Bases and Dimension - Duke Universitybfitzpat/teaching/218s20/lectures/... · Math 218 Brian D. Fitzpatrick Duke University March 1, 2020 MATH. Overview Geometric Motivation Visualizing

Bases and DimensionMath 218

Brian D. Fitzpatrick

Duke University

March 1, 2020

MATH

Overview

Geometric MotivationVisualizing Vector Spaces

Definitions and PropertiesDefinition of a BasisProperties of BasesDefinition of DimensionExamples

The Four Fundamental SubspacesBases of Column SpacesBases of Null SpacesBases of Row SpacesBases of Left Null Spaces

The Rank-Nullity TheoremStatementExample


QuestionWhat does L = Span{〈1, −7, 3〉 } “look like”?

AnswerL consists of all “multiples” of〈1, −7, 3〉 . L “looks like” a line in R3.

L

NoteWe can represent L in many different ways.

L = Col

rank=1

2−14

6

= Col

rank=1

−2 314 −21−6 9

= Null

nullity=1

[7 1 0−3 0 1

]

We only “need” one #‰v to define L = Span{ #‰v }.



AnswerL consists of all “multiples” of〈1, −7, 3〉 .

L “looks like” a line in R3.

L


L = Col

rank=1

2−14

6

= Col

rank=1

−2 314 −21−6 9

= Null

nullity=1

[7 1 0−3 0 1

]






L


L = Col

rank=1

2−14

6

= Col

rank=1

−2 314 −21−6 9

= Null

nullity=1

[7 1 0−3 0 1

]






L


L = Col

rank=1

2−14

6

= Col

rank=1

−2 314 −21−6 9

= Null

nullity=1

[7 1 0−3 0 1

]





L


L = Col

rank=1

2−14

6

= Col

rank=1

−2 314 −21−6 9

= Null

nullity=1

[7 1 0−3 0 1

]





L


L = Col

rank=1

2−14

6

= Col

rank=1

−2 314 −21−6 9

= Null

nullity=1

[7 1 0−3 0 1

]





L


L = Col

rank=1

2−14

6

= Col

rank=1

−2 314 −21−6 9

= Null

nullity=1

[7 1 0−3 0 1

]





L


L = Col

rank=1

2−14

6

= Col

rank=1

−2 314 −21−6 9

= Null

nullity=1

[7 1 0−3 0 1

]





L


L = Col

rank=1

2−14

6

= Col

rank=1

−2 314 −21−6 9

= Null

nullity=1

[7 1 0−3 0 1

]





L


L = Col

rank=1

2−14

6

= Col

rank=1

−2 314 −21−6 9

= Null

nullity=1

[7 1 0−3 0 1

]





L


L = Col

rank=1 2−14

6

= Col

rank=1

−2 314 −21−6 9

= Null

nullity=1

[7 1 0−3 0 1

]





L


L = Col

rank=1 2−14

6

= Col

rank=1 −2 314 −21−6 9

= Null

nullity=1

[7 1 0−3 0 1

]





L


L = Col

rank=1 2−14

6

= Col

rank=1 −2 314 −21−6 9

= Null

nullity=1[7 1 0−3 0 1

]



QuestionWhat does P = Span{〈3, −2, 1〉 , 〈−6, 5, 5〉 } “look like”?

AnswerP consists of all linear combinations of〈3, −2, 1〉 and 〈−6, 5, 5〉 . P “lookslike” a plane in R3.

P

NoteWe can represent P in many different ways.

P = Col

rank=2

6 18−4 −15

2 −15

= Col

rank=2

3 −6 −3−2 5 3

1 5 6

= Null

nullity=2

[5 7 −1

]

We only “need” two #‰v 1 and #‰v 2 to define P = Span{ #‰v 1,#‰v 2}.



AnswerP consists of all linear combinations of〈3, −2, 1〉 and 〈−6, 5, 5〉 .

P “lookslike” a plane in R3.

P


P = Col

rank=2

6 18−4 −15

2 −15

= Col

rank=2

3 −6 −3−2 5 3

1 5 6

= Null

nullity=2

[5 7 −1

]






P


P = Col

rank=2

6 18−4 −15

2 −15

= Col

rank=2

3 −6 −3−2 5 3

1 5 6

= Null

nullity=2

[5 7 −1

]






P


P = Col

rank=2

6 18−4 −15

2 −15

= Col

rank=2

3 −6 −3−2 5 3

1 5 6

= Null

nullity=2

[5 7 −1

]






P


P = Col

rank=2

6 18−4 −15

2 −15

= Col

rank=2

3 −6 −3−2 5 3

1 5 6

= Null

nullity=2

[5 7 −1

]





P


P = Col

rank=2

6 18−4 −15

2 −15

= Col

rank=2

3 −6 −3−2 5 3

1 5 6

= Null

nullity=2

[5 7 −1

]





P


P = Col

rank=2

6 18−4 −15

2 −15

= Col

rank=2

3 −6 −3−2 5 3

1 5 6

= Null

nullity=2

[5 7 −1

]





P


P = Col

rank=2

6 18−4 −15

2 −15

= Col

rank=2

3 −6 −3−2 5 3

1 5 6

= Null

nullity=2

[5 7 −1

]





P


P = Col

rank=2

6 18−4 −15

2 −15

= Col

rank=2

3 −6 −3−2 5 3

1 5 6

= Null

nullity=2

[5 7 −1

]





P


P = Col

rank=2

6 18−4 −15

2 −15

= Col

rank=2

3 −6 −3−2 5 3

1 5 6

= Null

nullity=2

[5 7 −1

]





P


P = Col

rank=2

6 18−4 −15

2 −15

= Col

rank=2

3 −6 −3−2 5 3

1 5 6

= Null

nullity=2

[5 7 −1

]We only “need” two #‰v 1 and #‰v 2 to define P = Span{ #‰v 1,

#‰v 2}.




