Eigenvalues and Eigenvectors (Tacoma Narrows Bridge video included)

Announcements

Ï Quiz 3 tomorrow on sections 3.1 and 3.2

Ï No calculators for this quiz. You have to �nd the determinantsusing the methods we have learned.

Section 5.1 Eigenvalues and Eigenvectors

Comes from the German word "eigen" meaning proper orcharacteristic.

Gives a better understanding of the linear transformation x 7→Ax interms of elements that can be easily visualized.

Tacoma Narrows Bridge Collapse- 1940

1. Bridge forming small waves like a river

2. Initially assumed safe but oscillations grew with time

3. One edge of the road 28 ft higher than the other edge

4. Finally collapsed

















Explanation:

1. Oscillation of bridge caused by the wind frequency being tooclose to the natural frequency of the bridge

2. Natural frequency of the bridge is the eigenvalue of thesmallest magnitude (based on a system that modeled thebridge)

3. Eigenvalues are very important to engineers when they designstructures.


Explanation:





Explanation:




Electrical Systems

Radio Tuning

1. Trying to match the frequency with which your station isbroadcasting.

2. Eigenvalues were used during the radio design by engineers.

Cars

1. Use Eigenvalues to damp out noise for a quiet ride.

2. Also used in design of car stereo systems so that sounds createlistening pleasure

3. To reduce the vibration of the car due to the music.

Cars

1. Use Eigenvalues to damp out noise for a quiet ride.

2. Also used in design of car stereo systems so that sounds createlistening pleasure

3. To reduce the vibration of the car due to the music.

Others

1. Test for cracks in a solid. (avoids testing each and every inchof a beam for example)

2. Used by oil companies to explore land for oil.

3. Applications in economics, physics, chemistry, biology (redblood cell production)

4. Discrete (and continuous) linear dynamical systems (part ofMA3521)

5. Pure and Applied Mathematics

Others






Others






Others






Others






Back to matrix-vector product

Let A=[3 −21 0

]. Find Au and Av where u=

[ −1−2

]and

v=[21

](the image of u and v after being "acted" on by A.)

Au=[3 −21 0

][ −1−2

]=

[ −3+4−1−0

]=

[1−1

]

Av=[3 −21 0

][21

]=

[6−22+0

]=

[42

]= 2v

Back to matrix-vector product

Let A=[3 −21 0

]. Find Au and Av where u=

[ −1−2

]and

v=[21

](the image of u and v after being "acted" on by A.)

Au=[3 −21 0

][ −1−2

]=

[ −3+4−1−0

]=

[1−1

]

Av=[3 −21 0

][21

]=

[6−22+0

]=

[42

]= 2v

y

x0

u

Au

v

Av

y

x0

u

Au

v

Av

y

x0

u

Au

v

Av

y

x0

u

Au

v

Av

y

x0

u

Au

v

Av

y

x0

u

Au

v

Av

y

x0

u

Au

v

Av

Observations

1. A transforms u into a new vector.

2. A "stretches" or "dilates" v in the same direction.

3. Au is not a scalar multiple of u

4. Av is a scalar multiple of v

5. We are interested in the second case, where the new vector(result of action by A) is a scalar multiple of the originalvector.

Observations






Observations






Observations






Observations






Eigenvalues and Eigenvectors

In other words we are interested in studying equations of the formAx= 2x or Ax=−3x.

We look for vectors that are scalar multiples of themselves (couldbe same as the original vector)

Eigenvalues and Eigenvectors

De�nitionAn eigenvector of an n×n matrix A is a NON-ZERO vector xsuch that Ax= λx for some scalar λ.

A scalar λ is called an eigenvalue of A if there is a nontrivial (ornonzero) solution x to Ax= λx; such an x is called an eigenvector

corresponding to λ.

Why nonzero?

Go back to Ax= λx. What is an obvious solution to this equation?

We know that A0= λ0 always. There is nothing interesting aboutthe zero (or the trivial) solution. So we need a nonzero vector foreigenvector.

An eigenvalue can be zero. (We will come back to this case later)

Why nonzero?




