Похожие презентации:
Span. Linear Algebra. Lecture 5
1.
SpanLinear Algebra
Lecture 5
Let V be a vector space, U ⊂ V a subset.
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2.
SpanLinear Algebra
Lecture 5
Let V be a vector space, U ⊂ V a subset.
Definition. span U is the set of all linear combinations of v1 , . . . , vm ∈ U
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3.
SpanLinear Algebra
Lecture 5
Let V be a vector space, U ⊂ V a subset.
Definition. span U is the set of all linear combinations of v1 , . . . , vm ∈ U
(with any m ).
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4.
SpanLinear Algebra
Lecture 5
Let V be a vector space, U ⊂ V a subset.
Definition. span U is the set of all linear combinations of v1 , . . . , vm ∈ U
(with any m ).
Theorem. span U is the smallest subspace which contains U .
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5.
SpanLinear Algebra
Lecture 5
Let V be a vector space, U ⊂ V a subset.
Definition. span U is the set of all linear combinations of v1 , . . . , vm ∈ U
(with any m ).
Theorem. span U is the smallest subspace which contains U .
What to prove?
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6.
SpanLinear Algebra
Lecture 5
Let V be a vector space, U ⊂ V a subset.
Definition. span U is the set of all linear combinations of v1 , . . . , vm ∈ U
(with any m ).
Theorem. span U is the smallest subspace which contains U .
What to prove?
span U is a subspace.
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7.
SpanLinear Algebra
Lecture 5
Let V be a vector space, U ⊂ V a subset.
Definition. span U is the set of all linear combinations of v1 , . . . , vm ∈ U
(with any m ).
Theorem. span U is the smallest subspace which contains U .
What to prove?
span U is a subspace.
span U is the smallest among subspaces.
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8.
Linear AlgebraLecture 5
Span
Let V be a vector space, U ⊂ V a subset.
Definition. span U is the set of all linear combinations of v1 , . . . , vm ∈ U
(with any m ).
Theorem. span U is the smallest subspace which contains U .
What to prove?
span U is a subspace.
span U is the smallest among subspaces.
Theorem. span U is the intersection of all subspaces which contain U .
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9.
Linear AlgebraLecture 5
Span
Let V be a vector space, U ⊂ V a subset.
Definition. span U is the set of all linear combinations of v1 , . . . , vm ∈ U
(with any m ).
Theorem. span U is the smallest subspace which contains U .
What to prove?
span U is a subspace.
span U is the smallest among subspaces.
Theorem. span U is the intersection of all subspaces which contain U .
Is the intersection of subspaces a subspace?
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10.
Linear dependence/independenceLinear Algebra
Lecture 5
Let V be a vector space, v1 , . . . , vm ∈ V be a list of vectors.
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11.
Linear dependence/independenceLinear Algebra
Lecture 5
Let V be a vector space, v1 , . . . , vm ∈ V be a list of vectors.
Definition. span(v1 , . . . , vm ) = {a1 v1 + · · · + am vm | a1 , . . . , am ∈ F}
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12.
Linear AlgebraLecture 5
Linear dependence/independence
Let V be a vector space, v1 , . . . , vm ∈ V be a list of vectors.
Definition. span(v1 , . . . , vm ) = {a1 v1 + · · · + am vm | a1 , . . . , am ∈ F}
Definition. A vector v ∈ V linearly depends of v1 , . . . , vm
if v ∈ span(v1 , . . . , vm ) .
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13.
Linear AlgebraLecture 5
Linear dependence/independence
Let V be a vector space, v1 , . . . , vm ∈ V be a list of vectors.
Definition. span(v1 , . . . , vm ) = {a1 v1 + · · · + am vm | a1 , . . . , am ∈ F}
Definition. A vector v ∈ V linearly depends of v1 , . . . , vm
if v ∈ span(v1 , . . . , vm ) .
v1 , . . . , vm ∈ V are called linearly independent if
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14.
Linear AlgebraLecture 5
Linear dependence/independence
Let V be a vector space, v1 , . . . , vm ∈ V be a list of vectors.
