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Ï Exam 2 will be on Feb 25 Thurs in class. Details later.Make-up exams will be given only if there is an excusedabsence from the Dean of Students or a Doctor's note aboutsudden serious illness. No exceptions on this. Travelplans/broken alarm clock are unacceptable excuses.

Last Class...

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.

Triangular Matrices

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

diagonal.

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

The eigenvalues of A are 4, 0 and 6.

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

The eigenvalues of A are 4, 0 and 6.

Zero Eigenvalue??

Zero eigenvalue means

1. The equation Ax= 0x has a nontrivial or nonzero solution

2. This means Ax= 0 has a nontrivial solution.

3. This means A is not invertible (or detA= 0) (by invertiblematrix theorem)

Zero is an eigenvalue of A if and only if A is not invertible

Zero Eigenvalue??

Zero eigenvalue means

1. The equation Ax= 0x has a nontrivial or nonzero solution

2. This means Ax= 0 has a nontrivial solution.

3. This means A is not invertible (or detA= 0) (by invertiblematrix theorem)

Zero is an eigenvalue of A if and only if A is not invertible

Zero Eigenvalue??

Zero eigenvalue means

1. The equation Ax= 0x has a nontrivial or nonzero solution

2. This means Ax= 0 has a nontrivial solution.

3. This means A is not invertible (or detA= 0) (by invertiblematrix theorem)

Zero is an eigenvalue of A if and only if A is not invertible

Zero Eigenvalue??

Zero eigenvalue means

1. The equation Ax= 0x has a nontrivial or nonzero solution

2. This means Ax= 0 has a nontrivial solution.

3. This means A is not invertible (or detA= 0) (by invertiblematrix theorem)

Zero is an eigenvalue of A if and only if A is not invertible

Important

If λ is an eigenvalue of a square matrix A, prove that λ2 is aneigenvalue of A2.

For any problem of this type, start with the equation that de�neseigenvalue and take it from there.

Solution: If λ is an eigenvalue of a square matrix A, we haveAx= λx.Multiply both sides by A. We get A2x=A(λx).This is same as writing A2x= λ(Ax)︸ ︷︷ ︸ since λ is a scalar.

Again Ax= λx. So, A2x= λ(λx)︸︷︷︸= λ2x. This equation means that

λ2 is an eigenvalue of A2.

Important

If λ is an eigenvalue of a square matrix A, prove that λ2 is aneigenvalue of A2.

For any problem of this type, start with the equation that de�neseigenvalue and take it from there.

Solution: If λ is an eigenvalue of a square matrix A, we haveAx= λx.

Multiply both sides by A. We get A2x=A(λx).This is same as writing A2x= λ(Ax)︸ ︷︷ ︸ since λ is a scalar.

Again Ax= λx. So, A2x= λ(λx)︸︷︷︸= λ2x. This equation means that

λ2 is an eigenvalue of A2.

Important

If λ is an eigenvalue of a square matrix A, prove that λ2 is aneigenvalue of A2.

For any problem of this type, start with the equation that de�neseigenvalue and take it from there.

Solution: If λ is an eigenvalue of a square matrix A, we haveAx= λx.Multiply both sides by A. We get A2x=A(λx).

This is same as writing A2x= λ(Ax)︸ ︷︷ ︸ since λ is a scalar.

Again Ax= λx. So, A2x= λ(λx)︸︷︷︸= λ2x. This equation means that

λ2 is an eigenvalue of A2.

Important

If λ is an eigenvalue of a square matrix A, prove that λ2 is aneigenvalue of A2.

For any problem of this type, start with the equation that de�neseigenvalue and take it from there.

Solution: If λ is an eigenvalue of a square matrix A, we haveAx= λx.Multiply both sides by A. We get A2x=A(λx).This is same as writing A2x= λ(Ax)︸ ︷︷ ︸ since λ is a scalar.

Again Ax= λx. So, A2x= λ(λx)︸︷︷︸= λ2x. This equation means that

λ2 is an eigenvalue of A2.