Why nonzero?




Example

Let A=[

7 2−4 1

]. Check whether u=

[1−1

]and v=

[1−3

]are

eigenvectors of A. If yes, �nd the corresponding eigenvalues.

Solution: To check whether a given vector u or v is an eigenvectorof A is easy.Find the product Au and check whether the new vector is a scalarmultiple of u. Also �nd Av and check whether the new vector is ascalar multiple of v.

Au=[

7 2−4 1

][1−1

]=

[5−5

]= 5u

Av=[

7 2−4 1

][1−3

]=

[1−7

]6= λv

So u is an eigenvector with λ= 5 and v is not an eigenvector.

Example

Let A=[

7 2−4 1

]. Check whether u=

[1−1

]and v=

[1−3

]are



Au=[

7 2−4 1

][1−1

]=

[5−5

]= 5u

Av=[

7 2−4 1

][1−3

]=

[1−7

]6= λv


Example

Let A=[

7 2−4 1

]. Check whether u=

[1−1

]and v=

[1−3

]are



Au=[

7 2−4 1

][1−1

]=

[5−5

]= 5u

Av=[

7 2−4 1

][1−3

]=

[1−7

]6= λv


Example

Let A=[

7 2−4 1

]. Check whether u=

[1−1

]and v=

[1−3

]are



Au=[

7 2−4 1

][1−1

]=

[5−5

]= 5u

Av=[

7 2−4 1

][1−3

]=

[1−7

]6= λv


Example

Let A=[

7 2−4 1

]. Show that 3 is an eigenvalue of A and �nd the

corresponding eigenvectors.

Solution: 3 is an eigenvalue of A if and only if the equation Ax= 3xhas a nontrivial solution.

This means

Ax−3x= 0

has a nontrivial solution. Or

(A−3I )x= 0

Back to homogeneous systems, trivial and nontrivial solutions.

Example

Let A=[

7 2−4 1

]. Show that 3 is an eigenvalue of A and �nd the

corresponding eigenvectors.

Solution: 3 is an eigenvalue of A if and only if the equation Ax= 3xhas a nontrivial solution. This means

Ax−3x= 0

has a nontrivial solution. Or

(A−3I )x= 0

Back to homogeneous systems, trivial and nontrivial solutions.

Revisiting Homogeneous Systems

1. Zero is always a solution (trivial solution).

2. For nontrivial solutions, we need linearly dependent columns

3. This means there are nonpivot columns or free variables

4. The free variables lead us to nontrivial solutions.











Example

Write the matrix

A−3I =[

7 2−4 1

]−

[3 00 3

]=

[4 2−4 −2

].

What do you think about the columns? Linearly dep/indep?

Theyare linearly dependent. So the homogeneous system has nontrivialsolutions. Thus 3 is an eigenvalue of A.

Example

Write the matrix

A−3I =[

7 2−4 1

]−

[3 00 3

]=

[4 2−4 −2

].

What do you think about the columns? Linearly dep/indep?Theyare linearly dependent. So the homogeneous system has

nontrivialsolutions. Thus 3 is an eigenvalue of A.

Example

Write the matrix

A−3I =[

7 2−4 1

]−

[3 00 3

]=

[4 2−4 −2

].

What do you think about the columns? Linearly dep/indep?Theyare linearly dependent. So the homogeneous system has nontrivialsolutions. Thus 3 is an eigenvalue of A.

Example

We still have to �nd the eigenvectors for this eigenvalue (λ= 3).Start with the matrix A−3I and do row operations. 4 2 0

−4 −2 0

R1+R2

4 2 0

0 0 0

So, 4x1 =−2x2 where x2 is free. Or, x1 =−1

2x2. Our solution is

thus, [x1x2

]=

[ −1

2x2

x2

]= x2

[ −1

2

1

].

Example


−4 −2 0

R1+R2

4 2 0

0 0 0


2x2.

Our solution isthus, [

x1x2

]=

[ −1

2x2

x2

]= x2

[ −1

2

1

].

