Chapter Two: Vector Spaces
I. Definition of Vector Space
II. Linear Independence
III. Basis and Dimension• Topic: Fields
• Topic: Crystals
• Topic: Voting Paradoxes
• Topic: Dimensional Analysis
Vector space ~ Linear combinations of vectors.
Ref: T.M.Apostol, “Linear Algebra”, Chap 3, Wiley (97)
I. Definition of Vector Space
I.1. Definition and Examples
I.2. Subspaces and Spanning Sets
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.
I.1. Definition and Examples
Definition 1.1: (Real) Vector Space ( V, ; )A vector space (over ) 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 , [ is the real number field (,+,) 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
Definition 1.1: (Real) Vector Space ( V, + ; )A vector space (over ) 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 , [ is the real number field (,+,) ]a) a v V ( Closure )b) ( a + b ) v = a v + b v ( Distributivity )c) a ( v + w ) = a v + a wd) ( a b ) v = a ( b v ) = a b v ( Associativity )e) 1 v = v
Definition in Conventional Notations
Example 1.3: 2
2 is a vector space if1 1
2 2
x ya b a b
x y
x y 1 1
2 2
ax by
ax by
,a b R
0
0
0with
Example 1.4: Plane in 3.
The plane through the origin 0
x
P y x y z
z
is a vector space.
P is a subspace of 3.
Proof it yourself / see Hefferon, p.81.
Proof it yourself / see Hefferon, p.82.
Example 1.5:
Let & be the (column) matrix addition & scalar multiplication, resp., then
( n, + ; ) is a vector space.
( n, + ; ) is not a vector space since closure is violated under scalar multiplication.
Example 1.6:
0
0
0
0
V
Let then (V, + ; ) is a vector space.
Definition 1.7: A one-element vector space is a trivial space.
Example 1.8: Space of Real Polynomials of Degree n or less, n
0
nk
n k kk
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 nk k
k kk k
a x b x
a b k kka b a b
Scalar multiplication:0
nk
kk
b b a x
a
Zero element:0
0n
k
k
x
0 0k
k 0i.e.,
n is a vector space with vectors 0
nk
kk
a x
a
0
nk
k kk
a b x
0
nk
kk
ba x
i.e.,
kkb baai.e.
,
E.g.,
n is isomorphic to n+1 with 10
0
~ , ,n
k nk n n
k
a x a a
P R
Inverse: 0
nk
kk
a x
a kka ai.e.
,
kkaa
The kth component of a is
Example 1.9: Function Space
The set { f | f : → } of all real valued functions of natural numbers is a vector space if
1 2 1 2f f n f n f n Vector addition:
a f n na fScalar multiplication:
nN
aR
f ( n ) is a vector of countably infinite dimensions: f = ( f(0), f(1), f(2), f(3), … )
E.g.,
2 1f n n ~ 1, 2, 5,10,f
Zero element:
( ) 0zero n
Inverse: ( )f n f n
Example 1.10: Space of All Real Polynomials,
0
, n
kk k
k
a x a n
P R N
is a vector space of countably infinite dimensions.
0 1 20
~ , , ,kk
k
a x a a a
P R
Example 1.11: Function Space
The set { f | f : → } of all real valued functions of real numbers is a vector space of uncountably infinite dimensions.
Example 13: Solution Space of a Linear Homogeneous Differential Equation
2
2: 0
d fS f f
d x
R R is a vector space with
f g x f x g x Vector addition:
a f x xa fScalar multiplication:
Zero element:
( ) 0zero x
Inverse: ( )f x f x
Closure:2 2
2 20 & 0
d f d gf g
d x d x
2
20
d a f bga f bg
d x
→
aR
Example 14: Solution Space of a System of Linear Homogeneous Equations
Remarks:
• Definition of a mathematical structure is not unique.
• The accepted version is time-tested to be most concise & elegant.
Lemma 1.16: Lose Ends
In any vector space V,
1. 0 v = 0 .
2. ( 1 ) v + v = 0 .
3. a 0 = 0 .
v V and a .
Proof:
1 0 0 v v v v 0 v v v 0 v1.
2.
1 1 1 v v v 0 v 0
3.
0a a0 v 0a v 0 v 0
Exercises 2.I.1.
1. At this point “the same” is only an intuition, but nonetheless for each vector space identify the k for which the space is “the same” as k.(a) The 23 matrices under the usual operations(b) The n m matrices (under their usual operations)(c) This set of 2 2 matrices
2.
(a) Prove that every point, line, or plane thru the origin in 3 is a vector space under the inherited operations.(b) What if it doesn’t contain the origin?