P


P = Col

rank=2 6 18−4 −15

2 −15

= Col

rank=2

3 −6 −3−2 5 3

1 5 6

= Null

nullity=2

[5 7 −1


#‰v 2}.




P


P = Col

rank=2 6 18−4 −15

2 −15

= Col

rank=2 3 −6 −3−2 5 3

1 5 6

= Null

nullity=2

[5 7 −1


#‰v 2}.




P


P = Col

rank=2 6 18−4 −15

2 −15

= Col

rank=2 3 −6 −3−2 5 3

1 5 6

= Nullnullity=2[

5 7 −1]



ObservationGeometrically, the “size” of L is one and the “size” of P is two.

QuestionIs there a useful way to measure the “size” of a vector space V ?

AnswerThe dimension of a vector space V measures its “size.” To definedim(V ), we need to define the concept of a basis of V .








AnswerThe dimension of a vector space V measures its “size.”

To definedim(V ), we need to define the concept of a basis of V .





Definitions and PropertiesDefinition of a Basis

DefinitionSuppose that V = Span{ #‰v 1,

#‰v 2, . . . ,#‰v d}. We say that the list

{ #‰v 1,#‰v 2, . . . ,

#‰v d} is a basis of V if it is linearly independent.


Example

Consider the vectors #‰e 1, #‰e 2, and #‰e 3 given by

#‰e 1 = 〈1, 0, 0〉 #‰e 2 = 〈0, 1, 0〉 #‰e 3 = 〈0, 0, 1〉

Note that every #‰x ∈ R3 may be written as

#‰x = 〈x1, x2, x3〉 = x1 · 〈1, 0, 0〉 + x2 · 〈0, 1, 0〉 + x3 · 〈0, 0, 1〉

This means that R3 = Span{ #‰e 1,#‰e 2,

#‰e 3}. Moreover, the equation

rank[

#‰e 1#‰e 2

#‰e 3

]= 3

implies that { #‰e 1,#‰e 2,

#‰e 3} is linearly independent. This means that{ #‰e 1,

#‰e 2,#‰e 3} is a basis of R3.


Example


#‰e 1 = 〈1, 0, 0〉 #‰e 2 = 〈0, 1, 0〉 #‰e 3 = 〈0, 0, 1〉


#‰x = 〈x1, x2, x3〉 = x1 · 〈1, 0, 0〉 + x2 · 〈0, 1, 0〉 + x3 · 〈0, 0, 1〉



rank[

#‰e 1#‰e 2

#‰e 3

]= 3





Example


#‰e 1 = 〈1, 0, 0〉 #‰e 2 = 〈0, 1, 0〉 #‰e 3 = 〈0, 0, 1〉


#‰x = 〈x1, x2, x3〉 = x1 · 〈1, 0, 0〉 + x2 · 〈0, 1, 0〉 + x3 · 〈0, 0, 1〉


#‰e 3}.

Moreover, the equation

rank[

#‰e 1#‰e 2

#‰e 3

]= 3





Example


#‰e 1 = 〈1, 0, 0〉 #‰e 2 = 〈0, 1, 0〉 #‰e 3 = 〈0, 0, 1〉


#‰x = 〈x1, x2, x3〉 = x1 · 〈1, 0, 0〉 + x2 · 〈0, 1, 0〉 + x3 · 〈0, 0, 1〉



rank[

#‰e 1#‰e 2

#‰e 3

]= 3


#‰e 3} is linearly independent.

This means that{ #‰e 1,



Example


#‰e 1 = 〈1, 0, 0〉 #‰e 2 = 〈0, 1, 0〉 #‰e 3 = 〈0, 0, 1〉


#‰x = 〈x1, x2, x3〉 = x1 · 〈1, 0, 0〉 + x2 · 〈0, 1, 0〉 + x3 · 〈0, 0, 1〉



rank[

#‰e 1#‰e 2

#‰e 3

]= 3





TheoremThe list { #‰e 1,

#‰e 2, . . . ,#‰e n} is a basis of Rn.

Definitions and PropertiesProperties of Bases

TheoremSuppose that { #‰v 1, . . . ,

#‰v d} is a basis of V and that { #‰w1, . . . ,#‰wk}

is linearly independent in V . Then k ≤ d .

Example

Recall that { #‰e 1,#‰e 2,

#‰e 3} is a basis of R3. This theorem imlies thatthe columns of 1 40 1 −1 24

1 −1 3 −4 40 −1 0 2 −2

form a linearly dependent list because 5 > 3.





Example


#‰e 3} is a basis of R3.

