This document discusses vector spaces and subspaces. It begins by defining a vector space as a set V with two operations, vector addition and scalar multiplication, that satisfy certain properties. Examples of vector spaces include R2 and the space of real polynomials of degree n or less.
It then defines a subspace as a subset of a vector space that is itself a vector space under the inherited operations. For a subset to be a subspace, it must be closed under vector addition and scalar multiplication, and contain the zero vector. Examples given include lines and planes through the origin in R3.
The span of a set S of vectors is defined as the set of all linear combinations of the vectors in S, and it
6. Algebraic Structures
Ref: Y.Choquet-Bruhat, et al, “Analysis, Manifolds & Physics”, Pt I.,
North Holland (82)
Structure Internal Operations Scalar Multiplication
Group * No
Ring, Field * , No
Module / Vector Space + Yes
Algebra + , * Yes
Field = Ring with idenity & all elements except 0 have inverses.
Vector space = Module over Field.
7. Definition : (Real) Vector Space ( V, ; R )
A vector space (over R) consists of a set V along with 2 operations ‘’ and ‘’ s.t.
(1) For the vector addition :
v, w, u V
a) v w V ( Closure )
b) v w = w v ( Commutativity )
c) ( v w ) u = v ( w u ) ( Associativity )
d) 0 V s.t. v 0 = v ( Zero element )
e) v V s.t. v (v) = 0 ( Inverse )
(2) For the scalar multiplication :
v, w V and a, b R, [ R is the real number field (R,+,)
f) a v V ( Closure )
g) ( a + b ) v = ( a v ) (b v ) ( Distributivity )
h) a ( v w ) = ( a v ) ( a w )
i) ( a b ) v = a ( b v ) ( Associativity )
j) 1 v = v
is always written as + so that one writes v + w instead of v w
and are often omitted so that one writes a b v instead of ( a b ) v
8.
9. Definition : (Real) Vector Space ( V, + ; R )
A vector space (over R) consists of a set V along with 2 operations ‘+’ and ‘ ’ s.t.
(1) For the vector addition + :
v, w, u V
a) v + w V ( Closure )
b) v + w = w + v ( Commutativity )
c) ( v + w ) + u = v + ( w + u ) ( Associativity )
d) 0 V s.t. v + 0 = v ( Zero element )
e) v V s.t. v v = 0 ( Inverse )
(2) For the scalar multiplication :
v, w V and a, b R, [ R is the real number field (R,+,) ]
f) a v V ( Closure )
g) ( a + b ) v = a v + b v ( Distributivity )
h) a ( v + w ) = a v + a w
i) ( a b ) v = a ( b v ) = a b v ( Associativity )
j) 1 v = v
Definition in Conventional
Notations
10. Example : R2
R2 is a vector space if
1 1
2 2
x y
a b a b
x y
x y 1 1
2 2
ax by
ax by
,a b R
0
0
0with
Example : Plane in R3.
The plane through the origin 0
x
P y x y z
z
is a vector space.
P is a subspace of R3.
11. Example :
Let & be the (column) matrix addition & scalar multiplication, resp., then
( Zn, + ; Z ) is a vector space.
( Zn, + ; R ) is not a vector space since closure is violated under scalar multiplication.
Example :
0
0
0
0
V
Let then (V, + ; R ) is a vector space.
Definition 1.7: A one-element vector space is a trivial space.
12. Example : Space of Real Polynomials of Degree n or less, Pn
0
n
k
n k k
k
a x a
P R 2 3
3 0 1 2 3 ka a x a x a x a P R
Vector addition:
0 0
n n
k k
k k
k k
a x b x
a b k kk
a b a b
Scalar multiplication:
0
n
k
k
k
b b a x
a
Zero element:
0
0
n
k
k
x
0 0k
k 0i.e.,
Pn is a vector space with vectors
0
n
k
k
k
a x
a
0
n
k
k k
k
a b x
0
n
k
k
k
ba x
i.e.,
kk
b baai.e.,
E.g.,
Pn is isomorphic to Rn+1 with 1
0
0
~ , ,
n
k n
k n n
k
a x a a
P R
Inverse:
0
n
k
k
k
a x
a kk
a ai.e.,
kk
aa
The kth component of a is
i.e.,
13. Definition : Subspaces
For any vector space, a subspace is a subset that is itself a vector space, under
the inherited operations.
0
x
P y x y z
z
is a subspace of R3.
Note: A subset of a vector space is a subspace iff it is closed under & .
→ It must contain 0. (c.f. Lemma 2.9.)
Proof: Let 1 1 1 1 2 2 2 2, , , , ,
T T
x y z x y z P r r
→ 1 1 1 2 2 20 , 0x y z x y z
1 2 1 2 1 2 1 2, ,
T
a b ax bx ay by az bz r r
with 1 2 1 2 1 2 1 1 1 2 2 2ax bx ay by az bz a x y z b x y z
→ 1 2a b P r r QED,a b R
0
Example : Plane in R3 0
x
P y x y z
z
is a subspace of R3.