Important

If λ is an eigenvalue of a square matrix A, prove that λ2 is aneigenvalue of A2.

For any problem of this type, start with the equation that de�neseigenvalue and take it from there.

Solution: If λ is an eigenvalue of a square matrix A, we haveAx= λx.Multiply both sides by A. We get A2x=A(λx).This is same as writing A2x= λ(Ax)︸ ︷︷ ︸ since λ is a scalar.

Again Ax= λx. So, A2x= λ(λx)︸︷︷︸= λ2x.

This equation means that

λ2 is an eigenvalue of A2.

Important

If λ is an eigenvalue of a square matrix A, prove that λ2 is aneigenvalue of A2.

For any problem of this type, start with the equation that de�neseigenvalue and take it from there.

Solution: If λ is an eigenvalue of a square matrix A, we haveAx= λx.Multiply both sides by A. We get A2x=A(λx).This is same as writing A2x= λ(Ax)︸ ︷︷ ︸ since λ is a scalar.

Again Ax= λx. So, A2x= λ(λx)︸︷︷︸= λ2x. This equation means that

λ2 is an eigenvalue of A2.

Important, see prob 25 sec 5.1

If λ is an eigenvalue of an invertible matrix A, prove that λ−1 is aneigenvalue of A−1.

Solution: If λ is an eigenvalue of a square matrix A, we haveAx= λx.Since A is invertible, multiply both sides by A−1. We getA−1A︸ ︷︷ ︸

I

x=A−1(λx)=⇒A−1(λx)= x

This is same as writing λ(A−1x)= x since λ is a scalar.Since λ 6= 0 (Why?) we can divide both sides by λ and we getA−1x= 1

λx.Thus 1

λ or λ−1 is an eigenvalue of A−1.

Important, see prob 25 sec 5.1

If λ is an eigenvalue of an invertible matrix A, prove that λ−1 is aneigenvalue of A−1.

Solution: If λ is an eigenvalue of a square matrix A, we haveAx= λx.

Since A is invertible, multiply both sides by A−1. We getA−1A︸ ︷︷ ︸

I

x=A−1(λx)=⇒A−1(λx)= x

This is same as writing λ(A−1x)= x since λ is a scalar.Since λ 6= 0 (Why?) we can divide both sides by λ and we getA−1x= 1

λx.Thus 1

λ or λ−1 is an eigenvalue of A−1.

Important, see prob 25 sec 5.1

If λ is an eigenvalue of an invertible matrix A, prove that λ−1 is aneigenvalue of A−1.

Solution: If λ is an eigenvalue of a square matrix A, we haveAx= λx.Since A is invertible, multiply both sides by A−1. We getA−1A︸ ︷︷ ︸

I

x=A−1(λx)

=⇒A−1(λx)= x

This is same as writing λ(A−1x)= x since λ is a scalar.Since λ 6= 0 (Why?) we can divide both sides by λ and we getA−1x= 1

λx.Thus 1

λ or λ−1 is an eigenvalue of A−1.

Important, see prob 25 sec 5.1

If λ is an eigenvalue of an invertible matrix A, prove that λ−1 is aneigenvalue of A−1.

Solution: If λ is an eigenvalue of a square matrix A, we haveAx= λx.Since A is invertible, multiply both sides by A−1. We getA−1A︸ ︷︷ ︸

I

x=A−1(λx)=⇒A−1(λx)= x

This is same as writing λ(A−1x)= x since λ is a scalar.Since λ 6= 0 (Why?) we can divide both sides by λ and we getA−1x= 1

λx.Thus 1

λ or λ−1 is an eigenvalue of A−1.

Important, see prob 25 sec 5.1

If λ is an eigenvalue of an invertible matrix A, prove that λ−1 is aneigenvalue of A−1.

Solution: If λ is an eigenvalue of a square matrix A, we haveAx= λx.Since A is invertible, multiply both sides by A−1. We getA−1A︸ ︷︷ ︸

I

x=A−1(λx)=⇒A−1(λx)= x

This is same as writing λ(A−1x)= x since λ is a scalar.