This theorem imlies thatthe columns of 1 40 1 −1 24

1 −1 3 −4 40 −1 0 2 −2






Example



1 −1 3 −4 40 −1 0 2 −2

form a linearly

dependent list because 5 > 3.





Example



1 −1 3 −4 40 −1 0 2 −2

form a linearly dependent list because 5 >

3.





Example



1 −1 3 −4 40 −1 0 2 −2




#‰v k} and { #‰w1, . . . ,#‰w`} are bases of V .

Then k = `.

NoteThis theorem says that any two bases of V have the same numberof vectors.



#‰v k} and { #‰w1, . . . ,#‰w`} are bases of V .

Then k = `.

NoteThis theorem says that any two bases of V have the same numberof vectors.

Definitions and PropertiesDefinition of Dimension

DefinitionSuppose that { #‰v 1, . . . ,

#‰v d} is a basis of V . The dimension of V isdim(V ) = d .

NoteThe dimension of V is unambiguous since any two bases of V havethe same number of vectors.

IntuitionThe dimension of V is a measurement of how “large” V is.

Example

dim(Rn) = n because { #‰e 1, . . . ,#‰e n} is a basis of Rn.






Example







Example







Example

dim(Rn) =

n because { #‰e 1, . . . ,#‰e n} is a basis of Rn.






Example

dim(Rn) = n

because { #‰e 1, . . . ,#‰e n} is a basis of Rn.






Example


Definitions and PropertiesExamples

ExampleConsider the vector space V given by

V = Col

rank=1

−231

What is the dimension of V ?

V

SolutionNote that V = Span{〈−2, 3, 1〉 } and {〈−2, 3, 1〉 } is linearlyindependent.

So, {〈−2, 3, 1〉 } is a basis of V and dim(V ) = 1.

Math-Speak

“V is a one-dimensional vector subspace of R3.”



V = Col

rank=1

−231


V



Math-Speak




V = Col

rank=1

−231


V



Math-Speak




V = Col

rank=1

−231


V



Math-Speak




V = Col

rank=1

−231


V

SolutionNote that V = Span{〈−2, 3, 1〉 } and {〈−2, 3, 1〉 } is linearlyindependent. So, {〈−2, 3, 1〉 } is a basis of V and dim(V ) = 1.

Math-Speak




V = Col

rank=1

−231


V


Math-Speak




V = Col

rank=1 −231


V


Math-Speak




V = Col

rank=2

−2 1

4 −113 −4−1 3


V

SolutionLet β = {〈−2, 4, 3, −1〉 , 〈1, −11, −4, 3〉 }.

Note that β is linearlyindependent and spans V . So, β is a basis of V and dim(V ) = 2.

Math-Speak

“V is a two-dimensional vector subspace of R4.”



V = Col

rank=2

−2 1

4 −113 −4−1 3


V

SolutionLet β = {〈−2, 4, 3, −1〉 , 〈1, −11, −4, 3〉 }.


Math-Speak




V = Col

rank=2

−2 1

4 −113 −4−1 3


V

SolutionLet β = {〈−2, 4, 3, −1〉 , 〈1, −11, −4, 3〉 }.


Math-Speak




V = Col

rank=2

−2 1

4 −113 −4−1 3


V

SolutionLet β = {〈−2, 4, 3, −1〉 , 〈1, −11, −4, 3〉 }.


Math-Speak




V = Col

rank=2

−2 1

4 −113 −4−1 3


V

SolutionLet β = {〈−2, 4, 3, −1〉 , 〈1, −11, −4, 3〉 }.


Math-Speak




V = Col

rank=2

−2 1

4 −113 −4−1 3


V

SolutionLet β = {〈−2, 4, 3, −1〉 , 〈1, −11, −4, 3〉 }. Note that β is linearlyindependent and spans V .

So, β is a basis of V and dim(V ) = 2.

Math-Speak




V = Col

rank=2

−2 1

4 −113 −4−1 3


V

SolutionLet β = {〈−2, 4, 3, −1〉 , 〈1, −11, −4, 3〉 }. Note that β is linearlyindependent and spans V . So, β is a basis of V and dim(V ) =

2.

Math-Speak




V = Col

rank=2

−2 1

4 −113 −4−1 3


V

SolutionLet β = {〈−2, 4, 3, −1〉 , 〈1, −11, −4, 3〉 }. Note that β is linearlyindependent and spans V . So, β is a basis of V and dim(V ) = 2.

Math-Speak




V = Col

rank=2

−2 1

4 −113 −4−1 3


V


Math-Speak




V = Col

rank=2−2 1

4 −113 −4−1 3


V


Math-Speak



Example

Consider the vector space V given by

V = Col

rank(A)=4

2 7 −30 179−3 −2 4 −21

1 6 −27 162−2 0 −1 12

1 4 −15 900 1 −2 20


SolutionThe columns of A are linearly independent and span V . So, thecolumns of A form a basis of V and dim(V ) = 4.

Math-Speak

“V is a four-dimensional vector subspace of R6.”


Example


V = Col

rank(A)=4

2 7 −30 179−3 −2 4 −21

1 6 −27 162−2 0 −1 12

1 4 −15 900 1 −2 20


SolutionThe columns of A are linearly independent and span V .

