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V is a vector space of dimension n. S = { v 1, v 2, v 3,..., v n } then S is INDEPENDENT if and only if S SPANS V. Return to outline

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KEY THEOREMS

KEY IDEAS KEY ALGORITHMS

LINKED TO EXAMPLES

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key theorems key ideas key algorithms n vectorsin an n dimensionalvector space

VECTOR SPACEindependentspan

Solve system equations

basis Find dot productcoordinates Take matrix times vector

dimensiondomain,null space, range of alinear mapping

LINEAR MAPPING Write matrix equationdomainnull spacerange

Find matrix for lin mapTake product of matrices

detA 0 matrix for Find inverse of matrixcompositioninverse

Find determinant of matrix

similarity ,eigenstuff

EIGENSTUFF Find eigenstuffsimilarity Similar diagonal matrix

V is a vector space of dimension n.

S = { v1 , v2 , v3 , . . . , vn } then

S is INDEPENDENT if and only if S SPANS V.

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If T is a LINEAR MAPPING then:

the dimension of the DOMAIN of T = the dimension of the NULL SPACE of T + the dimension of the RANGE of T

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0

A and B are SIMILAR matrices if and only if there exists a matrix P such that:

B = P –1 A PIf A is the matrix for T relative

tothe standard basis

then B is the matrix for T relative to

the columns of PIf B is diagonal then

the diagonal entries of B are eigenvalues andthe columns of P are eigenvectors

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3221424274263

zyxwzyxwzyxw

next

3221424274263

zyxwzyxwzyxw

next321214214274263

Reduces to:

3221424274263

zyxwzyxwzyxw

next321214214274263

000002210010021

Reduces to:

3221424274263

zyxwzyxwzyxw

next

000002210010021

zyxwzyxwzyxw

3221424274263

zyxwzyxwzyxw

next

zyxw

zyxwzyxwzyxw

2221

000002210010021

3221424274263

zyxwzyxwzyxw

nextzzzy

xxxw

122

121

zyxw

2221

3221424274263

zyxwzyxwzyxw

1200

0012

02

01

122

121

zx

zzzy

xxxw

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( 3 )-1 4 2

( 2 ) 5 3-2

• =

next

( 3 )-1 4 2

( 2 ) 5 3-2

• =

6 + -5 + 12 + -4 = 9

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( 3 -1 2 ) ( 1 ) = ( ) 2 1 -1 43

next

( 3 -1 2 ) ( 1 ) = ( 5 ) 2 1 -1 43

nextdot product of row 1 of matrix with vector

= entry 1 of answer

( 3 -1 2 ) ( 1 ) = ( 5 ) 2 1 -1 4 33

dot product of row 2 of matrix with vector

= entry 2 of answer return to outline

3221424274263

zyxwzyxwzyxw

347

212121424263

zyxw

System of linear equations:

Equivalent matrix equation:

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A toy maker manufactures bears and dolls.It takes 4 hours and costs $3 to make 1 bear.It takes 2 hours and costs $5 to make 1 doll.

cost totalrequired timetotal

dolls #bears #

TFind the matrix for T

next

A toy maker manufactures bears and dolls.It takes 4 hours and costs $3 to make 1 bear.It takes 2 hours and costs $5 to make 1 doll.

cost totalrequired timetotal

dolls #bears #

TFind the matrix for T

columnfirst 34

bear 1 make cost to bear 1 make torequired time

01

T

column second52

doll 1 make cost to doll 1 make torequired time

10

T

5324

for matrix the T

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( 3 -1 2 ) ( 1 2 ) = ( ) 2 1 -1 4 13 1

next

A B

( 3 -1 2 ) ( 1 2 ) = ( 5 ) 2 1 -1 4 13 1

dot product of row 1 of A with column 1 of B

= entry in row 1 column 1 of AB next

A B

( 3 -1 2 ) ( 1 2 ) = ( 5 7 ) 2 1 -1 4 13 1

dot product of row 1 of A with column 2 of B

= entry in row 1 column 2 of AB next

A B

( 3 -1 2 ) ( 1 2 ) = ( 5 7 ) 2 1 -1 4 1 33 1

dot product of row 2 of A with column 1 of B

= entry in row 2 column 1 of AB next

A B

( 3 -1 2 ) ( 1 2 ) = ( 5 7 ) 2 1 -1 4 1 3 43 1

dot product of row 2 of A with column 2 of B

= entry in row 2 column 2 of AB return to outline

100425010312001213

111100527010102001

Reduces to

next

100425010312001213

111100527010102001

Reduces to

A

A-1

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next

82

122A

To find eigenvalues for A, solve for :

08682

122det)det( 2

AI

next

82

122A

To find eigenvalues for A, solve for :

08682

122det)det( 2

AI

The eigenvalues are 2 and 4

next

82

122A

To find eigenvalues for A, solve for :

08682

122det)det( 2

AI

The eigenvalues are 2 and 4

An eigenvector belonging to 2 is in the null space of 2I - A

next

82

122A

To find eigenvalues for A, solve for :

08682

122det)det( 2

AI

The eigenvalues are 2 and 4

An eigenvector belonging to 2 is in the null space of 2I - A

62

124 2I - A

next

82

122A

To find eigenvalues for A, solve for :

08682

122det)det( 2

AI

The eigenvalues are 2 and 4

An eigenvector belonging to 2 is in the null space of 2I - A

62

124 2I - A

an eigenvector belonging to 2 is any nonzero multiple of

13

next

82

122A

To find eigenvalues for A, solve for :

08682

122det)det( 2

AI

The eigenvalues are 2 and 4

eigenvectors are:

13

12

next

82

122A

The eigenvalues are 2 and 4

eigenvectors are:

13

12

A is similar to the diagonal matrix B

4002

82

122A

The eigenvalues are 2 and 4

eigenvectors are:

13

12

B = P –1 A P

4002

=

1

1123

1123

82122

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bcaddbca

det

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columnor row a choose

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A is an nn matrix

detA 0 iff

A is nonsingular (invertible) iffThe columns of A are a basis for Rn iffThe null space of A contains only the zero vector iffA is the matrix for a 1-1 linear transformation

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