Definition. span(v1 , . . . , vm ) = {a1 v1 + · · · + am vm | a1 , . . . , am ∈ F}
Definition. A vector v ∈ V linearly depends of v1 , . . . , vm
if v ∈ span(v1 , . . . , vm ) .
v1 , . . . , vm ∈ V are called linearly independent if
none of v1 , . . . , vm linearly depends on the others.
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15.
Linear AlgebraLecture 5
Linear dependence/independence
Let V be a vector space, v1 , . . . , vm ∈ V be a list of vectors.
Definition. span(v1 , . . . , vm ) = {a1 v1 + · · · + am vm | a1 , . . . , am ∈ F}
Definition. A vector v ∈ V linearly depends of v1 , . . . , vm
if v ∈ span(v1 , . . . , vm ) .
v1 , . . . , vm ∈ V are called linearly independent if
none of v1 , . . . , vm linearly depends on the others.
2.17 (More symmetric) definition List v1 , . . . , vm ∈ V is linearly independent if
a1 v1 + · · · + am vm = 0 =⇒ a1 = · · · = am = 0 .
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16.
Linear Dependence LemmaLinear Algebra
Lecture 5
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17.
Linear Dependence LemmaLinear Algebra
Lecture 5
2.21 Linear Dependence Lemma
List v1 , . . . , vm ∈ V is linearly dependent
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18.
Linear AlgebraLecture 5
Linear Dependence Lemma
2.21 Linear Dependence Lemma
List v1 , . . . , vm ∈ V is linearly dependent
⇐⇒
∃j ∈ {1, 2, . . . , m} :
vj ∈ span(v1 , . . . , vj−1 ) .
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19.
Linear AlgebraLecture 5
Linear Dependence Lemma
2.21 Linear Dependence Lemma
List v1 , . . . , vm ∈ V is linearly dependent
⇐⇒
∃j ∈ {1, 2, . . . , m} :
vj ∈ span(v1 , . . . , vj−1 ) .
List v1 , . . . , vm ∈ V is linearly dependent ⇐⇒ ∃ a proper sublist vk1 , . . . , vkl with
the same span.
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20.
Independent ≤ spanningLinear Algebra
Lecture 5
In a finite-dimensional space,
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21.
Linear AlgebraLecture 5
Independent ≤ spanning
In a finite-dimensional space,
the length of every linearly
independent list of vectors
≤
the length of every
spanning list of vectors
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22.
Linear AlgebraLecture 5
Independent ≤ spanning
In a finite-dimensional space,
the length of every linearly
independent list of vectors
Proof
≤
the length of every
spanning list of vectors
Let u1 , . . . , up is linearly independent in V ,
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23.
Linear AlgebraLecture 5
Independent ≤ spanning
In a finite-dimensional space,
the length of every linearly
independent list of vectors
Proof
≤
the length of every
spanning list of vectors
Let u1 , . . . , up is linearly independent in V , and V = span(w1 , . . . , wq ) .
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24.
Linear AlgebraLecture 5
Independent ≤ spanning
In a finite-dimensional space,
the length of every linearly
independent list of vectors
≤
the length of every
spanning list of vectors
Proof
Let u1 , . . . , up is linearly independent in V , and V = span(w1 , . . . , wq ) .
Prove: p ≤ q .
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25.
Linear AlgebraLecture 5
Independent ≤ spanning
In a finite-dimensional space,
the length of every linearly
independent list of vectors
≤
the length of every
spanning list of vectors
Proof
Let u1 , . . . , up is linearly independent in V , and V = span(w1 , . . . , wq ) .
Plan:
Prove: p ≤ q .
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26.
Linear AlgebraLecture 5
Independent ≤ spanning
In a finite-dimensional space,
the length of every linearly
independent list of vectors
≤
the length of every
spanning list of vectors
Proof
Let u1 , . . . , up is linearly independent in V , and V = span(w1 , . . . , wq ) .
Plan: build up a list of length q
Prove: p ≤ q .
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27.
Linear AlgebraLecture 5
Independent ≤ spanning
In a finite-dimensional space,
the length of every linearly
independent list of vectors
≤
the length of every
spanning list of vectors
Proof
Let u1 , . . . , up is linearly independent in V , and V = span(w1 , . . . , wq ) .
