Download 4 Vector Spaces

Lecture 10
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4.1
Vector Spaces
Basic Definition and Examples
Throughout mathematics we come across many types objects which can be added and multiplied by scalars to arrive at similar types of objects. We have been doing this with R,
C, Mm,n (R), Mm,n (C) for last few days. In the last semester, you have come across with
the space of continuous functions, or the space of differentiable functions etc. What is the
common property to all of these innumerable spaces? If we study them individually, we will
never be able to conclude the study. In our future study or work, when we come across
objects which may be similar to the ones we have met, will we be able to recognize them?
Pattern recognition is one of the basic tenets of all sciences and more so of mathematics.
Thus, it is necessary to isolate the basic properties of these opearations ‘addition and
multiplication by scalars’ and study them independently of the rest of the structure of the
various objects which may differ from one to the other. This leads us to the abstract definition
of a vector space and its studies. Having discussed enough about the model Rn you will see
that there is nothing strange in this somewhat abstract approach that we are going to take.
On the other hand we will reap plenty of advantages.
Let K denote the field of real or complex numbers.
Definition 4.1 A nonemepty set V of objects (called elements or vectors) is called a vector
space over K if the following axioms are satisfied :
I. Closure axioms:
1. (closure under addition) For every pair of elements x, y ∈ V there is a unique element
x + y ∈ V called the sum of x and y.
2. (closure under multiplication by reals) For every x ∈ V and scalar α ∈ K there is a unique
element αx ∈ V called the product of α and x.
[In the modern terminaology, this is formualted by saying that V is a set together with
two operations
+ :V ×V →V; ·:K×V →V
respectively called addition and scalar multiplication. However, the functional notation
+(x, y), ·(α, x) are simplified to x + y, αx respectively, just to keep going with the age-old
practice.]
II. Axioms for addition:
3. (commutative law) x + y = y + x for all x, y ∈ V.
4. (associative law) x + (y + z) = (x + y) + z for all x, y, z ∈ V.
5. (existence of zero element) There exists a unique element 0 in V such that x+0 = 0+x = x
for all x ∈ V.
6. (existence of inverse or negatives) For x ∈ V there exists a unique element written as −x
such that x + (−x) = 0.
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III. Axioms for multiplication by scalars
7. (associativity) For all α, β ∈ K, x ∈ V,
α(βx) = (αβ)x.
8. (distributive law for addition in V ) For all x, y ∈ V and α ∈ K)
α(x + y) = αx + αy.
9. (distributive law for addition in K For all α, β ∈ K and x ∈ V,
(α + β)x = αx + βx
10. (existence of identity for multiplication) For all x ∈ V,
1x = x.
Remark 4.1 Observe that the elements of the field K are called scalars. Depending upon
whether we take K = R, C, in the definition above, we get a real vector space or a complex
vector space. Multiplication will also be referred as scalar multiplication.
Example 4.1
1. V = R with usual addition and multiplication.
2. V = C with usual addition of complex numbers and multiplication by real numbers as
multiplication. This makes C as a real vector space. C is also a complex vector space
under usual addition and scalar multiplication being multiplicaton o complex numbers.
3. Rn = {(a1 , a2 , . . . , an ) : a1 , . . . , an ∈ R}. We refer to Rn as n-dimensional euclidean
space. It is a real vector space. Likewise Cn is a complex vector space.
4. Let S be any set and F (S, K) denote the set of all functions from S to K. Given any
two functions f1 , f2 : S −→ K, and a scalar α ∈ K we define f1 + f2 and αf1 by
(f1 + f2 )(x) = f1 (x) + f2 (x); (αf1 )(x) = αf1 (x), x ∈ S.
Then it is easily verified that F (S, K) becomes a vector space. The zero map which
takes constant value zero is the zero element of this vector space. Clearly the negative
of a function f is the one which takes the value −f (x) for each x ∈ S.
5. Let a < b be real numbers and
V = {f : [a, b] −→ R : f is continuous}.
If the sum and scalar multiplication are defined as in the above example then we know
that V becomes a vector space. This is denoted by C[a, b].