Since λ 6= 0 (Why?) we can divide both sides by λ and we getA−1x= 1

λx.Thus 1

λ or λ−1 is an eigenvalue of A−1.

Important, see prob 25 sec 5.1

If λ is an eigenvalue of an invertible matrix A, prove that λ−1 is aneigenvalue of A−1.

Solution: If λ is an eigenvalue of a square matrix A, we haveAx= λx.Since A is invertible, multiply both sides by A−1. We getA−1A︸ ︷︷ ︸

I

x=A−1(λx)=⇒A−1(λx)= x

This is same as writing λ(A−1x)= x since λ is a scalar.Since λ 6= 0 (Why?) we can divide both sides by λ and we getA−1x= 1

λx.

Thus 1

λ or λ−1 is an eigenvalue of A−1.

Important, see prob 25 sec 5.1

If λ is an eigenvalue of an invertible matrix A, prove that λ−1 is aneigenvalue of A−1.

Solution: If λ is an eigenvalue of a square matrix A, we haveAx= λx.Since A is invertible, multiply both sides by A−1. We getA−1A︸ ︷︷ ︸

I

x=A−1(λx)=⇒A−1(λx)= x

This is same as writing λ(A−1x)= x since λ is a scalar.Since λ 6= 0 (Why?) we can divide both sides by λ and we getA−1x= 1

λx.Thus 1

λ or λ−1 is an eigenvalue of A−1.

Example 20 section 5.1

Without calculation, �nd one eigenvalue and 2 linearly independent

eigenvectors of A= 5 5 5

5 5 55 5 5

. Justify your answer.

Solution: What is special about this matrix? Invertible/NotInvertible?

Clearly not invertible (same rows, columns).What is an eigenvalue of A?0!!To �nd eigenvectors for this eigenvalue, we look at A−0I and rowreduce. Or row reduce A.

Example 20 section 5.1

Without calculation, �nd one eigenvalue and 2 linearly independent

eigenvectors of A= 5 5 5

5 5 55 5 5

. Justify your answer.

Solution: What is special about this matrix? Invertible/NotInvertible?Clearly not invertible (same rows, columns).What is an eigenvalue of A?

0!!To �nd eigenvectors for this eigenvalue, we look at A−0I and rowreduce. Or row reduce A.

Example 20 section 5.1

Without calculation, �nd one eigenvalue and 2 linearly independent

eigenvectors of A= 5 5 5

5 5 55 5 5

. Justify your answer.

Solution: What is special about this matrix? Invertible/NotInvertible?Clearly not invertible (same rows, columns).What is an eigenvalue of A?0!!To �nd eigenvectors for this eigenvalue, we look at A−0I and rowreduce. Or row reduce A.

Example 20 section 5.1We get (do the row reductions yourself) 1 1 1 0

0 0 0 00 0 0 0

.

Thus x1+x2+x3 = 0 where x2 and x3 are free. So x1 =−x2−x3.We have x1

x2x3

= −x2−x3

x2x3

= x2

−110

+x3

−101

Choose x2 = 1 and x3 = 1 (or anything nonzero) and two linearlyindependent eigenvectors are −1

10

,

−101

Example 20 section 5.1We get (do the row reductions yourself) 1 1 1 0

0 0 0 00 0 0 0

. Thus x1+x2+x3 = 0 where x2 and x3 are free. So x1 =−x2−x3.

