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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
Geometric MotivationVisualizing Vector Spaces
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 }.
Geometric MotivationVisualizing Vector Spaces
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 }.
Geometric MotivationVisualizing Vector Spaces
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 }.
Geometric MotivationVisualizing Vector Spaces
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 }.
Geometric MotivationVisualizing Vector Spaces
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 }.
Geometric MotivationVisualizing Vector Spaces
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 }.
Geometric MotivationVisualizing Vector Spaces
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 }.
Geometric MotivationVisualizing Vector Spaces
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 }.
Geometric MotivationVisualizing Vector Spaces
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 }.
Geometric MotivationVisualizing Vector Spaces
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 }.
Geometric MotivationVisualizing Vector Spaces
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 }.
Geometric MotivationVisualizing Vector Spaces
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 }.
Geometric MotivationVisualizing Vector Spaces
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 }.
Geometric MotivationVisualizing Vector Spaces
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}.
Geometric MotivationVisualizing Vector Spaces
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}.
Geometric MotivationVisualizing Vector Spaces
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}.
Geometric MotivationVisualizing Vector Spaces
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}.
Geometric MotivationVisualizing Vector Spaces
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}.
Geometric MotivationVisualizing Vector Spaces
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}.
Geometric MotivationVisualizing Vector Spaces
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}.
Geometric MotivationVisualizing Vector Spaces
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}.
Geometric MotivationVisualizing Vector Spaces
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}.
Geometric MotivationVisualizing Vector Spaces
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}.
Geometric MotivationVisualizing Vector Spaces
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}.
Geometric MotivationVisualizing Vector Spaces
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}.
Geometric MotivationVisualizing Vector Spaces
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}.
Geometric MotivationVisualizing Vector Spaces
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
= Nullnullity=2[
5 7 −1]
We only “need” two #‰v 1 and #‰v 2 to define P = Span{ #‰v 1,#‰v 2}.
Geometric MotivationVisualizing Vector Spaces
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 .
Geometric MotivationVisualizing Vector Spaces
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 .
Geometric MotivationVisualizing Vector Spaces
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 .
Geometric MotivationVisualizing Vector Spaces
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 .
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.
Definitions and PropertiesDefinition of a Basis
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.
Definitions and PropertiesDefinition of a Basis
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.
Definitions and PropertiesDefinition of a Basis
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.
Definitions and PropertiesDefinition of a Basis
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.
Definitions and PropertiesDefinition of a Basis
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.
Definitions and PropertiesDefinition of a Basis
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.
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.
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.
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.
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.
Definitions and PropertiesProperties of Bases
TheoremSuppose that { #‰v 1, . . . ,
#‰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 PropertiesProperties of Bases
TheoremSuppose that { #‰v 1, . . . ,
#‰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.
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.
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.
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.
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.
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.
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.”
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.”
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.”
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.”
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.”
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.”
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.”
Definitions and PropertiesExamples
ExampleConsider the vector space V given by
V = Col
rank=2
−2 1
4 −113 −4−1 3
What is the dimension of V ?
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.”
Definitions and PropertiesExamples
ExampleConsider the vector space V given by
V = Col
rank=2
−2 1
4 −113 −4−1 3
What is the dimension of V ?
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.”
Definitions and PropertiesExamples
ExampleConsider the vector space V given by
V = Col
rank=2
−2 1
4 −113 −4−1 3
What is the dimension of V ?
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.”
Definitions and PropertiesExamples
ExampleConsider the vector space V given by
V = Col
rank=2
−2 1
4 −113 −4−1 3
What is the dimension of V ?
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.”
Definitions and PropertiesExamples
ExampleConsider the vector space V given by
V = Col
rank=2
−2 1
4 −113 −4−1 3
What is the dimension of V ?
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.”
Definitions and PropertiesExamples
ExampleConsider the vector space V given by
V = Col
rank=2
−2 1
4 −113 −4−1 3
What is the dimension of V ?
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.”
Definitions and PropertiesExamples
ExampleConsider the vector space V given by
V = Col
rank=2
−2 1
4 −113 −4−1 3
What is the dimension of V ?
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.”
Definitions and PropertiesExamples
ExampleConsider the vector space V given by
V = Col
rank=2
−2 1
4 −113 −4−1 3
What is the dimension of V ?
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.”
Definitions and PropertiesExamples
ExampleConsider the vector space V given by
V = Col
rank=2
−2 1
4 −113 −4−1 3
What is the dimension of V ?
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.”
Definitions and PropertiesExamples
ExampleConsider the vector space V given by
V = Col
rank=2−2 1
4 −113 −4−1 3
What is the dimension of V ?
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.”
Definitions and PropertiesExamples
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
What is the dimension of V ?
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.”
Definitions and PropertiesExamples
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
What is the dimension of V ?
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.”
Definitions and PropertiesExamples
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
What is the dimension of V ?