6. Let U be an open subset of Rn . For a positive integer r, let
C r (U) = {f : U) −→ R : f has continuous derivatives of order r}.
This is a real vector space under the operations described in (4). Indeed, we have
· · · ⊂ C r+1 ⊂ C r ⊂ · · · ⊂ C
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7. Let t be an indeterminate. The set
K[t] = {a0 + a1 t + . . . + an tn : a0 , a1 , . . . , an ∈ K}
is a vector space over K under usual addition of polynomials and multiplication of
polynomials with real numbers.
′′
′
8. The set of all solutions to the differential equation y + ay + by = 0 where a, b ∈ R
form a real vector space.
9. Let V = Mm,n (K) denote the set of all m × n matrices with entries in K. Then V is
a vector space over K under usual matrix addition and multiplication of a matrix by
scalar.
The above examples indicate that the notion of a vector space is quite general. A result
proved for vector spaces will simultaneously apply to all the above different examples.
4.2
Subspaces and Linear Span
Definition 4.2 A nonempty subset W of a vector space V is called a subspace of V if it
is a vector space under the operations in V.
Theorem 4.1 A nonempty subset W of a vector space V is a subspace of V if W satisfies
the two closure axioms.
Proof: If W is a subspace of V then it satisfies the closure axioms. Suppose now that W
satisfies the closure axioms. We just need to prove existence of inverses and the zero element.
Let x ∈ W. By distributivity (0 + 0)x = 0x = 0x + 0x. Hence 0x = 0. By closure axioms
0 ∈ W. If x ∈ W then (−1)x is the negative of x which is in W by closure axioms.
♠
Example 4.2
1. The subset [0, ∞) ⊂ R is not a subspace. None of the sets N, Z, Q are (real) subspaces
of the vector space R. Neither is the set (−1, 1).
2. R is a subspace of the real vector space C. But it is not a subspace of the complex
vector space C.
3. C r [a, b] is a subspace of the vector space C s [a, b] for r ≥ s. All of them are subspaces
of F ([a, b]; R).
4. Mm,n (R) is a subspace of the real vector space Mm,n (C).
5. The set of points on the x-axis form a subspace of the plane. More generally, the set
of points on a line passing through the origin is a subspace of R2 . Likewise the set of
real solutions of a1 x1 + · · · + an xn = 0 form a subspace of Rn . It is called a hyperplane.
More generally, the set of solutions of a system of linear equations in n variables forms
a subspace of Kn .
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Lecture 11
Definition 4.3 Let S be a subset of a vector space V. The linear span of S is the subset
L(S) =
(
n
X
)
ci xi : x1 , x2 , . . . , xn ∈ S and c1 , c2 , . . . , cn are scalars
i=1
We set L(∅) = {0} by convention. A typical element
combination of xi ’s.
Pn
i=1 ci xi
of L(S) is called a linear
Theorem 4.2 The smallest subspace of V containing S is L(S).
Proof: If S ⊂ W ⊂ V and W is a subspace of V then by closure axioms L(S) ⊂ W. If we
show that L(S) itself is a subspace the proof will be completed. It is easy to verify that L(S)
is closed under addition and scalar multiplication and left to you as an exercise.
♠
Remark 4.2 (i) Different sets may span the same subspace. For example
L({î, ĵ}) = L({î, ĵ, î + ĵ}) = R2 .
The vector space R[t] is spanned by {1, t, t2 , . . . , tn } and also by {1, (1 + t), . . . , (1 + t)n }.
(ii) The linear span of any non zero element in a field K is the field itself.
(iii) While talking about the linear span or any other vector space notion, the underlying
field of scalars is understood. If we change this we get different objects and relations. For
instance, the real linear span of 1 ∈ C is the set of all real numbers where as the complex
linear span is the whole of C. In what follows, we often do not mention which field is involved
unless it is necessary to mention it specifically.
4.3
Linear Dependence
Definition 4.4 Let V be a vector space. A subset S of V is called linearly dependent (L.D.)
if there exist distinct elements v1 , . . . , vn ∈ S and αi ∈ K, not all zero, such that
n
X
αi vi = 0.
(43)
1
S is called linearly independent (L.I.) if it is not linearly dependent.
