Reduced echelon form Reduced echelon form Reduced echelon form

Matrix equations Matrix equations Matrix equations

Null space Null space Null space

Range Range Range

Determinant Determinant Determinant

Invertibility Invertibility Invertibility

Similar matrices Similar matrices Similar matrices

Eigenvalues Eigenvalues Eigenvalues

Eigenvectors Eigenvectors Eigenvectors

Diagonabilty Diagonabilty Diagonabilty

Power Power Power

72

41A

1100

1100

5311

4211

B

142

274

3106

C

72

41A Reduced echelon form

10

012411

10

41

10

4122

150

41

150

411222

72

41

151

rrr

rr

rrr

Because the reduced echelon form of A is the identity matrix,we know that the columns of A are a basis for R2

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72

41A Matrix equations

Because the reduced echelon form of A is the identity matrix:

solutionuniqueahascxA

solutionuniqueahasxA

0

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72

41A

0

0

Aofspacenullthe

solutionuniqueahasxABecause

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72

41A

Every vector in the range of A is of the form:

7

4

2

1

72

41yx

y

x

Is a linear combination of the columns of A.

The columns of A span R2 = the range of A

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72

41A

The determinant of A = (1)(7) – (4)(-2) = 15

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72

41A

Because the determinant of A is NOT ZERO, A is invertible (nonsingular)

151

152

154

157

1

10

01

1072

0141toreduces

AIIA

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72

41A

If A is the matrix for T relative to the standard basis,what is the matrix for T relative to the basis:

?1

2,

1

3

11

23

72

41

11

23

34

051

1 PAPQ

Q is similar to A.Q is the matrix for T relative to the basis, (columns of P)

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72

41A

053158871

72

41det)det(

2

AI

The eigenvalues for A are 3 and 5

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72

41A

1

2

42

42

732

413

3

ofnullspace

ofnullspace

AIofnullspace

1

1

22

44

752

415

5

ofnullspace

ofnullspace

AIofnullspace

1

1,

1

2rseigenvectoofbasisa Return to outline

50

03

72

41tosimilarisA

1

1,

1

2

50

03

rseigenvectoofbasisthe

torelativeTfirmatrixtheis

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11

12

72

41

11

12

50

031

72

41A

A square root of A =

72373

732732

11

12

70

03

11

121

1

10

10

11

12

70

03

11

12

A10 =

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1100

1100

5311

4211

B

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The reduced echelon form of B =

0000

0000

1100

2011

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0

0

0

0

1100

1100

5311

4211

0

z

y

x

w

iffxB

zz

zy

xx

zxw

2

0

0

0

0

0000

0000

1100

2011

1

1

0

2

,

0

0

1

1

nullspacetheforbasisa

4

3

2

1

1100

1100

5311

4211

c

c

c

c

z

y

x

w

cxB

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124

123

12

1

4

3

2

1

0000

0000

1100

4211

1100

1100

5311

4211

ccc

ccc

cc

c

toreduces

c

c

c

c

00 124123

cccandccc

ifonlysolutionahascxB

1100

1100

5311

4211

B

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The range of B is spanned by its columns. Because its null spacehas dimension 2 , we know that its range has dimension 2.(dim domain = dim range + dim null sp).Any two independent columns can serve as a basis for the range.

1

1

3

2

,

0

0

1

1

Bofrangetheforbasisa

0

0000

0000

1100

4211

det

1100

1100

5311

4211

det

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Because the determinant is 0, B has no inverse. ie. B is singular

1100

1100

5311

4211

B

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If P is a 4x4 nonsingular matrix, then B is similar toany matrix of the form P-1 BP

0)2(

110

110

421

det)1(

110

110

531

det)1(

1100

1100

5311

4211

det)det(

3

BI

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The eigenvalues are 0 and 2.

1100

1100

5311

4211

B

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The null space of (0I –B)= the null space of B.

The eigenspace belonging to 0= the null space of the matrix

1

1

0

2

,

0

0

1

1

The null space of (2I –B)=

The eigenspace belonging to 2

0

0

1

1

1100

1100

5311

4211

B

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There are not enough independent eigenvectors to make a basis for R4 .

The characteristic polynomial root 0 is repeated three times, but the eigenspace belonging to 0 is two dimensional.

B is NOT similar to a diagonal matrix.

142

274

3106

C

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The reduced echelon form of C is

000

010

01 21

142

274

3106

C

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becausesolutionnohasxC

1

1

1

1

6

0

010

000

132

1

1

1

142

274

3106

toreduces

toreduces

0

0

0

142

274

3106

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zz

y

zx

0

0

0

0

000

010

01 21

21

A basis for the null space is:

2

0

1

142

274

3106

C

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The columns of the matrix span the range.

The dimension of the null space is 1.

Therefore the dimension of the range is 2.

Choose 2 independent columns of C to form a basis for the range

1

2

3

,

4

7

10

142

274

3106

C

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The determinant of C is 0.

Therefore C has no inverse.

0

142

010

3106

detdet

142

010

3106

3 (-2)row 2 row with 2 row replace

142

274

3106

2

C

142

274

3106

C

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For any nonsingular 3x3 matrix P, C is similar to P-1 CP

0)1)(1(

)13(2)210(4)16)(6(

142

274

3106

det)det(

3

2

CI

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The eigenvalues are: 1, -1, and 0

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042

284

3105

1 CI

Its null space =

0

1

2

242

264

3107

1 CI

Its null space =

1

1

1

142

274

3106

0 CI

Its null space =

2

0

1

000

010

001

142

274

3106

tosimilarisC

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1

1

210

011

112

000

010

001

210

011

112

142

274

3106

PDPC

The columns of P are eigenvectors and the diagonal elements of D are eigenvalues.

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133

133

210

011

112

000

010

001

210

011

112

142

274

3106

PDPC

C