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.”
Definitions and PropertiesExamples
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
What is the dimension of V ?
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.”
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
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
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
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
The Four Fundamental SubspacesBases of Column Spaces
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).
The Four Fundamental SubspacesBases of Column Spaces
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).
The Four Fundamental SubspacesBases of Column Spaces
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)
The Four Fundamental SubspacesBases of Column Spaces
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)
The Four Fundamental SubspacesBases of Column Spaces
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)
The Four Fundamental SubspacesBases of Column Spaces
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.”
The Four Fundamental SubspacesBases of Column Spaces
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.”
The Four Fundamental SubspacesBases of Column Spaces
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.”
The Four Fundamental SubspacesBases of Column Spaces
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.”
The Four Fundamental SubspacesBases of Column Spaces
Example
Consider the vector space V given by
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 .
The Four Fundamental SubspacesBases of Column Spaces
Example
Consider the vector space V given by
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 .
The Four Fundamental SubspacesBases of Column Spaces
Example
Consider the vector space V given by
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 .
The Four Fundamental SubspacesBases of Column Spaces
Example
Consider the vector space V given by
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 .
The Four Fundamental SubspacesBases of Column Spaces
Example
Consider the vector space V given by
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 .
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
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
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
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
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
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
The Four Fundamental SubspacesBases of Null Spaces
ObservationWhen solving A #‰x =
#‰
O , the number of “degrees of freedom” is thenumber of free variables.
This suggests dim Null(A) = nullity(A).
The Four Fundamental SubspacesBases of Null Spaces
ObservationWhen solving A #‰x =
#‰
O , the number of “degrees of freedom” is thenumber of free variables. This suggests dim Null(A) =
nullity(A).
The Four Fundamental SubspacesBases of Null Spaces
ObservationWhen solving A #‰x =
#‰
O , the number of “degrees of freedom” is thenumber of free variables. This suggests dim Null(A) = nullity(A).
The Four Fundamental SubspacesBases of Null Spaces
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.
The Four Fundamental SubspacesBases of Null Spaces
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.
The Four Fundamental SubspacesBases of Null Spaces
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.
The Four Fundamental SubspacesBases of Null Spaces
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.
The Four Fundamental SubspacesBases of Null Spaces
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.
The Four Fundamental SubspacesBases of Null Spaces
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)
The Four Fundamental SubspacesBases of Null Spaces
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)
The Four Fundamental SubspacesBases of Null Spaces
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〉 }.
The Four Fundamental SubspacesBases of Null Spaces
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〉 }.
The Four Fundamental SubspacesBases of Null Spaces
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〉 }.
The Four Fundamental SubspacesBases of Null Spaces
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〉 }.
The Four Fundamental SubspacesBases of Null Spaces
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〉 }.
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
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
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
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
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
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
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
The Four Fundamental SubspacesBases of Row Spaces
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)
The Four Fundamental SubspacesBases of Row Spaces
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)
The Four Fundamental SubspacesBases of Row Spaces
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)
The Four Fundamental SubspacesBases of Row Spaces
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)
The Four Fundamental SubspacesBases of Row Spaces
Example
Consider the computations
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.”
The Four Fundamental SubspacesBases of Row Spaces
Example
Consider the computations
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.”
The Four Fundamental SubspacesBases of Row Spaces
Example
Consider the computations
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.”
The Four Fundamental SubspacesBases of Row Spaces
Example
Consider the computations
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.”
The Four Fundamental SubspacesBases of Row Spaces
Example
Consider the computations
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.”
The Four Fundamental SubspacesBases of Row Spaces
Example
Consider the computations
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.”
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
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
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
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
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
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
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
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
The Four Fundamental SubspacesBases of Left Null Spaces
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)
The Four Fundamental SubspacesBases of Left Null Spaces
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)
The Four Fundamental SubspacesBases of Left Null Spaces
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)
The Four Fundamental SubspacesBases of Left Null Spaces
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)
The Four Fundamental SubspacesBases of Left Null Spaces
Example
Consider the computation
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〉 }.
The Four Fundamental SubspacesBases of Left Null Spaces
Example
Consider the computation
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〉 }.
The Four Fundamental SubspacesBases of Left Null Spaces
Example
Consider the computation
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〉 }.
The Four Fundamental SubspacesBases of Left Null Spaces
Example
Consider the computation
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〉 }.
The Four Fundamental SubspacesBases of Left Null Spaces
Example
Consider the computation
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〉 }.
The Four Fundamental SubspacesBases of Left Null Spaces
Example
Consider the computation
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〉 }.
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
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
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
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
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
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
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
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
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
The Rank-Nullity TheoremStatement
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 TheoremStatement
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!
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!
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!
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!
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!
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!
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!
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!
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!
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!
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!
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!
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!
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!
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!
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!
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!
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!
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