvector for this eigenvalue.


A + 3I =

(−1 −33 9

)

(1 30 0

)So

(−31


A− 5I =

(−9 −33 1

)

(1 1/30 0

)

So

(−1/3

1

)or

(−13



A + 3I =

(−1 −33 9

)

(1 30 0

)

So

(−31


A− 5I =

(−9 −33 1

)

(1 1/30 0

)

So

(−1/3

1

)or

(−13



A + 3I =

(−1 −33 9

)

(1 30 0

)So

(−31

)is an eigenvector for this eigenvalue.

Also,

A− 5I =

(−9 −33 1

)

(1 1/30 0

)

So

(−1/3

1

)or

(−13



A + 3I =

(−1 −33 9

)

(1 30 0

)So

(−31


A− 5I =

(−9 −33 1

)

(1 1/30 0

)

So

(−1/3

1

)or

(−13



A + 3I =

(−1 −33 9

)

(1 30 0

)So

(−31


A− 5I =

(−9 −33 1

)

(1 1/30 0

)

So

(−1/3

1

)or

(−13



A + 3I =

(−1 −33 9

)

(1 30 0

)So

(−31


A− 5I =

(−9 −33 1

)

(1 1/30 0

)

So

(−1/3

1

)or

(−13



A + 3I =

(−1 −33 9

)

(1 30 0

)So

(−31


A− 5I =

(−9 −33 1

)

(1 1/30 0

)

So

(−1/3

1

)or

(−13


Summary

To find the eigenvalues of a matrix A, find the determinant ofA− λI. This will be a polynomial in λ, and its roots are theeigenvalues.


Find the eigenvalues of

A =

(−10 6−15 9

).

SolutionThe characteristic polynomial is∣∣∣∣−10− λ 6−15 9− λ

∣∣∣∣ = (−10−λ)(9−λ)−(−15)(6) = λ2+λ = λ(λ+1).

The eigenvalues are 0 and −1. A set of eigenvectors are

(35

)and(

23

).



A =

(−10 6−15 9

).


∣∣∣∣ = (−10−λ)(9−λ)−(−15)(6) = λ2+λ = λ(λ+1).


(35

)and(

23

).



A =

(−10 6−15 9

).


∣∣∣∣ = (−10−λ)(9−λ)−(−15)(6) = λ2+λ = λ(λ+1).

The eigenvalues are 0 and −1.

A set of eigenvectors are

(35

)and(

23

).



A =

(−10 6−15 9

).


∣∣∣∣ = (−10−λ)(9−λ)−(−15)(6) = λ2+λ = λ(λ+1).


(35

)and(

23

).

Applications

I In a Markov Chain, the steady-state vector is an eigenvectorcorresponding to the eigenvalue 1.

Date post:	23-Jun-2015
Category:	Technology
Upload:	matthew-leingang
View:	4,299 times
Download:	3 times

Lesson14: Eigenvalues And Eigenvectors

Technology