Example


−4 −2 0

R1+R2

4 2 0

0 0 0


2x2. Our solution is

thus, [x1x2

]=

[ −1

2x2

x2

]= x2

[ −1

2

1

].

Example

Choose any value for x2 other than zero. Preferably, choosesomething that will clear the fraction. So choose x2 = 2.

This gives,[x1x2

]= 2

[ −1

2

1

]=

[ −12

].

Remember that this is an eigenvector of A. Depending on your

choice of x2, any scalar multiple of

[ −1

2

1

]is an eigenvector of A.

Example

Choose any value for x2 other than zero. Preferably, choosesomething that will clear the fraction. So choose x2 = 2. This gives,[

x1x2

]= 2

[ −1

2

1

]=

[ −12

].

Remember that this is an eigenvector of A. Depending on your

choice of x2, any scalar multiple of

[ −1

2

1

]is an eigenvector of A.

Notes

1. Echelon form is used to �nd eigenvectors BUT NOTeigenvalues.

2. λ is an eigenvalue of A if and only if the equation(A−λI )x= 0 has nontrivial solution.

3. All solutions to (A−λI )x= 0 is nothing but the Null Space of(A−λI ).

4. This set is a subspace of Rn and is called the eigenspace of Acorresponding to λ.

5. Eigenspace consists of the zero vector and all eigenvectorscorresponding to λ.

Notes






Notes






Notes






Notes






Steps to check if λ is an eigenvalue

1. Write the matrix A−λI

2. Check whether the columns are linearly dependent (existenceof free variables)

3. If the columns are linearly dependent, λ is an eigenvalue.


1. Write the matrix A−λI2. Check whether the columns are linearly dependent (existence

of free variables)



1. Write the matrix A−λI2. Check whether the columns are linearly dependent (existence

of free variables)


Steps to �nd the eigenvectors for λ

Once we have an eigenvalue λ,

1. Start with the matrix A−λI

2. Row reduce to echelon form to identify the basic and freevariables.

3. Express the basic variables in terms of free variables, write theanswer in vector form.

4. Choose a convenient value for the free variable (NEVERchoose zero).



1. Start with the matrix A−λI2. Row reduce to echelon form to identify the basic and free

variables.






variables.






variables.



Example 6, section 5.1

Is

1−21

an eigenvector of A= 3 6 7

3 3 75 6 5

. If yes, �nd the

corresponding eigenvalue.

Solution: 3 6 73 3 75 6 5

1−21

= 3−12+7

3−6+75−12+5

= −2

4−2

The new vector is -2 times the old vector. So

1−21

is an

eigenvector of A with eigenvalue λ=−2.


Is

1−21


3 3 75 6 5

. If yes, �nd the


Solution: 3 6 73 3 75 6 5

1−21

= 3−12+7

3−6+75−12+5

= −2

4−2


1−21

is an



Is

1−21


3 3 75 6 5

. If yes, �nd the


Solution: 3 6 73 3 75 6 5

1−21

= 3−12+7

3−6+75−12+5

= −2

4−2


1−21

is an



Is λ= 3 an eigenvalue of A= 1 2 2

3 −2 10 1 1

? If yes, �nd the

corresponding eigenvector.

Solution:

A−3I = 1 2 2

3 −2 10 1 1

− 3 0 0

0 3 00 0 3

= −2 2 2

3 −5 10 1 −2

How to know whether the columns are linearly dependent? You cancheck whether the determinant of this matrix is zero. (Or you couldrow reduce and check if there are free variables)If det(A−3I )= 0, that means A−3I has linearly dependent columns(or there is a nontrivial solution).

Here det(A−3I )= 0 (Do this yourself), so the columns are linearlydependent and so λ= 3 is an eigenvalue.



3 −2 10 1 1

? If yes, �nd the


Solution:

A−3I = 1 2 2

3 −2 10 1 1

− 3 0 0

0 3 00 0 3

= −2 2 2

3 −5 10 1 −2





3 −2 10 1 1

? If yes, �nd the


Solution:

A−3I = 1 2 2

3 −2 10 1 1

− 3 0 0

0 3 00 0 3

= −2 2 2

3 −5 10 1 −2

How to know whether the columns are linearly dependent? You cancheck whether the determinant of this matrix is zero. (Or you couldrow reduce and check if there are free variables)

If det(A−3I )= 0, that means A−3I has linearly dependent columns(or there is a nontrivial solution).




