Download Chapter 4, General Vector Spaces Section 4.1, Real Vector Spaces

Chapter 4, General Vector Spaces
Section 4.1, Real Vector Spaces
In this chapter we will call objects that satisfy a set of axioms as vectors. This can be
thought as generalizing the idea of vectors to a class of objects.
Vector space axioms:
Definition: Let V be an arbitrary nonempty set of objects on which two operations
are defined on which two operations are defined addition, and multiplication by scalars.
By addition we mean a rule for associating with each pair of objects u and v in V an
object u+v, called the sum of u and v;
By scalar multiplication we mean a rule for associating with each scalar k and each object
u in V an object ku, called scalar multiplication of u by k.
If the following axioms are satisfied by all objects u,v,w in V and all scalars k and m, then
we call V a vector space and we call the objects in V vectors.
1. If u and v are vectors in V, then u+v is in V.
2. u+v=v+u
3. u+(v+w) = (u+v)+w
4. There is an object 0 in V, called a zero vector for V, such that 0 + u = u + 0 = u
for all u in V.
5. For each u in V, there is an object -u in V, called a negative of u such that u+(−u) =
(−u) + u = 0
6. If k is any scalar and u is any object in V, then ku is in V.
7. k(u+v)=ku+kv
8. (k+m)u=ku+mu
9. k(mu)=(km)u
10. 1u=u
Note:
• When the scalars are real numbers, V is called a real vector space.
• When the scalars are complex numbers, V is called a complex vector space.
• Addition and scalar multiplication do not need to be the operations we defined
before.
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Examples:
1. Zero vector space: V = {0}, 0+0=0, k0=0
2. The set of all n × m matrices with real entries, with
addition= matrix addition , and scalar multiplication= scalar matrix multiplication
forms a real vector space.
(a) If A and B are n × m matrices then A+B is also a n × m matrix
(b) A+B=B+A
(c) A+(B+C)=(A+B)+C
(d) There is zero matrix 0, such that A+0=0+A=A
(e) For each matrix A we have -A such that A+(-A)=0
(f) If k is any real scalar, then kA is in V
(g) k(A+B)=kA+kB
(h) (k+m)A=kA+mA
(i) k(mA)=(km)A
(j) 1A=A, here 1 is the scalar 1.
These properties hold by Theorem 1.4.1 and Theorem 1.4.2.
3. Every plane through the origin is a vector space:
Here our set is the set of points of the plane, addition and scalar multiplications are
the usual ones.
Then we have 0 in the plane, If ax + by + cz = 0 is the plane equation, then for
any u = (u1 , u2 , u3 ), v = (v1 , v2 , v3 ) on the plane, then u+v will be on the plane as
well. Other properties can easily be checked.
4. The set of all real-valued functions defined on the entire real line (−∞, ∞) with
addition (f + g)(x) = f (x) + g(x)
scalar multiplication (kf )(x) = kf (x)
is a real vector space. We have 1= identity function, 0=zero function
Example: Let V be the set of positive real numbers. Show that V is a vector space with
addition: u + v = uv, and scalar multiplication: ku = uk
Solution: We need to check that vector space axioms are satisfied by the objects of V.
1. Let u and v be objects in V, this means u and v are positive real numbers, then
u+v=uv must also be in V as uv will be positive real number.
2. u+v=v+u due to property of real numbers
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3. u+(v+w)=(u+v)+w this also holds as u,v,w are real numbers
4. There must be a zero vector 0 in V, such that 0+u=u+0=u,
Let us denote this zero vector by a, then we want u=a+u=au, dividing by u gives
a=1. That is zero vector of V is 1.
5. For each u in V there must be an object -u in V such that u+(-u)=(-u)+u=0. In
our vector space 0=1, so we want (-u) such that u+(−u) = u(−u) = 1 ⇒ −u = u1
6. if k is any scalar and u is an object in V, then ku must be in V. We have ku = uk
and since u is positive real number uk will be also a positive real number.
7. k(u+v)=ku+kv; Note that k(u + v) = (uv)k = uk v k = ku + kv
8. (k + m)u = uk+m = uk um = ku + km
9. k(mu) = k(um ) = (um )k = umk = (km)u
10. 1u = u1 = u
Examples that are not vector spaces:
1. Let V = R2 and define addition as v + w = (v1 + w1 , v2 + w2 ) and scalar multiplication as kv = (kv1 , 0).
Now consider 1v = (v1 , 0) 6= (v1 , v2 ). Hence V cannot be a vector space.
a 1
2. Consider the set of all 2 × 2 matrices of the form
with the standard matrix
1 b
addition and scalar multiplication. Consider
a1 1
a2 1
a1
a2
2
+