Remark 4.3 (i) Thus a relation such as (43) holds in a linearly independent set S with
distinct vi ’s iff all the scalars αi = 0.
(ii) Any subset which contains a L.D. set is again L.D.
(iii) The singleton set {0} is L.D. in every vector space.
(iv) Any subset of a L.I. set is L.I.
Example 4.3
(i) The set { ei = (0, . . . , 1, . . . , 0), 1 ≤ i ≤ n} is L.I. in Kn . This can be shown easily by
taking dot product with ei with a relation of the type (43).
(ii) The set S = {1, t, t2, . . . , tn } is L.I in K[t]. (Here 1 denotes the constant function 1.) To
prove this, let αi be arbitrary constants so that
n
X
αi ti = 0.
i=0
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(44)
Differentiate k times and put t = 0 to obtain αk = 0 for each k.
(iii) In the space C[a, b] for a < b ∈ R consider the set S = {1, cos2 t, sin2 t}. (Here again 1
denotes the constant function 1.) The familiar formula cos2 t + sin2 t = 1 tells you that S is
linearly dependent. What about the set {1, cos t, sin t}?
Lemma 4.1 Let V of a vector space V, S ⊂ V which is linearly independent. Then for any
vector v ∈ V \ L(S), S ∪ {v} is linearly independent.
Proof: Suppose ki=1 αi vi + βv = 0, with vi ∈ S. If β 6= 0, then this implies v =
P
−1 Pk
( i=1 αi vi ) ∈ L(S). Therefore β = 0. But then ki=1 αi vi = 0 and S is linearly indeβ
pendent and hence α1 = α2 = · · · αk = 0.
P
Theorem 4.3 Let S be a finite subset of a vector space and S1 be a subset of S which is
linearly independent. Then there exists a subset S2 such that S1 ⊂ S2 ⊂ S such that
(i) S2 is linearly independent and
(ii) L(S2 ) = L(S).
Proof: If L(S1 ) = L(S) take S2 = S1 . Otherwise it follows that S \ L(S1 ) 6= ∅. So, there
exists v1 ∈ S \ L(S1 ) and hence by the previous lemma, S1 ∪ {v1 } is linearly independent.
Replace S1 by this new set and repeat the above argument. Since S is a finite set this process
should terminate and yield the desired result.
♠
Theorem 4.4 Let S be a finite subset of a vector space and S1 be a subset of S which is
linearly independent. Then there exists a subset S2 such that S1 ⊂ S2 ⊂ S such that
(i) S2 is linearly independent and
(ii) L(S2 ) = L(S).
Proof: We shall first prove that if v ∈ V \ L(S1 ) then S1 ∪ {v} is linearly independent. If not
P
P
suppose ki=1 αi vi + βv = 0, with vi ∈ S1 . If β 6= 0 then this implies v = −1
( ki=1 αi vi ) ∈
β
P
L(S1 ). Therefore β = 0. But then ki=1 αi vi ) = 0 and S1 is linearly independent and hence
α1 = α2 = · · · αk = 0.
Now if L(S1 ) = L(S) take S2 = S1 . Otherwise it follows that S \ S1 6= ∅. So, there exists
v1 ∈ S \ S1 and hence S1 ∪ {v1 } is linearly independent. Replace S1 by this new set and
repeat the above argument. Since S is a finite set this process should terminate and yield
the desired result.
♠
Definition 4.5 Let S be a subset of a vector space. It is called a basis for L(S) if S is
linearly independent.
Thus a basis of a vector space V is a linearly independent subset S of V such that L(S) = V.
Recall that we define L(∅) = (0), the vector space consisting of single element.
Definition 4.6 A vector space V is called finite dimensional if there exists a finite set S ⊂ V
such that L(S) = V.
Remark 4.4 By the above theorem, it follows that every finite dimensional vector space has
a finite basis. To see this choose a finite set S such that L(S) = V and apply the theorem
with S1 = ∅. Indeed we can prove a slightly stronger result:
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Theorem 4.5 Let V be a finite dimensional vector space and S1 be a (finite) linearly independent subset of V. Then there exists a finite basis S2 of V such that S1 ⊂ S2 .