So, thecolumns of A form a basis of V and dim(V ) = 4.

Math-Speak



Example


V = Col

rank(A)=4

2 7 −30 179−3 −2 4 −21

1 6 −27 162−2 0 −1 12

1 4 −15 900 1 −2 20



Math-Speak



Example


V = Col

rank(A)=4

2 7 −30 179−3 −2 4 −21

1 6 −27 162−2 0 −1 12

1 4 −15 900 1 −2 20



Math-Speak


The Four Fundamental SubspacesBases of Column Spaces

QuestionsHow can we find a basis of Col(A)?

What is dim Col(A)?

Rn

Row(A)

Null(A)

Rm

Col(A)

dim=?

LNull(A)

A


QuestionsHow can we find a basis of Col(A)? What is dim Col(A)?

Rn

Row(A)

Null(A)

Rm

Col(A)

dim=?

LNull(A)

A



Rn

Row(A)

Null(A)

Rm

Col(A)

dim=?

LNull(A)

A



Rn

Row(A)

Null(A)

Rm

Col(A)dim=?

LNull(A)

A


ObservationThe columns of A span Col(A). However, the columns of A mightnot be linearly independent.

Idea“Purge” the nonpivot columns of A to obtain a basis of Col(A).


ObservationThe columns of A span Col(A). However, the columns of A mightnot be linearly independent.

Idea“Purge” the nonpivot columns of A to obtain a basis of Col(A).


Theorem (“Pivot Bases of Col(A)”)

The pivot columns of A form a basis of Col(A).

Theorem (“Reduced Bases of Col(A)”)

The nonzero rows of rref(Aᵀ) form a basis of Col(A).

Theoremdim Col(A) = rank(A)














Example

Consider the computations

rref

A1 1 −23 4 −7−3 −3 6−2 3 −1

=

1 0 −10 1 −10 0 00 0 0

rref

Aᵀ 1 3 −3 −21 4 −3 3−2 −7 6 −1

=

1 0 −3 −170 1 0 50 0 0 0

The “pivot basis” and the “reduced basis” of Col(A) are given by

“pivot basis”

{〈1, 3, −3, −2〉 , 〈1, 4, −3, 3〉 }“reduced basis”

{〈1, 0, −3, −17〉 , 〈0, 1, 0, 5〉 }

Note that dim Col(A) = rank(A) = 2.

Math-Speak

“Col(A) is a two-dimensional vector subspace of R4.”


Example


rref

A1 1 −23 4 −7−3 −3 6−2 3 −1

=

1 0 −10 1 −10 0 00 0 0

rref

Aᵀ 1 3 −3 −21 4 −3 3−2 −7 6 −1

=

1 0 −3 −170 1 0 50 0 0 0


“pivot basis”

{〈1, 3, −3, −2〉 , 〈1, 4, −3, 3〉 }“reduced basis”

{〈1, 0, −3, −17〉 , 〈0, 1, 0, 5〉 }


Math-Speak



Example


rref

A1 1 −23 4 −7−3 −3 6−2 3 −1

=

1 0 −10 1 −10 0 00 0 0

rref

Aᵀ 1 3 −3 −21 4 −3 3−2 −7 6 −1

=

1 0 −3 −170 1 0 50 0 0 0


“pivot basis”

{〈1, 3, −3, −2〉 , 〈1, 4, −3, 3〉 }“reduced basis”

{〈1, 0, −3, −17〉 , 〈0, 1, 0, 5〉 }


Math-Speak



Example


rref

A1 1 −23 4 −7−3 −3 6−2 3 −1

=

1 0 −10 1 −10 0 00 0 0

rref

Aᵀ 1 3 −3 −21 4 −3 3−2 −7 6 −1

=

1 0 −3 −170 1 0 50 0 0 0


“pivot basis”

{〈1, 3, −3, −2〉 , 〈1, 4, −3, 3〉 }“reduced basis”

{〈1, 0, −3, −17〉 , 〈0, 1, 0, 5〉 }


Math-Speak



Example


V = Span

−5

20

, −20

80

, −3

1−3

, 22−812

, −16

6−6

Note that V = Col(A) where

rref

A −5 −20 −3 22 −162 8 1 −8 60 0 −3 12 −6

=

1 4 0 −2 20 0 1 −4 20 0 0 0 0

So dim(V ) = 2 and {〈−5, 2, 0〉 , 〈−3, 1, −3〉 } is a basis of V .


Example


V = Span

−5

20

, −20

80

, −3

1−3

, 22−812

, −16

6−6


rref

A −5 −20 −3 22 −162 8 1 −8 60 0 −3 12 −6

=

1 4 0 −2 20 0 1 −4 20 0 0 0 0



Example


V = Span

−5

20

, −20

80

, −3

1−3

, 22−812

, −16

6−6


rref

A −5 −20 −3 22 −162 8 1 −8 60 0 −3 12 −6

=

1 4 0 −2 20 0 1 −4 20 0 0 0 0

So dim(V ) =

2 and {〈−5, 2, 0〉 , 〈−3, 1, −3〉 } is a basis of V .


