Chapter Two: Vector Spaces
Vector space ~ Linear combinations of vectors.
I. Definition of Vector Space
II. Linear Independence
III. Basis and Dimension
•
•
•
•
Topic: Fields
Topic: Crystals
Topic: Voting Paradoxes
Topic: Dimensional Analysis
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,  ; 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
Definition in Conventional Notations
Definition 1.1: (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
Example 1.3: R2
R2
is a vector space if
with
x 
 y   ax  by1 
ax  by  a  1   b  1    1

x
y
ax

by
 2
 2  2
2
 a, b  R
 0
0 
 0
Proof it yourself / see Hefferon, p.81.
Example 1.4: Plane in R3.
The plane through the origin


P


 x
 y x  y  z  0
 
z
 
P is a subspace of R3.
Proof it yourself / see Hefferon, p.82.





is a vector space.
Example 1.5:
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 1.6:
Let



V 


 0
 0
 
 0
 
 0






then (V, + ; R ) 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, Pn

Pn  


ak x ak  R 

k 0


n
k
n
The kth component of a is
k 0
 ak  ak
n
a  b   ak x   bk x 
k
k 0
k
k 0
n
 a
k 0
0
n
 0x
 bk  x k i.e.,  a  bk  ak  bk
k 0
 a     ak  x k
i.e.,
k 0
n
Pn is isomorphic to Rn+1 with
i.e.,
b ak  bak
i.e.,  0k  0 k
k
n
Inverse:
k
n
 n
k 
b a  b   a k x    b ak x k
k 0
 k 0
Scalar multiplication:
Zero element:
n
a   ak x k
Pn is a vector space with vectors
Vector addition:

P3  a0  a1 x  a2 x 2  a3 x 3 ak  R
E.g.,
a
k 0
k
x
 Pn
k
 ak  ak
~
 a0 ,
, an   R n 1
Example 1.9: Function Space
The set { f | f : N → R } of all real valued functions of natural numbers
is a vector space if
Vector addition:
Scalar multiplication:
Zero element:
Inverse:
 f1  f 2  n   f1  n   f 2  n 
n N
 a f  n   a f  n 
a R
zero(n )  0
  f  (n)   f  n 
f ( n ) is a vector of countably infinite dimensions: f = ( f(0), f(1), f(2), f(3), … )
E.g.,
f  n   n2  1
~
f  1, 2, 5, 10,

Example 1.10: Space of All Real Polynomials, P

P 


a k x a k  R, n  N 

k 0

n
k
P is a vector space of countably infinite dimensions.

a
k 0
k
xk P
~
 a0 , a1, a2 ,   R
Example 1.11: Function Space
The set { f | f : R → R } 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

S  f :RR


d2 f
 f 0 
d x2

Vector addition:
Scalar multiplication:
Zero element:
Inverse:
Closure:
f
is a vector space with
 g  x   f  x   g  x 
 a f  x   a f  x 
a R
zero( x )  0
  f  ( x)   f  x 
d 2  a f  bg 
d2 f
d2g
 f 0 &
g 0 →
  a f  bg   0
2
2
dx
dx
d x2
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.
0v=0.
2.
( 1 ) v + v = 0 .
3.
a0=0.
 v V and a  R.
Proof:
1.
2.
3.
0  v  v  1  0 v  v  v  0v  v  0v
 1 v  v   1  1 v  0v  0
a 0  a  0 v    a 0 v  0 v  0
Exercises 2.I.1.
1.
(a)
(b)
(c)
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 Rk.
The 23 matrices under the usual operations
The n  m matrices (under their usual operations)
This set of 2  2 matrices



 a 0
b c a  b  c  0





2.
(a)
(b)
Prove that every point, line, or plane thru the origin in R3 is a
vector space under the inherited operations.
What if it doesn’t contain the origin?
I.2. Subspaces and Spanning Sets
Definition 2.1: Subspaces
For any vector space, a subspace is a subset that is itself a vector space,
under the inherited operations.
Note: A subset of a vector space is a subspace iff it is closed under  & .
→ It must contain 0.
Example 2.2: Plane in R3
Proof:
→

with
→
Let
(c.f. Lemma 2.9.)


