MAC 2103
Module 8
General Vector Spaces I
Learning Objectives
Upon completing this module, you should be able to:
1. Recognize from the standard examples of vector
spaces, that a vector space is closed under vector
addition and scalar multiplication.
2. Determine if a subset W of a vector space V is a
subspace of V.
3. Find the linear combination of a finite set of vectors.
4. Find W = span(S), a subspace of V, given a set of
vectors S in a vector space V.
5. Determine if a finite set of non-zero vectors in V is a
linearly dependent set or linearly independent set.
6. Use the Wronskian to determine if a set of vectors
that are differentiable functions is linearly
independent.
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1
General Vector Spaces I
There are three major topics in this module:
Real Vector Spaces or Linear Spaces
Subspaces
Linear Independence
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What are the Standard Examples of
Vector Spaces?
We have seen some of them before; some standard
examples of vector spaces are as follows:
R1 , R 2 , R 3 , R n , M m,n , Pn ,C(!", "),C[a,b]
Can you identify them? We will look at some of them
later in this module.
For now, know that we can always add any two vectors
and multiply all vectors by a scalar within any vector
space.
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2
What are the Standard Examples of
Vector Spaces? (Cont.)
Since we can always add any two vectors and multiply all
vectors by a scalar in any vector space, we say that a
vector space is closed under vector addition and scalar
multiplication. In other words, it is closed under linear
combinations.
A vector space is also called a linear space. In fact, a
linear space is a better name.
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What is a Vector Space?
Let V be a non-empty set of objects u, v, and w, on which
two operations, vector addition and scalar multiplication,
are defined. If V can satisfy the following ten axioms,
then V is a vector space. (Please pay extra attention to
axioms 1 and 6.)
1. If u, v ∈ V, then u + v ∈ V ~ Closure under addition
2. u + v = v + u ~ Commutative property
3. u + (v + w)= (u + v)+ w ~ Associative property
4. There is a unique zero vector such that u + 0 = 0 + u = u, for
all u in V. ~ Additive identity
5. For each u, there is a unique vector -u such that u + (-u) = 0.
~ Additive inverse
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What is a Vector Space? (Cont.)
Here are the next five properties:
6. If k is in a field (ℜ), k is a scalar and u ∈ V, then ku ∈ V
~ Closure under scalar multiplication
7. k(u + v) = ku + kv ~ Distributive property
8. (k + m)u = ku + mu ~ Distributive property
9. k(mu)= (km)u ~ Associative property
10. 1u = u ~ Scalar identity
Looks familiar. You have used them in ℜ ℜ²,and ℜ³
before.
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What is a Vector Space? (Cont.)
Example: Show that the set of all 4 x 3 matrices with the
operations of matrix addition and scalar multiplication is a
vector space.
If A and B are 4 x 3 matrices and s is a scalar, then A + B
and sA are also 4 x 3 matrices. Since the resulting
matrices have the same form, the set is closed under
matrix addition and matrix multiplication. We know from
the previous modules that the other vector space axioms
hold as well. Thus, we can conclude that the set is a
vector space.
Similarly, we can show that the set of all m x n matrices,
Mm,n, is a vector space.
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What is a Subspace?
A subspace is a non-empty subset of a vector space;
it is a subset that satisfies all the ten axioms of a
vector space, including axioms 1 and 6:
• Closure under addition, and
• Closure under scalar multiplication.
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How to Determine if a Subset W of a Vector Space V
is Subspace of V?
Since a subset inherits the ten axioms from its larger
vector space, to determine if a subset W of a vector
space V is a subspace of V, we only need to check
the following two axioms:
1. If u , v ∈ W, then u + v ∈ W ~ Closure under addition
2. If k is a scalar and u ∈ W, then ku ∈ W ~ Closure
under scalar multiplication
Note that the zero subspace = {0} and V itself are both valid
subspaces of V. One is the smallest subspace of V, and
one is the largest subspace of V.
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How to Determine if a Subset W of a Vector Space V is
Subspace of V? (Cont.)
Example: Is the following set of vectors a subspace of ℜ³?
u = (3, -2, 0) and v = (4, 5, 0).