Example


V = Span

−5

20

, −20

80

, −3

1−3

, 22−812

, −16

6−6


rref

A −5 −20 −3 22 −162 8 1 −8 60 0 −3 12 −6

=

1 4 0 −2 20 0 1 −4 20 0 0 0 0

So dim(V ) = 2

and {〈−5, 2, 0〉 , 〈−3, 1, −3〉 } is a basis of V .


Example


V = Span

−5

20

, −20

80

, −3

1−3

, 22−812

, −16

6−6


rref

A −5 −20 −3 22 −162 8 1 −8 60 0 −3 12 −6

=

1 4 0 −2 20 0 1 −4 20 0 0 0 0


The Four Fundamental SubspacesBases of Null Spaces

QuestionsHow can we find a basis of Null(A)?

What is dim Null(A)?

Rn

Row(A)

Null(A)

dim=?

Rm

Col(A)

dim= rank(A)

LNull(A)

A


QuestionsHow can we find a basis of Null(A)? What is dim Null(A)?

Rn

Row(A)

Null(A)

dim=?

Rm

Col(A)

dim= rank(A)

LNull(A)

A



Rn

Row(A)

Null(A)

dim=?

Rm

Col(A)

dim= rank(A)

LNull(A)

A



Rn

Row(A)

Null(A)

dim=?

Rm

Col(A)dim=

rank(A)

LNull(A)

A



Rn

Row(A)

Null(A)

dim=?

Rm

Col(A)dim= rank(A)

LNull(A)

A



Rn

Row(A)

Null(A)dim=?

Rm

Col(A)dim= rank(A)

LNull(A)

A


ObservationWhen solving A #‰x =

#‰

O , the number of “degrees of freedom” is thenumber of free variables.

This suggests dim Null(A) = nullity(A).



#‰

O , the number of “degrees of freedom” is thenumber of free variables. This suggests dim Null(A) =

nullity(A).



#‰

O , the number of “degrees of freedom” is thenumber of free variables. This suggests dim Null(A) = nullity(A).


Example

Consider the general solution #‰x to A #‰x =#‰

O given by

A =

1 −2 0 −20 0 1 30 0 0 0

#‰x =

2 c1 + 2 c2

c1−3 c2

c2

= c1

2100

+ c2

20−3

1

This shows that

Null

A 1 −2 0 −20 0 1 30 0 0 0

= Col

rank(B)=22 21 00 −30 1

The columns of B form a basis of Null(A). So dim Null(A) = 2.


Example


O given by

A =

1 −2 0 −20 0 1 30 0 0 0

#‰x =

2 c1 + 2 c2

c1−3 c2

c2

= c1

2100

+ c2

20−3

1

This shows that

Null

A 1 −2 0 −20 0 1 30 0 0 0

= Col

rank(B)=22 21 00 −30 1



Example


O given by

A =

1 −2 0 −20 0 1 30 0 0 0

#‰x =

2 c1 + 2 c2

c1−3 c2

c2

= c1

2100

+ c2

20−3

1

This shows that

Null

A 1 −2 0 −20 0 1 30 0 0 0

= Col

rank(B)=22 21 00 −30 1

The columns of B form a basis of Null(A).

So dim Null(A) = 2.


Example


O given by

A =

1 −2 0 −20 0 1 30 0 0 0

#‰x =

2 c1 + 2 c2

c1−3 c2

c2

= c1

2100

+ c2

20−3

1

This shows that

Null

A 1 −2 0 −20 0 1 30 0 0 0

= Col

rank(B)=22 21 00 −30 1

The columns of B form a basis of Null(A). So dim Null(A) =

2.


Example


O given by

A =

1 −2 0 −20 0 1 30 0 0 0

#‰x =

2 c1 + 2 c2

c1−3 c2

c2

= c1

2100

+ c2

20−3

1

This shows that

Null

A 1 −2 0 −20 0 1 30 0 0 0

= Col

rank(B)=22 21 00 −30 1



Theorem (“Pivot Bases of Null(A)”)

Use rref(A) to write the general solution #‰x to A #‰x =#‰

O as

#‰x = c1 · #‰v 1 + c2 · #‰v 2 + · · ·+ cd · #‰v d

Then { #‰v 1,#‰v d , . . . ,

#‰v d} form a basis of Null(A).

Theoremdim Null(A) = nullity(A)


Theorem (“Pivot Bases of Null(A)”)

Use rref(A) to write the general solution #‰x to A #‰x =#‰

O as

#‰x = c1 · #‰v 1 + c2 · #‰v 2 + · · ·+ cd · #‰v d

Then { #‰v 1,#‰v d , . . . ,

#‰v d} form a basis of Null(A).

Theoremdim Null(A) = nullity(A)


Example

Consider the computation

rref

A 3 −24 18−2 16 −12

8 −64 48

=

1 −8 60 0 00 0 0

So dim Null(A) = nullity(A) = 2. Solving A #‰x =#‰

O gives

#‰x =

x1x2x3

=

8 c1 − 6 c2c1c2

= c1

810

+ c2

−601

The “pivot basis” of Null(A) is {〈8, 1, 0〉 , 〈−6, 0, 1〉 }.


Example


rref

A 3 −24 18−2 16 −12

8 −64 48

=

1 −8 60 0 00 0 0

So dim Null(A) = nullity(A) =

2. Solving A #‰x =#‰

O gives

#‰x =

x1x2x3

=

8 c1 − 6 c2c1c2

= c1

810

+ c2

−601



Example


rref

A 3 −24 18−2 16 −12

8 −64 48

=

1 −8 60 0 00 0 0

So dim Null(A) = nullity(A) = 2.