P


 x
 y x  y  z  0
 
z
 





is a subspace of R3.
r1   x1 , y1 , z1  , r2   x2 , y2 , z2   P
T
T
x1  y1  z1  0 , x2  y2  z2  0
a r1  br2   ax1  bx2 , ay1  by2 , az1  bz2 
T
 ax1  bx2    ay1  by2    az1  bz2   a  x1  y1  z1   b  x2  y2  z2 
ar1  br2  P  a, b  R
QED
0
Example 2.3: The x-axis in Rn is a subspace.
Proof follows directly from the fact that
r   x,0,
,0   x-axis
T
Example 2.4:
• { 0 } is a trivial subspace of Rn.
• Rn is a subspace of Rn.
Both are improper subspaces.
All other subspaces are proper.
Example 2.5: Subspace is only defined wrt inherited operations.
({1}, ; R) is a vector space if we define 11 = 1 and a1=1 aR.
However, neither ({1}, ; R) nor ({1},+ ; R) is a subspace of the vector space
(R,+ ; R).
Example 2.6: Polynomial Spaces.
Pn is a proper subspace of Pm if n < m.
Example 2.7: Solution Spaces.
The solution space of any real linear homogeneous ordinary
differential equation, L f = 0,
is a subspace of the function space of 1 variable { f : R → R }.
Example 2.8: Violation of Closure.
R+ is not a subspace of R since (1) v  R+  v R+.
Lemma 2.9:
Let S be a non-empty subset of a vector space ( V, + ; R ).
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 R3


S












 x
 y x  2y  z  0
 
z
 
2y  z
 y  y, z  R


 z 


is a 2-D subspace of R3.
   2
 1 
   
 
   y 1  z  0 
1
   0


 
 


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.

L

 a 0
b c a  b  c  0








 b  c 0 
 b
 b, c  R
c



  1 0 
 1 0 
  b 
  c  0 1  b, c  R
1
0




 
is a subspace of the space of 22 matrices.



Definition 2.13: 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.,

span S  

n
 ck sk
k 1

sk  S , ck  R 

with
span    0 
Lemma 2.15: The span of any subset of a vector space is a subspace.
Proof:
Let S = { s1 , …, sn | sk  ( V,+,R ) }

and
n
n
k 1
k 1
n
n
k 1
k 1
u   uk sk , v   vk sk  span S
w  au  bv    auk  bvk  sk   wk sk  span 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.
QED
Example 2.16:
For any vV, span{v} = { a v | a R } is a 1-D subspace.
Example 2.17:
  1  1  
span    ,     R2
 1  1 
Proof:
The problem is tantamount to showing that for all x, y R,  unique a,b R s.t.
 x
 1
1

a

b
 y
 1
 1
 
 
 
i.e.,
Since
ab x
ab  y
a
1
 x  y
2
has a unique solution for arbitrary x & y.
b
1
 x  y
2
 x, y  R
QED
Example 2.18: P2
Let

2
S  span  3x  x 2 , 2 x   a  3x  x   2bx a, b  R
Question:
?
S  P2
c0 0




k
c
x


k
k1

2

= subspace of P2 ?
Answer is yes since
c1  3a  2b
c2  a
and
a  c2
b
1
1
c

3
a

1

 c1  3c2 
2
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 R3
Planes thru 0
Lines thru 0
See next section for proof.
Exercises 2.I.2
1. Consider the set






 x
 y x  y  z 1 

 
z

 

under these operations.
 x1   x2   x1  x2  1
 y  y   y  y 
2 
 1  2  1
z  z   z z 
 1  2  1 2 
(a)
(b)
(c)
 x   rx  r  1
r  y    r y 
 z   rz 
  

Show that it is not a subspace of R3. (Hint. See Example 2.5).
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.)
Show that any subspace of R3 must pass thru the origin, and so any
subspace of R3 must involve zero in its description.
Does the converse hold?
Does any subset of R3 that contains the origin become a subspace
when given the inherited operations?
2.
(a)
(b)
(c)
(d)
3.
Because ‘span of’ is an operation on sets we naturally consider how it
interacts with the usual set operations. Let [S]  Span S.
If S  T are subsets of a vector space, is [S]  [T] ?
Always? Sometimes? Never?
If S, T are subsets of a vector space, is [ S  T ] = [S]  [T] ?
If S, T are subsets of a vector space, is [ S  T ] = [S]  [T] ?
Is the span of the complement equal to the complement of the span?
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.)