Since a subset inherits the ten axioms from its larger vector
space, to determine if a subset W of a vector space V is a
subspace of V, we only need to verify the following two axioms:
1. If u , v ∈ W, then u + v ∈ W .
2. If k is any scalar and u ∈ W, then ku ∈ W.
Check:
u + v = (3+4, -2+5, 0+0) = (7, 3, 0) ∈ W .
ku = (3k, -2k, 0) ∈ W .
Thus, W is a subspace of ℜ³ and is the xy-plane in ℜ³.
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What is a Linear Combination of Vectors?
By definition, a vector w is called a linear combination of the
vectors v1, v2, …, vr if it can be expressed in the form
! ! ! ! r
!
w = k1v1 + k2 v2 + ... + kr vr = ! ki vi
i =1
where k1, k2, …, kr are scalars.
For example, if we have a set of vectors in ℜ³, S = {v1, v2, v3 },
where v1 = (2, 4, 3), v2 = (-1, 3, 1), and v3 = (8, 23, 17),
we can see that v3 is a linear combination of v1 and v2, since
v3 = 5v1 + 2v2 = 5(2, 4, 3) + 2(-1, 3, 1) = (8, 23, 17).
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How to Find a Linear Combination of
a Finite Set of Vectors?
Note:
Example: Let S = {u, v, w} ℜ³=V. Express
If u, v, and w
p = (-3,8,4 ) as linear combination of u = are vectors in a
(1,1,2), v = (-1,3,0), and w = (0,1,2).
! ! ! ! vector space V
p = k1u1 + k2 v + k3w , then the set W
(!3, 8, 4) = k1 (1,1, 2) + k2 (!1, 3, 0) + k3 (0,1, 2) = span(S) of all
linear
(!3, 8, 4) = (k1 ! k2 , k1 + 3k2 + k3 , 2k1 + 2k3 ) combinations of
In order to solve for the scalars k1, k2, and k3, we u, v, and w is a
equate the corresponding components and subspace of V;
obtain the system as follows: p = (-3, 8, 4) is
! !
k1 ! k2 = !3 just one of the
linear
Ak = p k1 + 3k2 + k3 = 8 combinations in
2k1 + 2k3 = 4 the set W =
span(S).
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How to Find a Linear Combination of
a Finite Set of Vectors? (Cont.)
We can solve this system using Gauss-Jordan Elimination.
" 1 !1 0 !3 %
r1 $ ' " %
! $ 1 !1 0 !3 '
[ A | p ] = r2 $ 1 3 1 8 ' r1
$ '
$ ' 1 11
r3 $# 2 0 2 4 '&
r2 $ 0 1 4 4 '
!2r2 + r3 $ '
" 1 !1 0 !3 % $# 0 0 23 92 '&
r1 $ '
!r1 + r2 ( r2 $ 0 4 1 11 ' " 1 !1 0 !3 %
$ ' r1 $ '
!2r1 + r3 ( r3 $# 0 2 2 10 '& r2 $ 0 1 14 114 '
$ '
3 r3 ( r3
2
" 1 !1 0 !3 % $ 0 0 1 3 '
r1 $ ' # &
4 r2 ( r2
1 $ 0 1 14 114 '
$ '
r3 $ 0 2 2 10 '
# &
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How to Find a Linear Combination of
a Finite Set of Vectors? (Cont.)
# 1 "1 0 "3 & Thus, the system is
r1 % (
r2 % 0 1 14 114 ( consistent and p can be
% ( expressed as a linear
2
3 r3 ! r3 % 0 0 1 3 (
$ ' combination of u, v, and
# 1 "1 0 "3 & w as follows:
r1 ! r1 % (
" 14 r3 + r2 ! r2 % 0 1 0 2 ( p = -u + 2v + 3w
% (
r3 %$ 0 0 1 3 ('
Note: If the system is
# 1 0 0 "1 & inconsistent, we will not be
r2 + r1 ! r1 % (
% 0 1 0 2 ( able to express p as a linear
r2
% ( combination of u, v, and w.
r3 %$ 0 0 1 3 (' Then, p is not a linear
k3 = 3, k2 = 2, k1 = "1 combination of u, v, and w.
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What is the Spanning Set?
