Module 4

Vector Spaces

Definition-

Let ‘F’ be any given field, then a given set V is said to be a vector space if-

1. There is a defined composition in ‘V’. This composition called addition of vectors which is denoted by ‘+’

2. There is a defined an external composition in ‘V’ over ‘F’. That will be denoted by scalar multiplication.

3. The two compositions satisfy the following conditions-

(a)

(b)

(c)

(d) If then 1 is the unity element of the field F.

If V is a vector space over the field F, then we will denote vector space as V(F).

Note that each member of V(F) will be referred as a vector and each member of F as a scalar

Note- Here we have denoted the addition of the vectors by the symbol ‘+’. This symbol also be used for the addition of scalars.

If then represents addition in vectors(V).

If then a+b represents the addition of the scalars or addition in the field F.

Note- For V to be an abelian group with respect to addition vectors, the following conditions must be followed-

1.

2.

3.

4. There exists an element which is also called the zero vector. Such that-

It is the additive identity in V.

5. There exists a to every vector such that

This is the additive inverse of V

Note- All the properties of an abelian group will hold in V if (V,+) is an abelian group.

These properties are-

1.

2.

3.

4.

5. The additive identity 0 will be unique.

6. The additive inverse will be unique.

7.

Note- The Greek letters are used to denote elements of vectors and Latin letters a , b , c etc. for scalars that means the elements of field F.

Note- Suppose C is the field of complex numbers and R is the field of real numbers, then C is a vector space over R, here R is the subfield of C.

Here note that R is not a vector space over C.

Example-: The set of all matrices with their elements as real numbers is a vector space over the field F of real numbers with respect to addition of matrices as addition of vectors and multiplication of a matrix by a scalar as scalar multiplication.

Sol. V is an abelian group with respect to addition of matrices in groups.

The null matrix of m by n is the additive identity of this group.

If then as is also the matrix of m by n with elements of real numbers.

So that V is closed with respect to scalar multiplication.

We conclude that from matrices-

1.

2.

3.

4. Where 1 is the unity element of F.

Therefore we can say that V (F) is a vector space.

Example-2: The vector space of all polynomials over a field F.

Suppose F[x] represents the set of all polynomials in indeterminate x over a field F. The F[x] is vector space over F with respect to addition of two polynomials as addition of vectors and the product of a polynomial by a constant polynomial.

Example-3: The vector space of all real valued continuous (differentiable or integrable) functions defined in some interval [0,1]

Suppose f and g are two functions, and V denotes the set of all real valued continuous functions of x defined in the interval [0,1]

Then V is a vector space over the filed R with vector addition and multiplication as:

And

Here we will first prove that V is an abelian group with respect to addition composition as in rings.

V is closed with respect to scalar multiplication since af is also a real valued continuous function

We observe that-

1. If and f,g , then

2. If and , then

3. If and , then

4. If 1 is the unity element of R and , then

So that V is a vector space over R.

Some important theorems on vector spaces-

Theorem-1: Suppose V(F) is a vector space and 0 be the zero vector of V. Then-

1.

2.

3.

4.

5.

6.

Proof-

1. We have

We can write

So that,

Now V is an abelian group with respect to addition.

So that by right cancellation law, we get-

2. We have

We can write

So that,

Now V is an abelian group with respect to addition.

So that by right cancellation law, we get-

3. We have

is the additive inverse of

4. We have

is the additive inverse of

5. We have-

6. Suppose the inverse of ‘a’ exists because ‘a’ is non-zero element of the field.

Again let-

, then to prove let then inverse of ‘a’ exists

We get a contradiction that must be a zero vector. Therefore ‘a’ must be equal to zero.

So that

Definition

Suppose V is a vector space over the field F and . Then W is called subspace of V if W itself a vector space over F with respect to the operations of vector addition and scalar multiplication in V.

Theorem-1- The necessary and sufficient conditions for a non-empty sub-set W of a vector space V(F) to be a subspace of V are-

1.

2.

Proof:

Necessary conditions-

W is an abelian group with respect to vector addition If W is a subspace of V.

So that

Here W must be closed under a scalar multiplication so that the second condition is also necessary.

Sufficient conditions-

Let W is a non-empty subset of V satisfying the two given conditions.

From first condition-

So that we can say that zero vector of V belongs to W. It is a zero vector of W as well.

Now

So that the additive inverse of each element of W is also in W.