3 −2 10 1 1

? If yes, �nd the


Solution:

A−3I = 1 2 2

3 −2 10 1 1

− 3 0 0

0 3 00 0 3

= −2 2 2

3 −5 10 1 −2





3 −2 10 1 1

? If yes, �nd the


Solution:

A−3I = 1 2 2

3 −2 10 1 1

− 3 0 0

0 3 00 0 3

= −2 2 2

3 −5 10 1 −2



To �nd an eigenvector, start with A−3I and row reduce

A−3I = −2 2 2

3 −5 10 1 −2

Divide R1 by -2 and we get

A−3I = 1 −1 −1

3 −5 10 1 −2

1 −1 −1 0

3 −5 1 0

0 1 −2 0

R2-3R1

1 −1 −1 0

0 −2 4 0

0 1 −2 0

R3+0.5R2

1 −1 −1 00 −2 4 00 0 0 0

We have x2 = 2x3 and x1 = x2+x3 = 3x3.Our solution is thus, x1

x2x3

= 3x3

2x3x3

= x3

321

.

Choose x3 = 1 and we have x1x2x3

= 3

21

.


Find a basis for the eigenspace of A=[

7 4−3 −1

]corresponding to

λ= 1,5.

Solution: For λ= 1,

A− I =[

7 4−3 −1

]−

[1 00 1

]=

[6 4−3 −2

] 6 4 0

−3 −2 0

R2+0.5R1

6 4 0

0 0 0



7 4−3 −1

]corresponding to

λ= 1,5.


A− I =[

7 4−3 −1

]−

[1 00 1

]=

[6 4−3 −2

]

6 4 0

−3 −2 0

R2+0.5R1

6 4 0

0 0 0



7 4−3 −1

]corresponding to

λ= 1,5.


A− I =[

7 4−3 −1

]−

[1 00 1

]=

[6 4−3 −2

] 6 4 0

−3 −2 0

R2+0.5R1

6 4 0

0 0 0


From the �rst row, 6x1 =−4x2 or x1 =−2

3x2.

Our solution is thus,[x1x2

]=

[ −2

3x2

x2

]= x2

[ −2

3

1

].

Choose a convenient value for x2. Pick x2 = 3 to clear the fraction.This gives [

x1x2

]= 3

[ −2

3

1

]=

[ −23

].



3x2. Our solution is thus,[

x1x2

]=

[ −2

3x2

x2

]= x2

[ −2

3

1

].


x1x2

]= 3

[ −2

3

1

]=

[ −23

].



3x2. Our solution is thus,[

x1x2

]=

[ −2

3x2

x2

]= x2

[ −2

3

1

].


x1x2

]= 3

[ −2

3

1

]=

[ −23

].

Example 12, section 5.1For λ= 5,

A−5I =[

7 4−3 −1

]−

[5 00 5

]=

[2 4−3 −6

]

Divide R1 by 2 and we get [1 2−3 −6

] 1 2 0

−3 −6 0

R2+3R1

1 2 0

0 0 0


A−5I =[

7 4−3 −1

]−

[5 00 5

]=

[2 4−3 −6

]Divide R1 by 2 and we get [

1 2−3 −6

]

1 2 0

−3 −6 0

R2+3R1

1 2 0

0 0 0


A−5I =[

7 4−3 −1

]−

[5 00 5

]=

[2 4−3 −6

]Divide R1 by 2 and we get [

1 2−3 −6

] 1 2 0

−3 −6 0

R2+3R1

1 2 0

0 0 0


From the �rst row, x1 =−2x2. Our solution is thus,

[x1x2

]=

[ −2x2x2

]= x2

[ −21

].

Choose a convenient value for x2. Pick x2 = 1. This gives[x1x2

]=

[ −21

].