Let S = {v1, v2,…, vr } be a set of vectors in a vector
space V, then there exists a subspace W of V
consisting of all linear combinations of the vectors in
S.
W is called the space spanned by v1, v2,…, vr.
Alternatively, we say that the vectors v1, v2,…, vr
span W.
Thus, W = span(S) = span {v1, v2,…, vr } and the set S
is the spanning set of the subspace W.
In short, if every vector in V can be expressed as a
linear combinations of the vectors in S, then S is the
spanning set of the vector space V.
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How to Find the Space Spanned by a Set of Vectors?
In our previous example, S = {u, v, w } = {(1,1,2),(-1,3,0),(0,1,2)}
is a set of vectors in the vector space ℜ³, and
! ! !
p = Ak !W , k = (k1 , k2 , k 3 ) = ("1, 2, 3)
! ! !
Is (x1 , x2 , x 3 ) = x !W ? Or can we solve x = Ak for any x?
Yes, if A-1 exists. Find det(A) to see if there is a unique solution?
If we let W be the subspace of ℜ³ consisting of all linear
combinations of the vectors in S, then x ∈ W for any x ∈ ℜ³.
Thus, W = span(S) = ℜ³.
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How to Determine if a Finite Set of Non-Zero Vectors is
a Linearly Dependent Set or Linearly Independent Set?
Let S = {v1, v2,…, vr } be a set of finite non-zero vectors in a
vector space V. The vector equation
! ! ! !
k1v1 + k2 v2 + ... + kr vr = 0
has at least one solution, namely the trivial solution , 0 = k1=
k2= … = kr. If the only solution is the trivial solution, then S is a
linearly independent set. Otherwise, S is a linearly dependent
set. If v1, v2,…, vr ∈ ℜⁿ , then the vector equation
! ! ! ! ! ! ! !
k1v1 + k2 v2 + ... + kr vr = 0 = Ak, A = [ v1 v2 ... vr ]
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How to Use the Wronskian to Determine if a Set of
Vectors that are Differentiable Functions is Linearly
Independent?
Let S = { f1, f2, …, fn } be a set of vectors in C(n-1)(-∞,∞). The
Wronskian is
f1 (x) f2 (x) . . . fn (x)
f1' (x) f2' (x) . . . f1' (x)
w(x) = . . .
. . .
. . .
f1(n !1) (x) f2(n !1) (x) . . . fn(n !1) (x)
If the functions f1, f2, …, fn have n-1 continuous derivatives on
the interval (-∞,∞), and if w(x) ≠ 0 on the interval (-∞,∞), then
we can say that S is a linearly independent set of vectors in
C(n-1)(-∞,∞). http://faculty.valenciacc.edu/ashaw/
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How to Use the Wronskian to Determine if a Set of
Vectors that are Differentiable Functions is Linearly
Independent? (Cont.)
Example: Let S = { f1 , f2, f3 } = {5, e2x, e3x }. Show that S is a
linearly independent set of vectors in C2(-∞,∞).
The Wronskian is
5 e2 x e3x
2e2 x 3e3x
w(x) = 0 2e2 x 3e3x =5
4e2 x 9e3x
0 4e2 x 9e3x
= 5(18e2 x e3x ! 12e2 x e3x ) = 5(6e2 x e3x ) = 30e5 x " 0
Since w(x) ≠ 0 on the interval (-∞,∞), we can say that S is a linearly
independent set of vectors in C2(-∞,∞), the linear space of twice
continuously differentiable functions on (-∞,∞).
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What have we learned?
We have learned to:
1. Recognize from the standard examples of vector
spaces, that a vector space is closed under vector
addition and scalar multiplication.
2. Determine if a subset W of a vector space V is a
subspace of V.
3. Find the linear combination of a finite set of vectors.
4. Find W = span(S), a subspace of V, given a set of
vectors S in a vector space V.
5. Determine if a finite set of non-zero vectors in V is a
linearly dependent set or linearly independent set.
6. Use the Wronskian to determine if a set of vectors that
are differentiable functions is linearly independent.
http://faculty.valenciacc.edu/ashaw/
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Credit
Some of these slides have been adapted/modified in part/whole from the
following textbook:
• Anton, Howard: Elementary Linear Algebra with Applications, 9th Edition
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