Proof: Let
→ 1 1 1 2 2 20 , 0x y z x y z
with
→ 1 2a b P r r ,a b R
0
14. Example : The x-axis in Rn is a subspace.
,0, ,0 -axis
T
x x rProof follows directly from the fact that
Example :
• { 0 } is a trivial subspace of Rn.
• Rn is a subspace of Rn.
Both are improper subspaces.
All other subspaces are proper.
Example : Subspace is only defined wrt inherited operations.
({1}, ; R) is a vector space if we define 11 = 1 and a1=1 aR.
However, neither ({1}, ; R) nor ({1},+ ; R) is a subspace of the vector space
(R,+ ; R).
15. Definition : Span
Let S = { s1 , …, sn | sk ( V,+,R ) } be a set of n vectors in vector space V.
The span of S is the set of all linear combinations of the vectors in S, i.e.,
1
,
n
k k k k
k
span S c S c
s s R span 0with
Lemma : The span of any subset of a vector space is a subspace.
Proof:
Let S = { s1 , …, sn | sk ( V,+,R ) }
1 1
,
n n
k k k k
k k
u v span S
u s v sand
1
n
k k k
k
a b au bv
w u v s
1
n
k k
k
w span S
s ,a b R
Converse: Any vector subspace is the span of a subset of its members.
Also: span S is the smallest vector space containing all members of S.
16. Example :
For any vV, span{v} = { a v | a R } is a 1-D subspace.
Example :
Proof:
The problem is tantamount to showing that for all x, y R, unique a,b R s.t.
1 1
1 1
x
a b
y
i.e.,
a b x
a b y
has a unique solution for arbitrary x & y.
Since
1
2
a x y
1
2
b x y ,x y R
2
1 1
,
1 1
span
R
17. Example : P2
Let 2
3 , 2S span x x x 2
3 2 ,a x x bx a b R
Question:
0
2 0
?
c
S
P
Answer is yes since
1 3 2c a b 2c a
2a c 1
1
3
2
b c a
and
1 2
1
3
2
c c
2
1
k
k
k
c x
= subspace of P2 ?
Lesson: A vector space can be spanned by different sets of vectors.
18. Example 2.19: All Possible Subspaces of R3
Planes thru 0
Lines thru 0
21. Definition: A subspace of a vector space V is a subset H of V that has
three properties:
a. The zero vector of V is in H.
b. H is closed under vector addition. That is, for each u and v in H,
the sum is in H.
c. H is closed under multiplication by scalars. That is, for each u in
H and each scalar c, the vector cu is in H.
u v
II. Subspaces
22. Properties (a), (b), and (c) guarantee that a subspace H of V is itself a
vector space, under the vector space operations already defined in V.
Every subspace is a vector space.
Conversely, every vector space is a subspace (of itself and possibly of
other larger spaces).
23. A Subspace Spanned By A Set
The set consisting of only the zero vector in a vector space V is a subspace
of V, called the zero subspace and written as {0}.
As the term linear combination refers to any sum of scalar multiples of
vectors, and Span {v1,…,vp} denotes the set of all vectors that can be
written as linear combinations of v1,…,vp.
Example 2: Given v1 and v2 in a vector space V, let
. Show that H is a subspace of V.
Solution: The zero vector is in H, since .
To show that H is closed under vector addition, take two arbitrary vectors in
H, say,
and .
By Axioms 2, 3, and 8 for the vector space V,
24. So is in H.
Furthermore, if c is any scalar, then by Axioms 7 and 9,
which shows that cu is in H and H is closed under scalar multiplication.
Thus H is a subspace of V.
u w
1 1 2 2 1 1 2 2
u ( v v ) ( )v ( )vc c s s cs cs
Theorem 1: If v1,…,vp are in a vector space V, then Span {v1,…,vp} is a
subspace of V.
We call Span {v1,…,vp} the subspace spanned (or generated) by
{v1,…,vp}.
Give any subspace H of V, a spanning (or generating) set for H is a set
{v1,…,vp} in H such that
1
Span{v ,...v }p
H
26. Basis
Definition 1.1: Basis
A basis of a vector space V is an ordered set of linearly
independent (non-zero) vectors that spans V.
Notation:
1 , , nβ β
Example 1.2:
2 1
,
4 1
B
is a basis for R2
B is L.I. :
2 1 0
4 1 0
a b
→
2 0
4 0
a b
a b
→
0
0
a
b
B spans R2:
2 1
4 1
x
a b
y
→
2
4
a b x
a b y
→
1
2
2
a y x
b x y
L.I. → Minimal
Span → Complete
27. Example 1.3:
1 2
,
1 4
B
is a basis for R2 that differs from B only in order.
Definition 1.5: Standard / Natural Basis for Rn
1 0 0
0 1 0
, , ,
0 0 1
n
E 1 2, , , n e e e
kth component of ei =
1
0
i k
for
i k
28. Example 1.6:
For the function space
cos sin ,a b a b R
a natural basis is cos , sin
Another basis is cos sin , 2cos 3sin
Proof is straightforward.