Plan: build up a list of length q by gradual substituting wi for uj .
Prove: p ≤ q .
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28.
Linear AlgebraLecture 5
Independent ≤ spanning
In a finite-dimensional space,
the length of every linearly
independent list of vectors
≤
the length of every
spanning list of vectors
Proof
Let u1 , . . . , up is linearly independent in V , and V = span(w1 , . . . , wq ) .
Plan: build up a list of length q by gradual substituting wi for uj .
Prove: p ≤ q .
w1 , . . . , wq
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29.
Linear AlgebraLecture 5
Independent ≤ spanning
In a finite-dimensional space,
the length of every linearly
independent list of vectors
≤
the length of every
spanning list of vectors
Proof
Let u1 , . . . , up is linearly independent in V , and V = span(w1 , . . . , wq ) .
Plan: build up a list of length q by gradual substituting wi for uj .
Prove: p ≤ q .
w1 , . . . , wq
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30.
Linear AlgebraLecture 5
Independent ≤ spanning
In a finite-dimensional space,
the length of every linearly
independent list of vectors
≤
the length of every
spanning list of vectors
Proof
Let u1 , . . . , up is linearly independent in V , and V = span(w1 , . . . , wq ) .
Plan: build up a list of length q by gradual substituting wi for uj .
Prove: p ≤ q .
w1 , . . . , wq
u1 , w 1 , . . . , w q
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31.
Linear AlgebraLecture 5
Independent ≤ spanning
In a finite-dimensional space,
the length of every linearly
independent list of vectors
≤
the length of every
spanning list of vectors
Proof
Let u1 , . . . , up is linearly independent in V , and V = span(w1 , . . . , wq ) .
Plan: build up a list of length q by gradual substituting wi for uj .
Prove: p ≤ q .
w1 , . . . , wq
u1 , w 1 , . . . , w q
u1 , w1 , . . . , wq is linear dependent as u1 ∈ V = span(w1 , . . . , wq )
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32.
Linear AlgebraLecture 5
Independent ≤ spanning
In a finite-dimensional space,
the length of every linearly
independent list of vectors
≤
the length of every
spanning list of vectors
Proof
Let u1 , . . . , up is linearly independent in V , and V = span(w1 , . . . , wq ) .
Plan: build up a list of length q by gradual substituting wi for uj .
Prove: p ≤ q .
w1 , . . . , wq
u1 , w 1 , . . . , w q
u1 , w1 , . . . , wq is linear dependent as u1 ∈ V = span(w1 , . . . , wq )
∃j : wj ∈ span(u1 , w1 , . . . , wj−1 )
by 2.21
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33.
Linear AlgebraLecture 5
Independent ≤ spanning
In a finite-dimensional space,
the length of every linearly
independent list of vectors
≤
the length of every
spanning list of vectors
Proof
Let u1 , . . . , up is linearly independent in V , and V = span(w1 , . . . , wq ) .
Plan: build up a list of length q by gradual substituting wi for uj .
Prove: p ≤ q .
w1 , . . . , wq
u1 , w 1 , . . . , w q
u1 , w1 , . . . , wq is linear dependent as u1 ∈ V = span(w1 , . . . , wq )
∃j : wj ∈ span(u1 , w1 , . . . , wj−1 )
by 2.21
Through wj away from the list u1 , w1 , . . . , wq .
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34.
Linear AlgebraLecture 5
Independent ≤ spanning
In a finite-dimensional space,
the length of every linearly
independent list of vectors
≤
the length of every
spanning list of vectors
Proof
Let u1 , . . . , up is linearly independent in V , and V = span(w1 , . . . , wq ) .
Plan: build up a list of length q by gradual substituting wi for uj .
Prove: p ≤ q .
w1 , . . . , wq
u1 , w 1 , . . . , w q
u1 , w1 , . . . , wq is linear dependent as u1 ∈ V = span(w1 , . . . , wq )
∃j : wj ∈ span(u1 , w1 , . . . , wj−1 )
by 2.21
Through wj away from the list u1 , w1 , . . . , wq .
and continue.