=
1 b1
1 b2
2 b1 + b2
this sum is not in the set so this set with the standart matrix addition and scalar
multiplication is not a vector space.
Theorem 4.1.1: Let V be a vector, u a vector in V, and k a scalar; then:
(a) 0u=0
(b) k0=0
(c) (-1)u=-u
(d) If ku=0, then k=0 or u=0.
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Section 4.2 Subspaces
Definition: A subset W of a vector space V is called a subspace of V if W is itself
a vector space under the addition and scalar multiplication defined on V.
• There is a quick way to decide if a subset of a vector space is a subspace or not.
Theorem 4.2.1: If W is a set of one or more vectors from a vector space V, then
W is a subspace of V if and only if the following conditions hold
(a) If u and v are vectors in W, then u+v is in W
(b) If k is any scalar and u is any vector in W, then ku is in W.
• If (a) is satisfied we call W to be closed under addition
• If (b) is satisfied we call W to be closed under scalar multiplication
Example:
1. {0} is a subspace of Rn , since 0 + 0 = 0 (closed under addition), k0 = 0(closed
under scalar multiplication)
2. Rn is a subspace of Rn
• {0} and Rn are called trivial subspaces of Rn
3. Let W be a plane through origin determined by two vectors v1 and v2 that is we
can express any vector x in W as x = t1 v1 + t2 v2 . Then W is a subspace of Rn .
Note: Every subspace of Rn must contain the vector 0. Why?
Example: Determine whether the given sets are subspaces of R3 . If it is not indicate
which closure properties fail.
(a) All vectors of the form (a, 0, 0)
(b) All vectors with integer coefficients
(c) All vectors (a, b, c) for which b = a + c
Solution: For each set we need to check for any two vectors v, w in the set, if v + w
in the set and kv is in the set for any scalar k ∈ R.
(a) Let v = (a, 0, 0), w = (b, 0, 0) then v + w = (a + b, 0, 0), kv = (ka, 0, 0) and these
are both in the set hence it is a subspace of R3 .
(b) Let v = (a1 , a2 , a3 ), w = (b1 , b2 , b3 ) where ai , bi ∈ Z, then v + w = (a1 + b1 , a2 +
b2 , a3 + b3 ), kv = (ka1 , ka2 , ka3 ). kv will not be in the set if k ∈ R − Z. Hence this
set is not a subspace of R3 .
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(c) Let v = (a1 , b1 , c1 ), w = (a2 , b2 , c2 ) we have v+w and kv for any real number k are
in the set. Hence this set is a subspace of R3 .
Theorem 4.2.3: If S = {w1 , w2 , . . . , wr } is a nonempty set of vectors in a vector
space V, then:
1. The set W of all possible linear combinations of the vectors in S is a subspace of V
2. The set W in part (a) is the smallest subspace of V that contains all of the vectors
in S in the sense that any other subspace that contains those vectors contains W.
Idea of the proof: Let us form a subspace from the vectors v1 , . . . , vr . For a subspace we need a subset that is closed under scalar multiplication and addition. So if vi is
in the set where i = 1, . . . , r, then kvi must be in the set for any scalar k, so we need
t1 v1 , t2 v2 , . . . , tr vr to be in the set. Hence t1 v1 + t2 v2 + . . . + tr vr must be in the set.
Definition: If S = {v1 , . . . , vr } is a set of vectors in a vector space V, then the
subspace W of V consisting of all linear combinations of the vectors in S is called the
space spanned by v1 , . . . , vr and we say that the vectors v1 , . . . , vr span W. We denote
this by W=span(S) or W=span{v1 , . . . , vr }.
Example:
1. {0} = Span{0}
2. Rn = Span{e1 , e2 , . . . , en }
3. Line passing through origin and parallel to the vector v = Span{v}
4. Plane passing through origin and parallel to the noncolliner vectors v1 , v2 =Span{v1 , v2 }
5. Set of polynomials with degree less than or equal to n, denoted by Pn with ususal
polynomial addition and scalar multiplication forms a vector space and Pn =Span{1, x, x2 , x3 , . . . , xn }
Example: Determine whether the polynomials p1 = 1 − x + 2x2 , p2 = 3 + x, p3 =
5 − x + 4x2 , p4 = −2 − 2x + 2x2 span P2 ?
Solution: If these polynomials span P2 then any arbitrary vector ax2 + bx + c in P2
can be expressed as a linear combination of p1 , p2 , p3 , p4 . That is
ax2 + bx + c = k1 p1 + k2 p2 + k3 p3 + k4 p4 where k1 , k2 , k3 , k4 ∈ R
Expressing this equation in terms of components we get
a = 2k1 + 4k3 + 2k4
b = −k1 + k2 − k3 − 2k4
c = k1 + 3k2 + 5k3 − 2k4
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Hence our question is
a,b,c values. Form the