From the �rst row, x1 =−2x2. Our solution is thus,[x1x2

]=

[ −2x2x2

]= x2

[ −21

].


]=

[ −21

].


From the �rst row, x1 =−2x2. Our solution is thus,[x1x2

]=

[ −2x2x2

]= x2

[ −21

].


]=

[ −21

].


Find a basis for the eigenspace of A= 1 0 −1

1 −3 04 −13 1

corresponding to λ=−2.

Solution: For λ=−2,

A+2I = 1 0 −1

1 −3 04 −13 1

+ 2 0 0

0 2 00 0 2

= 3 0 −1

1 −1 04 −13 3

3 0 −1 0

1 −1 0 0

4 −13 3 0

Swap


Find a basis for the eigenspace of A= 1 0 −1

1 −3 04 −13 1

corresponding to λ=−2.

Solution: For λ=−2,

A+2I = 1 0 −1

1 −3 04 −13 1

+ 2 0 0

0 2 00 0 2

= 3 0 −1

1 −1 04 −13 3

3 0 −1 0

1 −1 0 0

4 −13 3 0

Swap

1 −1 0 0

3 0 −1 0

4 −13 3 0

R2-3R1

R3-4R1

1 −1 0 0

0 3 −1 0

0 −9 3 0

R3+3R2

1 −1 0 0

3 0 −1 0

4 −13 3 0

R2-3R1

R3-4R1

1 −1 0 0

0 3 −1 0

0 −9 3 0

R3+3R2


1 −1 0 00 3 −1 00 0 0 0

Here x3 is a free variable.From the second row, 3x2 = x3 or x2 = 1

3x3.

From �rst row, x1 = x2 and so x1 = 1

3x3.

Our solution is thus, x1x2x3

= 1

3x3

1

3x3x3

= x3

1

31

3

1

.


= 1

13

.


1 −1 0 00 3 −1 00 0 0 0


3x3.


3x3.


= 1

3x3

1

3x3x3

= x3

1

31

3

1

.


= 1

13

.


1 −1 0 00 3 −1 00 0 0 0


3x3.


3x3.


= 1

3x3

1

3x3x3

= x3

1

31

3

1

.


= 1

13

.


1 −1 0 00 3 −1 00 0 0 0


3x3.


3x3.


= 1

3x3

1

3x3x3

= x3

1

31

3

1

.


= 1

13

.

Any Surprises to Expect?

Try problems 13 and 15 in your homework. There might besomething interesting going on in these cases. If you get stuck, wewill go over them in class tomorrow.

Triangular Matrices

TheoremThe eigenvalues of a triangular matrix are the entries on its main

diagonal.

Proof.For simplicity, consider a 3×3 upper triangular matrix. Then

A−λI = a11 a12 a13

0 a22 a230 0 a33

− λ 0 0

0 λ 00 0 λ

= a11−λ a12 a13

0 a22−λ a230 0 a33−λ

(A−λI )x= 0 has a free variable if and only if atleast one of thediagonal entries is zero. This happens if and only if λ equals one ofthe entries a11, a22, a33 of A.

Triangular Matrices

TheoremThe eigenvalues of a triangular matrix are the entries on its main

diagonal.

Proof.For simplicity, consider a 3×3 upper triangular matrix. Then

A−λI = a11 a12 a13

0 a22 a230 0 a33

− λ 0 0

0 λ 00 0 λ

= a11−λ a12 a13

0 a22−λ a230 0 a33−λ

(A−λI )x= 0 has a free variable if and only if atleast one of thediagonal entries is zero. This happens if and only if λ equals one ofthe entries a11, a22, a33 of A.

Triangular Matrices

Example

1. Let

A= 5 1 9

0 2 30 0 6

The eigenvalues of A are 5, 2 and 6.

2. Let

A= 4 1 9

0 0 30 0 6


Triangular Matrices

Example

1. Let

A= 5 1 9

0 2 30 0 6


2. Let

A= 4 1 9

0 0 30 0 6


Date post:	07-May-2015
Category:	Education
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Eigenvalues and Eigenvectors (Tacoma Narrows Bridge video included)

Education