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35.
Subspaces of a finite-dimensional spaceLinear Algebra
Lecture 5
2.26 A subspace of a finite-dimensional space is finite-dimensional.
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36.
Subspaces of a finite-dimensional spaceLinear Algebra
Lecture 5
2.26 A subspace of a finite-dimensional space is finite-dimensional.
Proof
Let U ⊂ V = span(v1 , . . . , vp ) .
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37.
Subspaces of a finite-dimensional spaceLinear Algebra
Lecture 5
2.26 A subspace of a finite-dimensional space is finite-dimensional.
Proof
Let U ⊂ V = span(v1 , . . . , vp ) .
Let us build a linear independent list ⊂ U .
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38.
Subspaces of a finite-dimensional spaceLinear Algebra
Lecture 5
2.26 A subspace of a finite-dimensional space is finite-dimensional.
Proof
Let U ⊂ V = span(v1 , . . . , vp ) .
Let us build a linear independent list ⊂ U .
Lemma. If a list w1 , . . . , wn ⊂ U is linear independent, but U 6= span(w1 , . . . , wn ) ,
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39.
Subspaces of a finite-dimensional spaceLinear Algebra
Lecture 5
2.26 A subspace of a finite-dimensional space is finite-dimensional.
Proof
Let U ⊂ V = span(v1 , . . . , vp ) .
Let us build a linear independent list ⊂ U .
Lemma. If a list w1 , . . . , wn ⊂ U is linear independent, but U 6= span(w1 , . . . , wn ) ,
then ∃w ∈ U such that w1 , . . . , wn , w is linear independent.
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40.
Subspaces of a finite-dimensional spaceLinear Algebra
Lecture 5
2.26 A subspace of a finite-dimensional space is finite-dimensional.
Proof
Let U ⊂ V = span(v1 , . . . , vp ) .
Let us build a linear independent list ⊂ U .
Lemma. If a list w1 , . . . , wn ⊂ U is linear independent, but U 6= span(w1 , . . . , wn ) ,
then ∃w ∈ U such that w1 , . . . , wn , w is linear independent.
Reformulation. A linear independent list can be increased, unless it spans.
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41.
Subspaces of a finite-dimensional spaceLinear Algebra
Lecture 5
2.26 A subspace of a finite-dimensional space is finite-dimensional.
Proof
Let U ⊂ V = span(v1 , . . . , vp ) .
Let us build a linear independent list ⊂ U .
Lemma. If a list w1 , . . . , wn ⊂ U is linear independent, but U 6= span(w1 , . . . , wn ) ,
then ∃w ∈ U such that w1 , . . . , wn , w is linear independent.
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42.
Subspaces of a finite-dimensional spaceLinear Algebra
Lecture 5
2.26 A subspace of a finite-dimensional space is finite-dimensional.
Proof
Let U ⊂ V = span(v1 , . . . , vp ) .
Let us build a linear independent list ⊂ U .
Lemma. If a list w1 , . . . , wn ⊂ U is linear independent, but U 6= span(w1 , . . . , wn ) ,
then ∃w ∈ U such that w1 , . . . , wn , w is linear independent.
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43.
Subspaces of a finite-dimensional spaceLinear Algebra
Lecture 5
2.26 A subspace of a finite-dimensional space is finite-dimensional.
Proof
Let U ⊂ V = span(v1 , . . . , vp ) .
Let us build a linear independent list ⊂ U .
Lemma. If a list w1 , . . . , wn ⊂ U is linear independent, but U 6= span(w1 , . . . , wn ) ,
then ∃w ∈ U such that w1 , . . . , wn , w is linear independent.
Reformulation. A linear independent list can be increased, unless it spans.
Dual statement. A span of a vector space can be decreased, unless it is linearly
independent.
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44.
Linear AlgebraLecture 5
Bases
2.27 Definition
A basis of V is a list of vectors in V
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45.
Bases2.27 Definition
Linear Algebra
Lecture 5
A basis of V is a list of vectors in V
that is linearly independent and spans V .
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46.