Solving A #‰x =#‰

O gives

#‰x =

x1x2x3

=

8 c1 − 6 c2c1c2

= c1

810

+ c2

−601



Example


rref

A 3 −24 18−2 16 −12

8 −64 48

=

1 −8 60 0 00 0 0

So dim Null(A) = nullity(A) = 2. Solving A #‰x =

#‰

O gives

#‰x =

x1x2x3

=

8 c1 − 6 c2c1c2

= c1

810

+ c2

−601



Example


rref

A 3 −24 18−2 16 −12

8 −64 48

=

1 −8 60 0 00 0 0

So dim Null(A) = nullity(A) = 2. Solving A #‰x =

#‰

O gives

#‰x =

x1x2x3

=

8 c1 − 6 c2c1c2

= c1

810

+ c2

−601


The Four Fundamental SubspacesBases of Row Spaces

QuestionsHow can we find a basis of Row(A)?

What is dim Row(A)?

Rn

Row(A)

dim=?

Null(A)

dim=n−rank(A)

Rm

Col(A)

dim=rank(A)

LNull(A)

A


QuestionsHow can we find a basis of Row(A)? What is dim Row(A)?

Rn

Row(A)

dim=?

Null(A)

dim=n−rank(A)

Rm

Col(A)

dim=rank(A)

LNull(A)

A



Rn

Row(A)

dim=?

Null(A)

dim=n−rank(A)

Rm

Col(A)

dim=rank(A)

LNull(A)

A



Rn

Row(A)

dim=?

Null(A)

dim=n−rank(A)

Rm

Col(A)dim=rank(A)

LNull(A)

A



Rn

Row(A)

dim=?

Null(A)dim=

n−rank(A)

Rm

Col(A)dim=rank(A)

LNull(A)

A



Rn

Row(A)

dim=?

Null(A)dim=n−rank(A)

Rm

Col(A)dim=rank(A)

LNull(A)

A



Rn

Row(A)dim=?


Rm

Col(A)dim=rank(A)

LNull(A)

A


RecallRow(A) = Col(Aᵀ)

Theorem (“Pivot Basis” of Row(A))

The pivot columns of Aᵀ form a basis of Row(A).

Theorem (“Reduced Basis” of Row(A))

The nonzero rows of rref(A) form a basis of Row(A).

Theoremdim Row(A) = rank(A)























Example


rref

A4 9 50 1 13 4 13 7 4

=

1 0 −10 1 10 0 00 0 0

rref

Aᵀ 4 0 3 39 1 4 75 1 1 4

=

1 0 3/4 3/40 1 −11/4 1/40 0 0 0

The “pivot basis” and the “reduced basis” of Row(A) are given by

“pivot basis”

{〈4, 9, 5〉 , 〈0, 1, 1〉}“reduced basis”

{〈1, 0, −1〉 , 〈0, 1, 1〉 }

Note that dim Row(A) = rank(A) = 2.

Math-Speak

“Row(A) is a two-dimensional vector subspace of R3.”


Example


rref

A4 9 50 1 13 4 13 7 4

=

1 0 −10 1 10 0 00 0 0

rref

Aᵀ 4 0 3 39 1 4 75 1 1 4

=

1 0 3/4 3/40 1 −11/4 1/40 0 0 0


“pivot basis”

{〈4, 9, 5〉 , 〈0, 1, 1〉}“reduced basis”

{〈1, 0, −1〉 , 〈0, 1, 1〉 }


Math-Speak



Example


rref

A4 9 50 1 13 4 13 7 4

=

1 0 −10 1 10 0 00 0 0

rref

Aᵀ 4 0 3 39 1 4 75 1 1 4

=

1 0 3/4 3/40 1 −11/4 1/40 0 0 0


“pivot basis”

{〈4, 9, 5〉 , 〈0, 1, 1〉}“reduced basis”

{〈1, 0, −1〉 , 〈0, 1, 1〉 }

Note that dim Row(A) =

rank(A) = 2.

Math-Speak



Example


rref

A4 9 50 1 13 4 13 7 4

=

1 0 −10 1 10 0 00 0 0

rref

Aᵀ 4 0 3 39 1 4 75 1 1 4

=

1 0 3/4 3/40 1 −11/4 1/40 0 0 0


“pivot basis”

{〈4, 9, 5〉 , 〈0, 1, 1〉}“reduced basis”

{〈1, 0, −1〉 , 〈0, 1, 1〉 }

Note that dim Row(A) = rank(A) =

2.

