Lecture 4: Linear independence, span, and bases
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Travis Schedler
Tue, Sep 20, 2011 (version: Wed, Sep 21, 6:30 PM)
Goals (2)
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Understand linear independence and examples
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Understand span and examples
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Understand basis and examples
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Preview dimension
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Prove the main theorem on linear independence and span, for
the finite-dimensional situation
Warm-up exercise 1 (3)
What are all of the subspaces of R2 ? Try to define them also using
set notation; e.g., {(a, b) ∈ R2 | ab = 1} or {(a, 1a ) | a ∈ R, a 6= 0}
would describe the hyperbola xy = 1.
Answer: The subspaces are {0} (otherwise denoted by 0), R2
itself, and all of the lines through the origin,
Lλ,µ = {(a, b) ∈ R2 | µa + λb = 0} = {(λt, µt) | t ∈ R}, where λ
and µ are not both zero. (Note that Lλ,µ = Lcλ,cµ for nonzero
c ∈ R).
Warm-up exercise 2 (4)
True or false: If true, explain why. If false, give a counterexample:
Suppose V = V1 + V2 + V3 . Then, V = V1 ⊕ V2 ⊕ V3 if and only
if 0 = V1 ∩ V2 = V1 ∩ V3 = V2 ∩ V3 .
Answer: False: one counterexample is V = R2 , with V1 = the
x-axis, V2 = the y -axis, and V3 = the line x = y . Then,
V = R2 = V1 + V2 + V3 , but Vi ∩ Vj = 0 for all i 6= j.
Linear (in)dependence (5)
Recall from the book:
Definition
A list (v1 , . . . , vm ) is linearly independent if the only choice of
a1 , . . . , am ∈ F such that
a1 v1 + · · · + am vm = 0
(0.1)
is a1 = · · · = am = 0.
The list is linearly dependent if there exists a1 , . . . , am ∈ F, not all
zero, such that (0.1) holds.
Examples (6)
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Any sublist of a linearly independent list is still linearly
independent.
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For any nonzero vector v ∈ V , the list (v ), just containing the
vector v only, is linearly independent.
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Any list containing zero is linearly dependent.
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A list (v , w ) of two elements is linearly independent if and
only if v 6= 0 and w is not a scalar multiple of v .
The case of R3 :
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The list ((1, 0, 0), (0, 1, 0), (0, 0, 1)) is linearly independent.
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The list ((1, 1, 2), (2, 2, 4)) is linearly dependent.
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The list ((1, 0, 0), (0, 1, 1), (1, 2, 2)) is linearly dependent.
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In general, (u, v , w ) is linearly independent if and only if they
do not all lie in the same plane through the origin.
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Any list of length four or more turns out to be linearly
dependent (why?)
Span (7)
Definition
The span of a list (v1 , . . . , vm ) of vectors in V is the vector
subspace
Span(v1 , . . . , vm ) := {a1 v1 + · · · + am vm | a1 , . . . , am ∈ F}. (0.2)
Given a vector space V , a list (v1 , . . . , vm ) of elements of V is a
spanning list for V if V = Span(v1 , . . . , vm ).
Note that, in a definition, “if” really means “if and only if,” since
we are making a definition.
Exercises:
I Show that the set Span(v1 , . . . , vm ) is always a vector
subspace.
I Show that Span(v1 , . . . , vm ) = Span(v1 ) + · · · + Span(vm ).
This relates span to sum!
I Show that (v1 , . . . , vm ) is linearly independent if and only if
Span(v1 , . . . , vm ) = Span(v1 ) ⊕ · · · ⊕ Span(vm ) and all the vi
are nonzero.
Examples (8)
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If we append any vectors to a spanning list, the result is still
spanning.
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Span(0) = {0}. In general, Span(v1 , . . . , vm ) = {0} if and
only if all of v1 , . . . , vm are zero.
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For nonzero v , Span(v ) = {λv | λ ∈ F} is the line through v
and the origin.
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For nonzero v and for w not a multiple of v ,
Span(v , w ) = {av + bw | a, b ∈ F} is the plane through the
origin containing v and w .
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Span(x, x 2 , . . . , x n ) = the space of degree ≤ n polynomials
which are multiples of x (i.e., which vanish at 0).
The linear dependence lemma (9)
We strengthen (a) from Lemma 2.4 and state it differently:
Lemma
(a) A list (v1 , . . . , vm ) is linearly independent if and only if v1 6= 0
and, for all 2 ≤ j ≤ m, vj ∈
/ Span(v1 , . . . , vj−1 ).
(b) If the list is linearly dependent and vj ∈ Span(v1 , . . . , vj−1 ),
then the span of the list (v1 , . . . , vj−1 , vj+1 , . . . , vm ) obtained by
removing vj is the same as the span of (v1 , . . . , vm ).
Intuitive idea: As we saw, a list (v1 , . . . , vm ) in Rn is linearly
independent if and only if v1 spans a line, (v1 , v2 ) spans a plane,
etc., and (v1 , . . . , vj ) spans a j-dimensional space for all 1 ≤ j ≤ m,
i.e., each vj is not in the span of the previous v1 , . . . , vj−1 .