Example 1.7:
For the function space of cubic polynomials P3 ,
a natural basis is
Other choices can be
Proof is again straightforward.
Rule: Set of L.C.’s of a L.I. set is L.I. if each L.C. contains a different vector.
29. Example 1.8:
The trivial space { 0 } has only one basis, the empty one .
Note: By convention, 0 does not count as a basis vector. ( Any set of
vectors containing 0 is linearly dependent. )
Example 1.9:
The space of all finite degree polynomials has a basis with infinitely many
elements 1, x, x2, … .
Example 1.10: Solution Set of Homogeneous Systems
The solution set of
0
0
x y w
z w
1 1
1 0
,
0 1
0 1
y w y w
Ris
1 1
1 0
,
0 1
0 1
Span
( Proof of L.I. is
left as exercise )
30. Example 1.11: Matrices
Find a basis for this subspace of M22 : 2 0
0
a b
a b c
c
S
2
,
0
b c b
b c
c
S R
Solution:
1 1 2 0
,
0 0 1 0
b c b c
R
∴ Basis is
1 1 2 0
,
0 0 1 0
( Proof of L.I. is
left as exercise )
Theorem 1.12:
In any vector space, a subset is a basis if and only if each vector in the space can be
expressed as a linear combination of elements of the subset in a unique way.
Proof: A basis is by definition spanning
→ every vector can be expressed as a linear combination of the basis vectors.
Let i i i i
i i
c d βv β then i i i
i
c d β 0
∴ L.I. uniqueness
31. Definition 1.13: Representation wrt a Basis
Let B = β1 , …, βn be a basis of vector space V and
Then the representation of v wrt B is
1
i i
n
i
c
βv
1
2
Rep
n
c
c
c
vB
B
cj are called the coordinates (components) of v wrt B.
Subscript B is
often omitted
Example 1.14: P3
Let 2 3
1, 2 , 2 , 2x x xB 2 3
1 ,1 , ,x x x x x x D
Then
2
0
1/ 2
Rep
1/ 2
0
x x
B
B
2
0
0
Rep
1
0
x x
D
D
vB
32. Dimension
Definition 2.1
A vector space is finite-dimensional if it has a basis with only finitely many vectors.
Lemma 2.2: Exchange Lemma
Assume that B = β1 , …, βn is a basis for a vector space, and that for the vector v
the relationship holds:
To be proved:
All bases for a vector space have the same number of elements.
→ Dimension Number of vectors in basis.
→ Basis = Minimal spanning = L.I. set = Smallest set.
Then exchanging βj for v yields another basis for the space.
1 1 j j n nc c c v β β β where cj 0.
Proof: See Hefferon p.120.
33. Theorem 2.3:
In any finite-dimensional vector space, all of the bases have the
same number of elements.
Proof:
Let B = β1 , …, βn be a basis of n elements.
Any other basis D = δ1 , …, δm must have m n.
1 1 1 k k n nc c c δ β β βLet with ck 0.
By lemma 2.2, D1 = β1 , …, βk 1 ,δ1 ,βk + 1 , …, βn is a basis.
Next, replacing βj in D1 begets
D2 = β1 , …, βk 1 ,δ1 ,βk + 1 , …, βj 1 ,δ2 ,βj + 1 , …, βn
Repeating the process n times results in a basis Dn = δ1 , …, δn that spans V.
Which contradicts with the assumption that D is L.I.
1 1 1n n nc c δ δ δ with at least one ck 0.If m > n, then we can write
Hence m = n.
34. Definition 2.4: Dimension
The dimension of a vector space is the number of vectors in any of its bases.
Example 2.5: Rn
Any basis for Rn has n vectors since the standard basis En has n vectors.
→ Rn is n-D.
Example 2.6: Pn
dim Pn = n+1.
since its natural basis, 1, x, x2, …, xn , has n+1 elements.
Example 2.7:
A trivial space is 0-D since its basis is empty.
Comments:
All results in this book are applicable to finite-D vectors spaces.
Most of them are also applicable to countably infinite-D vectors spaces.
For uncountably infinite-D vectors spaces, e.g., Hilbert spaces, convergence most be
taken into account.
35. Corollary 2.8:
No L.I. set can have a size greater than the dimension of the enclosing space.
Example 2.9 : Only subspaces in R3.
2-D: Planes thru 0
1-D: Lines thru 0
0-D: {0}
36. Corollary 2.10:
Any L.I. set can be expanded to make a basis.
Corollary 2.11:
Any spanning set can be shrunk to a basis.
Corollary 2.12:
In an n-D space, a set of n vectors is L.I. iff it spans the space.
Remark 2.13:
The statement ‘any infinite-dimensional vector space has a basis’ is known to be
equivalent to a statement called the Axiom of Choice.
Mathematicians differ philosophically on whether to accept or reject this statement
as an axiom on which to base mathematics (although, the great majority seem to
accept it).