Math-Speak



Example


rref

A4 9 50 1 13 4 13 7 4

=

1 0 −10 1 10 0 00 0 0

rref

Aᵀ 4 0 3 39 1 4 75 1 1 4

=

1 0 3/4 3/40 1 −11/4 1/40 0 0 0


“pivot basis”

{〈4, 9, 5〉 , 〈0, 1, 1〉}“reduced basis”

{〈1, 0, −1〉 , 〈0, 1, 1〉 }


Math-Speak



Example


rref

A4 9 50 1 13 4 13 7 4

=

1 0 −10 1 10 0 00 0 0

rref

Aᵀ 4 0 3 39 1 4 75 1 1 4

=

1 0 3/4 3/40 1 −11/4 1/40 0 0 0


“pivot basis”

{〈4, 9, 5〉 , 〈0, 1, 1〉}“reduced basis”

{〈1, 0, −1〉 , 〈0, 1, 1〉 }


Math-Speak


The Four Fundamental SubspacesBases of Left Null Spaces

QuestionsHow can we find a basis of LNull(A)?

What is dim LNull(A)?

Rn

Row(A)

dim= rank(A)

Null(A)

dim=n−rank(A)

Rm

Col(A)

dim=rank(A)

LNull(A)

dim=?

A


QuestionsHow can we find a basis of LNull(A)? What is dim LNull(A)?

Rn

Row(A)

dim= rank(A)

Null(A)

dim=n−rank(A)

Rm

Col(A)

dim=rank(A)

LNull(A)

dim=?

A



Rn

Row(A)

dim= rank(A)

Null(A)

dim=n−rank(A)

Rm

Col(A)

dim=rank(A)

LNull(A)

dim=?

A



Rn

Row(A)

dim= rank(A)

Null(A)

dim=n−rank(A)

Rm

Col(A)dim=rank(A)

LNull(A)

dim=?

A



Rn

Row(A)

dim= rank(A)


Rm

Col(A)dim=rank(A)

LNull(A)

dim=?

A



Rn

Row(A)dim=

rank(A)


Rm

Col(A)dim=rank(A)

LNull(A)

dim=?

A



Rn

Row(A)dim= rank(A)


Rm

Col(A)dim=rank(A)

LNull(A)

dim=?

A



Rn

Row(A)dim= rank(A)


Rm

Col(A)dim=rank(A)

LNull(A)dim=?

A


RecallLNull(A) = Null(Aᵀ)

Theorem (“Pivot Basis” of LNull(A))

Use rref(Aᵀ) to write the general solution #‰x to Aᵀ #‰x =#‰

O as

#‰x = c1 · #‰v 1 + c2 · #‰v 2 + · · ·+ cd · #‰v d

Then { #‰v 1,#‰v d , . . . ,

#‰v d} form a basis of LNull(A).

Theoremdim LNull(A) = nullity(Aᵀ) = corank(A)





O as

#‰x = c1 · #‰v 1 + c2 · #‰v 2 + · · ·+ cd · #‰v d

Then { #‰v 1,#‰v d , . . . ,







O as

#‰x = c1 · #‰v 1 + c2 · #‰v 2 + · · ·+ cd · #‰v d

Then { #‰v 1,#‰v d , . . . ,


Theoremdim LNull(A) = nullity(Aᵀ)

= corank(A)





O as

#‰x = c1 · #‰v 1 + c2 · #‰v 2 + · · ·+ cd · #‰v d

Then { #‰v 1,#‰v d , . . . ,




Example


rref

Aᵀ3 −21 33−6 42 −66

5 −35 552 −14 22

=

1 −7 110 0 00 0 00 0 0

So dim LNull(A) = nullity(Aᵀ) = 2. Solving Aᵀ #‰x =#‰

O gives

#‰x =

x1x2x3

=

7 c1 − 11 c2c1c2

= c1

710

+ c2

−1101

The “pivot basis” of LNull(A) is {〈7, 1, 0〉 , 〈−11, 0, 1〉 }.


Example


rref

Aᵀ3 −21 33−6 42 −66

5 −35 552 −14 22

=

1 −7 110 0 00 0 00 0 0

So dim LNull(A) =

nullity(Aᵀ) = 2. Solving Aᵀ #‰x =#‰

O gives

#‰x =

x1x2x3

=

7 c1 − 11 c2c1c2

= c1

710

+ c2

−1101



Example


rref

Aᵀ3 −21 33−6 42 −66

5 −35 552 −14 22

=

1 −7 110 0 00 0 00 0 0

So dim LNull(A) = nullity(Aᵀ) =

2. Solving Aᵀ #‰x =#‰

O gives

#‰x =

x1x2x3

=

7 c1 − 11 c2c1c2

= c1

710

+ c2

−1101



Example


rref

Aᵀ3 −21 33−6 42 −66

5 −35 552 −14 22

=

1 −7 110 0 00 0 00 0 0

So dim LNull(A) = nullity(Aᵀ) = 2.

Solving Aᵀ #‰x =#‰

O gives

#‰x =

x1x2x3

=

7 c1 − 11 c2c1c2

= c1

710

+ c2

−1101



Example


rref

Aᵀ3 −21 33−6 42 −66

5 −35 552 −14 22

=

1 −7 110 0 00 0 00 0 0

So dim LNull(A) = nullity(Aᵀ) = 2. Solving Aᵀ #‰x =

#‰

O gives

#‰x =

x1x2x3

=

7 c1 − 11 c2c1c2

= c1

710

+ c2

−1101



Example


rref

Aᵀ3 −21 33−6 42 −66

5 −35 552 −14 22

=

1 −7 110 0 00 0 00 0 0

So dim LNull(A) = nullity(Aᵀ) = 2. Solving Aᵀ #‰x =

#‰

O gives

#‰x =

x1x2x3

=

7 c1 − 11 c2c1c2

= c1

710

+ c2

−1101


The Rank-Nullity TheoremStatement

ObservationThe dimensions of the four fundamental subspaces of A can beinferred from rank(A).