For part (b), the idea is: if vj ∈ Span(v1 , . . . , vj−1 ), then we can
replace vj by a linear combination of v1 , . . . , vj−1 , so the span is
unchanged if we discard vj .
Proof of lemma (10)
Proof.
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Axler proves half of (a) and all of (b):
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If (v1 , . . . , vm ) is linearly dependent, then either v1 = 0 or, for
some j, vj ∈ Span(v1 , . . . , vj−1 ) [half of part (a)].
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In the latter case, the span of (v1 , . . . , vm ) is the same as the
span of the list obtained by removing vj [part (b)].
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Using this, it remains to prove the other half of part (a).
Suppose that v1 = 0. Then (0, v2 , . . . , vm ) is linearly
dependent: 1 · v1 + 0 · v2 + · · · + 0 · vm = 0.
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Suppose that j ≥ 2 and vj ∈ Span(v1 , . . . , vj−1 ). Write
vj = a1 v1 + · · · + aj−1 vj−1 .
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Then a1 v1 + · · · + aj−1 vj−1 + (−1) · vj = 0. So (v1 , . . . , vm ) is
linearly dependent.
Bases (11)
Definition
A basis of V is a list (v1 , . . . , vm ) of vectors in V that is both
linearly independent and a spanning list.
Main (characterizing) property:
Proposition (Proposition 2.8)
A list (v1 , . . . , vm ) in V is a basis if and only if, for every vector
v ∈ V , there exist unique a1 , . . . , am ∈ F such that
v = a1 v1 + · · · + am vm .
Proof: Read it in Axler and understand it!
I Idea of proof: Break into parts: (a) The list (v1 , . . . , vm ) is
spanning if and only if every vector v can be written in at
least one way as v = a1 v1 + · · · + am vm .
I (b) The list (v1 , . . . , vm ) is linearly independent if and only if
every vector v can be written in at most one way as
v = a1 v1 + · · · + am vm .
Exercise: Read the proof in Axler and use it to prove (a) and (b)
separately.
Examples (12)
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A basis of R2 : ((1, 0), (0, 1)). More generally, (u, v ) where
u 6= 0 and v is not a multiple of u.
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The standard basis of Rn : (v1 , . . . , vn ), where
vi = (0, 0, . . . , 0, 1, 0, 0, . . . , 0), with 1 in the i-th place.
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Functions {1, 2, . . . , n} → F: a basis is the collection of delta
functions, (δ1 , . . . , δn ), where δi (j) = 0 if i 6= j, and δi (i) = 1.
Deep result we will prove: For Fn or the space of functions
{1, . . . , n} → F, a spanning list is a basis if and only if it has
length n.
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Similarly, again for Fn or the space of functions
{1, . . . , n} → F, a linearly independent list is a basis if and
only if it has length n.
Preview on Dimension (13)
The above deep properties are saying that Fn and {1, . . . , n} → F
have dimension n.
Definition
A vector space V is finite-dimensional if and only if there is a
(finite) spanning list (v1 , . . . , vm ).
A deep result we will prove next time:
Proposition
If V is finite-dimensional, then the following numbers coincide: (a)
The minimum length of a spanning list; (b) the maximum length
of a linearly independent list; (c) the length of every basis.
This number is called the dimension of V .
A vector space which is not finite-dimensional is called
infinite-dimensional. This includes P(F) and F∞ .
Main theorem on finite spanning and linearly independent
lists (14)
The key ingredient in the proof of the previous proposition on
dimension is our main theorem:
Theorem
In a finite-dimensional vector space, the length of every linearly
independent list is less than or equal to the length of every
spanning list.
Proof:
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Let (w1 , . . . , wn ) be a spanning list and (u1 , . . . , um ) be a
linearly independent list. We have to show m ≤ n.
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In fact, we will show that (u1 , . . . , um ) can be extended to a
spanning list of length n by adding some wj ’s, so in particular
m ≤ n.
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To do so, we go backwards, beginning with (w1 , . . . , wn ), and
at each step replacing a wj with a ui .
Proof of main theorem continued (15)
First, let us append u1 . Form the list (u1 , w1 , . . . , wm ). Since
u1 ∈ Span(w1 , . . . , wn ), this list is linearly dependent.
I By the linear dependence lemma, since u1 6= 0, for some j,
wj ∈ Span(u1 , w1 , . . . , wj−1 ).
I Then the list (u1 , w1 , . . . , wj−1 , wj+1 , . . . , wm ) is also
spanning.
I Inductively, suppose that we have a spanning list of length n,
beginning with u1 , . . . , ui , and ending with n − i of the w ’s.
I Append ui+1 to the list. Since ui+1 is in the span of the other
vectors, the result is linearly dependent.
I Reorder the list to begin with u1 , . . . , ui+1 . By the linear
dependence lemma, either uj ∈ Span(u1 , . . . , uj−1 ) (impossible
since (u1 , . . . , ui+1 ) is linearly independent), or else one of the
v ’s, say vk , is in the span of the previous elements in the list.
I Now discard vk , completing the induction. We get a spanning
list of length n, beginning with u1 , . . . , um . So m ≤ n.
For next time: Read all of Chapter 2, and understand the
proofs of Theorems 2.10 and 2.12; bring questions to class.
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