Linear AlgebraLecture 5
Bases
2.27 Definition
A basis of V is a list of vectors in V
that is linearly independent and spans V .
2.28 Examples
• Standard base in Fn :
(1, 0, . . . , 0), (0, 1, 0 . . . , 0), . . . , (0, 0, . . . , 0, 1)
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47.
Linear AlgebraLecture 5
Bases
2.27 Definition
A basis of V is a list of vectors in V
that is linearly independent and spans V .
2.28 Examples
• Standard base in Fn :
(1, 0, . . . , 0), (0, 1, 0 . . . , 0), . . . , (0, 0, . . . , 0, 1)
• 1, z, . . . , z m is a basis in Pm (F) .
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48.
Linear AlgebraLecture 5
Bases
2.27 Definition
A basis of V is a list of vectors in V
that is linearly independent and spans V .
2.28 Examples
• Standard base in Fn :
(1, 0, . . . , 0), (0, 1, 0 . . . , 0), . . . , (0, 0, . . . , 0, 1)
• 1, z, . . . , z m is a basis in Pm (F) .
2.29 Criterion for basis
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49.
Linear AlgebraLecture 5
Bases
2.27 Definition
A basis of V is a list of vectors in V
that is linearly independent and spans V .
2.28 Examples
• Standard base in Fn :
(1, 0, . . . , 0), (0, 1, 0 . . . , 0), . . . , (0, 0, . . . , 0, 1)
• 1, z, . . . , z m is a basis in Pm (F) .
2.29 Criterion for basis
v1 , . . . , vn is a basis of V ⇐⇒
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50.
Linear AlgebraLecture 5
Bases
2.27 Definition
A basis of V is a list of vectors in V
that is linearly independent and spans V .
2.28 Examples
• Standard base in Fn :
(1, 0, . . . , 0), (0, 1, 0 . . . , 0), . . . , (0, 0, . . . , 0, 1)
• 1, z, . . . , z m is a basis in Pm (F) .
2.29 Criterion for basis
v1 , . . . , vn is a basis of V ⇐⇒ ∀v ∈ V
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51.
Linear AlgebraLecture 5
Bases
2.27 Definition
A basis of V is a list of vectors in V
that is linearly independent and spans V .
2.28 Examples
• Standard base in Fn :
(1, 0, . . . , 0), (0, 1, 0 . . . , 0), . . . , (0, 0, . . . , 0, 1)
• 1, z, . . . , z m is a basis in Pm (F) .
2.29 Criterion for basis
v1 , . . . , vn is a basis of V ⇐⇒ ∀v ∈ V ∃ unique a1 , . . . , an ∈ F
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52.
Linear AlgebraLecture 5
Bases
2.27 Definition
A basis of V is a list of vectors in V
that is linearly independent and spans V .
2.28 Examples
• Standard base in Fn :
(1, 0, . . . , 0), (0, 1, 0 . . . , 0), . . . , (0, 0, . . . , 0, 1)
• 1, z, . . . , z m is a basis in Pm (F) .
2.29 Criterion for basis
v1 , . . . , vn is a basis of V ⇐⇒ ∀v ∈ V ∃ unique a1 , . . . , an ∈ F
v = a1 v1 + · · · + an vn
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53.
Spanning list contains a basisLinear Algebra
Lecture 5
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54.
Spanning list contains a basisLinear Algebra
Lecture 5
2.31
Every spanning list in a vector space
can be reduced to a basis of the vector space.
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55.
Spanning list contains a basisLinear Algebra
Lecture 5
2.31
Every spanning list in a vector space
can be reduced to a basis of the vector space.
2.32 Every finite-dimensional vector space has a basis.
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56.
Linearly independent list extends to a basisLinear Algebra
Lecture 5
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57.
Linearly independent list extends to a basisLinear Algebra
Lecture 5
2.33
Every linearly independent list of vectors in a finite-dimensional vector space
can be extended to a basis of the vector space.
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58.
Linearly independent list extends to a basisLinear Algebra
Lecture 5
2.33
Every linearly independent list of vectors in a finite-dimensional vector space
can be extended to a basis of the vector space.
2.34 Every subspace of V is part of a direct sum equal to V .
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