Rn

Row(A)

dim= rank(A)

Null(A)

dim=n−rank(A)

Rm

Col(A)

dim= rank(A)

LNull(A)

dim=m−rank(A)

A



Rn

Row(A)

dim= rank(A)

Null(A)

dim=n−rank(A)

Rm

Col(A)dim=

rank(A)

LNull(A)

dim=m−rank(A)

A



Rn

Row(A)

dim= rank(A)

Null(A)

dim=n−rank(A)

Rm

Col(A)dim= rank(A)

LNull(A)

dim=m−rank(A)

A



Rn

Row(A)

dim= rank(A)

Null(A)dim=

n−rank(A)

Rm

Col(A)dim= rank(A)

LNull(A)

dim=m−rank(A)

A



Rn

Row(A)

dim= rank(A)


Rm

Col(A)dim= rank(A)

LNull(A)

dim=m−rank(A)

A



Rn

Row(A)dim=

rank(A)


Rm

Col(A)dim= rank(A)

LNull(A)

dim=m−rank(A)

A



Rn

Row(A)dim= rank(A)


Rm

Col(A)dim= rank(A)

LNull(A)

dim=m−rank(A)

A



Rn

Row(A)dim= rank(A)


Rm

Col(A)dim= rank(A)

LNull(A)dim=

m−rank(A)

A



Rn

Row(A)dim= rank(A)


Rm

Col(A)dim= rank(A)

LNull(A)dim=m−rank(A)

A


RecallEvery m × n matrix A satisfies rank(A) + nullity(A) = n.

Theorem (The Rank-Nullity Theorem)

Every m × n matrix A satisfies dim Col(A) + dim Null(A) = n.


RecallEvery m × n matrix A satisfies rank(A) + nullity(A) = n.

Theorem (The Rank-Nullity Theorem)

Every m × n matrix A satisfies dim Col(A) + dim Null(A) = n.

The Rank-Nullity TheoremExample

Example

Construct a matrix A satisfying

〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)

Note that A is 3× 3.

Rn

=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm

=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A

SolutionThe lists {〈1, −3, 2〉 , 〈−4, 6, 8〉 } and {〈2, 3, 9〉 , 〈7, −8, 4〉 } arelinearly independent, so dim Col(A) ≥ 2 and dim Null(A) ≥ 2. Butnow

dim Col(A) + dim Null(A) ≥ 2 + 2 = 4 > 3

which contradicts the Rank-Nullity theorem. No such A exists!


Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn

=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm

=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A





Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn

=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm

=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A





Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn

=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm

=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A





Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn

=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A





Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A





Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A





Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A

SolutionThe lists {〈1, −3, 2〉 , 〈−4, 6, 8〉 } and {〈2, 3, 9〉 , 〈7, −8, 4〉 } arelinearly independent

, so dim Col(A) ≥ 2 and dim Null(A) ≥ 2. Butnow




Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A

SolutionThe lists {〈1, −3, 2〉 , 〈−4, 6, 8〉 } and {〈2, 3, 9〉 , 〈7, −8, 4〉 } arelinearly independent, so dim Col(A) ≥

2 and dim Null(A) ≥ 2. Butnow




Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A

SolutionThe lists {〈1, −3, 2〉 , 〈−4, 6, 8〉 } and {〈2, 3, 9〉 , 〈7, −8, 4〉 } arelinearly independent, so dim Col(A) ≥ 2

and dim Null(A) ≥ 2. Butnow




Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A

SolutionThe lists {〈1, −3, 2〉 , 〈−4, 6, 8〉 } and {〈2, 3, 9〉 , 〈7, −8, 4〉 } arelinearly independent, so dim Col(A) ≥ 2 and dim Null(A) ≥

2. Butnow




Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A

SolutionThe lists {〈1, −3, 2〉 , 〈−4, 6, 8〉 } and {〈2, 3, 9〉 , 〈7, −8, 4〉 } arelinearly independent, so dim Col(A) ≥ 2 and dim Null(A) ≥ 2.

Butnow




Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A


dim Col(A) + dim Null(A) ≥

2 + 2 = 4 > 3



Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A


dim Col(A) + dim Null(A) ≥ 2 + 2 =

4 > 3



Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A


dim Col(A) + dim Null(A) ≥ 2 + 2 = 4 >

3



Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A





Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A



which contradicts the Rank-Nullity theorem.

No such A exists!


Example


〈1, −3, 2〉 , 〈−4, 6, 8〉 ∈ Col(A)

〈2, 3, 9〉 , 〈7, −8, 4〉 ∈ Null(A)


Rn=3

Row(A)

Null(A)

〈2, 3, 9〉〈7, −8, 4〉

Rm=3

Col(A)

〈1, −3, 2〉〈−4, 6, 8〉

LNull(A)

A




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Bases and Dimension - Duke Universitybfitzpat/teaching/218s20/lectures/... · Math 218 Brian D. Fitzpatrick Duke University March 1, 2020 MATH. Overview Geometric Motivation Visualizing