Proof: Choose a finite set S ′ such that L(S ′ ) = V. Put S = S ′ ∪ S1 . Then S1 ⊂ S which is
linearly independent and L(S) = V. By the previous result, there is a finite L.I. subset S2 of
S such that S1 ⊂ S2 and L(S2 ) = V.
♠
Remark 4.5 We need not even assume that S1 is finite set the above theorem. The finiteness
of S1 is a consequence of the following theorem.
Lecture 12
Theorem 4.6 For any S = {v1 , . . . , vk } ⊂ V every subset S ′ of L(S) with more than k
elements is linearly dependent.
Proof: Choose k + 1 elements ui of S ′ in some order, express them as linear combinations of
P
elements of S say, ui = kj=1 aij vj . Consider the (k + 1) × k matrix A = ((aij )). To show that
ui are linearly dependent, we will show that the row vectors Ri of A are linearly dependent.
P
P
For a relation of the type i αi Ri = 0 implies that i αi aij = 0 for each j and hence,
X
αi ui =
i
X
i
αi
X
aij vj =
X X
(
j
j
αi aij )vj = 0.
i
Apply GEM to A. Since the number of columns is one less than the number of rows, in
G(A) the last row must be identically zero. We know that G(A) = RA for some invertible
matrix R. If z = (α1 , . . . , αk+1) denotes the last row of R then it cannot be identically zero.
Moreover we have zA = (0, . . . , 0) the last row of G(A). This means that row vectors of A
are linearly dependent.
♠
Corollary 4.1 For a finite dimensional vector space, the number of elements in any two
bases will be the same.
Definition 4.7 Given a finite dimensional vector space V , its dimension is defined to be the
number of elements in any basis for V. This is well defined because of the above corollary.
Example 4.4
1. A basis for the vector space {0} is an emptyset. It follows that the
dimnesion of {0} is 0.
2. dim K = 1 for any field considered as a vector space over itself.
3. dimC C = 1; however, dimR C= 2.
4. dim Rn = n.
5. dim Mm,n (K) = mn.
6. Let Symn (K) denote the space of all symmetric n×n matrices with entries in K. What
is the dimension of Symn (K)? (Answer: n(n + 1)/2.) Indeed, check that
{Eii , 1 ≤ i ≤ n} ∪ {Eij + Eji , i < j}
forms a basis.
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7. Let Hermn (C) be the space of all n × n matrices A with complex entries such that
A = A∗ (= ĀT ). This is a vector space over R and not over C. (Why?) What is the
value of dimR [Hermn (C)]? (Answer:n2 .) Indeed,
{Eii , 1 ≤ i ≤ n} ∪ {Eij + Eji , i < j} ∪ {ı(Eij − Eji), i < j}
forms a basis.
Exercise:
(i) What is the dimension of K[t]? Note that if S is any fintie subset of K[t] then there is
an integer n such that the degree of any element in S is less than n. It follows that tn+1
cannot be in L(S). Thus no finite subset can span the whole of K[t] and hence it is not finite
dimensional. In such a case, we say the vector space is infinite dimensional.
(ii) Show that in any vector space of dimension n any subset S such that L(S) = V has
at least n elements.
(iii) Let V be a finite dimensional vector space. Suppose S is a subset which is linearly
independent with maximal possible number of elements, i.e., if you put any more elements
V in S then it will not remain linearly independent. Show that S is a basis for V. (iv) Show
that a subspace W of a finite dimensional space V is finite dimensional. Further prove that
dim W ≤ dim V and equality holds iff W = V. (Most of these results are true for infinite
dimensional case also. But this last mentioned result is an exception: A subspace of a finite
dimensional space need not be finite dimensional. Given any infinite dimensional vector
space, there are proper subspaces W of V with dim W = dim V. You are welocme to find
one such example inside K[t].
(v)(Optional) Let V and W be vector subspaces of another vector spaceU. Show that
L(V ∪ W ) = {v + w : v ∈ V, w ∈ W }.
This subspace is denoted by V + W and is called the sum of V and W. Show that V ∩ W is
a subspace and
dim (V + W ) = dim V + dim W − dim (V ∩ W ).
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