We have x1x2x3

= −x2−x3

x2x3

= x2

−110

+x3

−101

Choose x2 = 1 and x3 = 1 (or anything nonzero) and two linearlyindependent eigenvectors are −1

10

,

−101

Example 20 section 5.1We get (do the row reductions yourself) 1 1 1 0

0 0 0 00 0 0 0

. Thus x1+x2+x3 = 0 where x2 and x3 are free. So x1 =−x2−x3.We have x1

x2x3

= −x2−x3

x2x3

= x2

−110

+x3

−101

Choose x2 = 1 and x3 = 1 (or anything nonzero) and two linearlyindependent eigenvectors are −1

10

,

−101

Example 20 section 5.1We get (do the row reductions yourself) 1 1 1 0

0 0 0 00 0 0 0

. Thus x1+x2+x3 = 0 where x2 and x3 are free. So x1 =−x2−x3.We have x1

x2x3

= −x2−x3

x2x3

= x2

−110

+x3

−101

Choose x2 = 1 and x3 = 1 (or anything nonzero) and two linearlyindependent eigenvectors are −1

10

,

−101

Observations

1. The eigenvalue λ= 0 has 2 linearly independent eigenvectors

2. We say that this eigenspace is a two-dimensional subspace ofR3.

3. Examples with 2 linearly independent eigenvectors are veryimportant for section 5.3 when we do diagonalization and indi�erential equations where eigenvalues repeat.

Observations

1. The eigenvalue λ= 0 has 2 linearly independent eigenvectors

2. We say that this eigenspace is a two-dimensional subspace ofR3.

3. Examples with 2 linearly independent eigenvectors are veryimportant for section 5.3 when we do diagonalization and indi�erential equations where eigenvalues repeat.

Observations

1. The eigenvalue λ= 0 has 2 linearly independent eigenvectors

2. We say that this eigenspace is a two-dimensional subspace ofR3.

3. Examples with 2 linearly independent eigenvectors are veryimportant for section 5.3 when we do diagonalization and indi�erential equations where eigenvalues repeat.

Example 16 section 5.1

Let A=

3 0 2 01 3 1 00 1 1 00 0 0 4

. Find a basis for eigenspace corresponding

to λ= 4.

Solution: To �nd an eigenvector, start with A−4I and row reduce

A−4I =

3 0 2 01 3 1 00 1 1 00 0 0 4

−

4 0 0 00 4 0 00 0 4 00 0 0 4

=

−1 0 2 01 −1 1 00 1 −3 00 0 0 0

−1 0 2 0 0

1 −1 1 0 0

0 1 −3 0 0

0 0 0 0 0

R2+R1

−1 0 2 0 0

0 −1 3 0 0

0 1 −3 0 0

0 0 0 0 0

R3+R2

−1 0 2 0 0

1 −1 1 0 0

0 1 −3 0 0

0 0 0 0 0

R2+R1

−1 0 2 0 0

0 −1 3 0 0

0 1 −3 0 0

0 0 0 0 0

R3+R2

−1 0 2 0 0

0 −1 3 0 0

0 0 0 0 0

0 0 0 0 0

Thus, x2 = 3x3 and x1 = 2x3 with x3 and x4 free.x1x2x3x4

=

2x33x3x3x4

= x3

2310

+x3

0001

.

A basis for eigenspace will be thus

2310

,

0001

.

−1 0 2 0 0

0 −1 3 0 0

0 0 0 0 0

0 0 0 0 0

Thus, x2 = 3x3 and x1 = 2x3 with x3 and x4 free.

x1x2x3x4

=

2x33x3x3x4

= x3

2310

+x3

0001

.

A basis for eigenspace will be thus

2310

,

0001

.

−1 0 2 0 0

0 −1 3 0 0

0 0 0 0 0

0 0 0 0 0

Thus, x2 = 3x3 and x1 = 2x3 with x3 and x4 free.

x1x2x3x4

=

2x33x3x3x4

= x3

2310

+x3

0001

.

A basis for eigenspace will be thus

2310

,

0001

.

Theorem

TheoremEigenvectors corresponding to distinct eigenvalues of an n×n

matrix A are linearly independent.

Next week...

1. How to �nd eigenvalues of a 2×2 and 3×3 matrix?

2. The process of diagonalization (uses eigenvalues andeigenvectors)

3. Finding "complex" eigenvalues

4. Quick look at eigenvalues and eigenvectors being used to learnlong-term behavior of a dynamical system.

5. Quiz 4 (last quiz) will be on thurs Feb 18 based on sections3.3, 5.1 and 5.2.