00
aa b c
b c
I.2. Subspaces and Spanning Sets
Definition 2.1: SubspacesFor any vector space, a subspace is a subset that is itself a vector space, under the inherited operations.
Example 2.2: Plane in 3 0
x
P y x y z
z
is a subspace of 3.
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 2.3: The x-axis in n is a subspace.
,0, ,0 -axisT
x x r Proof follows directly from the fact that
Example 2.4:
• { 0 } is a trivial subspace of n.
• n is a subspace of n.
Both are improper subspaces.
All other subspaces are proper.
Example 2.5: Subspace is only defined wrt inherited operations.
({1}, ; ) is a vector space if we define 11 = 1 and a1=1 a.
However, neither ({1}, ; ) nor ({1},+ ; ) is a subspace of the vector space (,+ ; ).
Example 2.6: Polynomial Spaces.
n is a proper subspace of m if n < m.
Example 2.7: Solution Spaces.
The solution space of any real linear homogeneous ordinary differential equation, f = 0,
is a subspace of the function space of 1 variable { f : → }.
Example 2.8: Violation of Closure.
+ is not a subspace of since (1) v + v +.
Lemma 2.9:
Let S be a non-empty subset of a vector space ( V, + ; ).
W.r.t. the inherited operations, the following statements are equivalent:
1. S is a subspace of V.
2. S is closed under all linear combinations of pairs of vectors.
3. S is closed under arbitrary linear combinations.
Proof: See Hefferon, p.93.
Remark: Vector space = Collection of linear combinations of vectors.
Example 2.11: Parametrization of a Plane in 3
2 0
x
S y x y z
z
is a 2-D subspace of 3.
2
,
y z
y y z
z
R
2 1
1 0 ,
0 1
y z y z
R
i.e., S is the set of all linear combinations of 2 vectors (2,1,0)T, & (1,0,1)T.
Example 2.12: Parametrization of a Matrix Subspace.
00
aL a b c
b c
is a subspace of the space of 22 matrices.
0,
b cb c
b c
R1 0 1 0
,1 0 0 1
b c b c
R
Definition 2.13: Span
Let S = { s1 , …, sn | sk ( V,+, ) } 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 kk
span S c S c
s s R span 0with
Lemma 2.15: The span of any subset of a vector space is a subspace.
Proof:
Let S = { s1 , …, sn | sk ( V,+, ) }1 1
,n n
k k k kk k
u v span S
u s v sand
1
n
k k kk
a b au bv
w u v s1
n
k kk
w span S
s ,a b RQED
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.
Example 2.16:
For any vV, span{v} = { a v | a } is a 1-D subspace.
Example 2.17:
Proof:
The problem is tantamount to showing that for all x, y , unique a,b s.t.
1 1
1 1
xa b
y
i.e.,a b x
a b y
has a unique solution for arbitrary x & y.
Since 1
2a x y 1
2b x y ,x y R QED
21 1,
1 1span
R
Example 2.18: 2
Let 23 , 2S span x x x 23 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
13
2b c a
and
1 2
13
2c c
2
1
kk
k
c x
= subspace of 2 ?
Lesson: A vector space can be spanned by different sets of vectors.
Definition: Completeness
A subset S of a vector space V is complete if span S = V.
Example 2.19: All Possible Subspaces of 3
See next section for proof.
Planes thru 0
Lines thru 0
Exercises 2.I.2
(a) Show that it is not a subspace of 3. (Hint. See Example 2.5).(b) Show that it is a vector space.
( To save time, you need only prove axioms (d) & (j), and closure under all linear combinations of 2 vectors.)
(c) Show that any subspace of 3 must pass thru the origin, and so any subspace of 3 must involve zero in its description. Does the converse hold?Does any subset of 3 that contains the origin become a subspace when given the inherited operations?
1. Consider the set 1
x
y x y z
z
under these operations.
1 2 1 2
1 2 1 2
1 2 1 2
1x x x x
y y y y
z z z z
1x rx r
r y r y
z rz
2. Because ‘span of’ is an operation on sets we naturally consider how it interacts with the usual set operations. Let [S] Span S.(a) If S T are subsets of a vector space, is [S] [T] ?
Always? Sometimes? Never?(b) If S, T are subsets of a vector space, is [ S T ] = [S] [T] ?(c) If S, T are subsets of a vector space, is [ S T ] = [S] [T] ?(d) Is the span of the complement equal to the complement of the span?
3. Find a structure that is closed under linear combinations, and yet is not a vector space. (Remark. This is a bit of a trick question.)