2 0
−1 1
1 3
reduced to determining whether
augmented matrix;


1 0
4
2 a
Row operations


0 1
−1 −2 b
→
5 −2 c
0 0
this system is consistent for all
2 1
1 −1
0 0
a
2

b + a2 
c−2a−3b
3
From the last row we see that linear system is consistent if c = 3b + 2a which gives restrictions on the scalars a,b,c. Hence p1 , p2 , p3 , p4 cannot span P2 . For example x2 +x+6
is in P2 but not in the span of p1 , p2 , p3 , p4 .
Theorem 4.2.4: The solution set of a homogenous linear system Ax = 0 in n unknowns is a subspace of Rn .
Exercise: Prove this theorem
Example: Find a general solution of the linear system and list a set of vectors that
span the solution space
x1 + 6x2 + 2x3 − 5x4 = 0
−x1 − 6x2 − x3 − 3x4 = 0
2x1 + 12x2 + 5x3 − 18x4 = 0
Solution:




1
6
2 −5 0
1 6 2 −5 0
−1 −6 −1 −3 0 Row operations
0 0 1 −8 0
→
2 12 5 −18 0
0 0 0 0 0
x1 , x3 are leading variables, x2 , x4 are free variables, let x2 = s, x4 = t ⇒ x3 = 8t, x1 =
−6s − 11t
  

 


x1
−6s − 11t
−6
−11
x2  

 


s
 =
 = s 1  + t 0 
x3  

0
 8 
8t
x4
t
0
1

  
−11
−6
1  0 

 
⇒ Solution space=Span{
 0  ,  8 }
1
0
Theorem 4.2.5: If S = {v1 , v2 , . . . , vr } and S 0 = {w1 , w2 , . . . , wk } are two sets of
vectors in a vector space V, then
span {v1 , v2 , . . . , vr } = span {w1 , w2 , . . . , wk }
if and only if each vector in S is a linear combination of those in S’ and each vector in S’
is a linear combination of those in S.
Importance: Different number of vectors and different vectors can span the same
space. We will see that one needs a minimum number of such vectors and that number
is unique. For that we need linear independency.
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