So that-

Thus W is closed with respect to vector addition.

Theorem-2:

The intersection of any two subspaces and of a vector space V(F) is also a subspace of V(F).

Proof: As we know that therefore is not empty.

Suppose and

Now,

And

Since is a subspace, therefore-

and then

Similarly,

then

Now

Thus,

And

Then

So that is a subspace of V(F).

Example-1: If are fixed elements of a field F, then set of W of all ordered triads ( of elements of F,

Such that-

Is a subspace of

Sol. Suppose are any two elements of W.

Such that

And these are the elements of F, such that-

…………………….. (1)

……………………. (2)

If a and b are two elements of F, we have

Now

=

=

=

So that W is subspace of

Example-2: Let R be a field of real numbers. Now check whether which one of the following are subspaces of

1. {

2. {

3. {

Solution- 1. Suppose W = {

Let be any two elements of W.

If a and b are two real numbers, then-

Since are real numbers.

So that and or

So that W is a subspace of

2. Let W = {

Let be any two elements of W.

If a and b are two real numbers, then-

So that W is a subspace of

3. Let W ={

Now is an element of W. Also is an element of R.

But which do not belong to W.

Since are not rational numbers.

So that W is not closed under scalar multiplication.

W is not a subspace of

Suppose V(F) be a vector space. If , then any vector-

Where

Is called a linear combination of the vectors

Linear Span-

Definition- Let V(F) be a vector space and S be a non-empty subset of V. Then the linear span of S is the set of all linear combinations of finite sets of elements of S and it is denoted by L(S). Thus we have-

L(S) =

Theorem:

The linear span L(S) of any subset S of a vector space V(F) is s subspace of V generated by S.

Proof: Suppose be any two elements of L(S).

Then

And

Where a and b are the elements of F and are the elements of S.

If a,b be any two elements of F, then-

((

((

Thus has been expressed as a linear combination of a finite set of the elements of S.

Consequently

Thus, and so that

Hence L(S) is a subspace of V(F).

Linear dependence-

Let V(F) be a vector space.

A finite set of vector of V is said to be linearly dependent if there exists scalars (not all of them as some of them might be zero)

Such that-

Linear independence-

Let V(F) be a vector space.

A finite set of vector of V is said to be linearly independent if every relation if the form-

Theorem:

The set of non-zero vectors of V(F) is linearly dependent if some

Is a linear combination of the preceding ones.

Proof: if some

Is a linear combination of the preceding ones, then ∃ scalars such that-

…… (1)

The set {} is linearly dependent because the linear combination (1) the scalar coefficient

Hence the set {} of which {} is a subset must be linearly dependent.

Example-1: If the set S = of vector V(F) is linearly independent, then none of the vectors can be zero vector.

Sol. Let be equal to zero vector where

Then

For any in F.

Here therefore from this relation we can say that S is linearly dependent. This is the contradiction because it is given that S is linearly independent.

Hence none of the vectors can be zero.

We can conclude that a set of vectors which contains the zero vector is necessarily linear dependent.

Example-2: Show that S = {(1 , 2 , 4) , (1 , 0 , 0) , (0 , 1 , 0) , (0 , 0 , 1)} is linearly dependent subset of the vector space . Where R is the field of real numbers.

Sol. Here we have-

1 (1 , 2 , 4) + (-1) (1 , 0 , 0) + (-2) (0 ,1, 0) + (-4) (0, 0, 1)

= (1, 2 , 4) + (-1 , 0 , 0) + (0 ,-2, 0) + (0, 0, -4)

= (0, 0, 0)

That means it is a zero vector.

In this relation the scalar coefficients 1 , -1 , -2, -4 are all not zero.

So that we can conclude that S is linearly dependent.

Example-3: If are linearly independent vectors of V(F).where F is the field of complex numbers, then so also are .

Sol. Suppose a, b, c are scalars such that-

…………….. (1)

But are linearly independent vectors of V(F), so that equations (1) implies that-

The coefficient matrix A of these equations will be-

Here we get rank of A = 3, so that a = 0, b = 0, c = 0 is the only solution of the given equations, so that are also linearly independent.

Definition:

A subset S of a vector space V(F) is said to be a basis of V(F), if-

1. S consists of linearly independent vectors

2. Each vector in V is a linear combination of a finite number of elements of S.

Finite dimensional vector spaces

The vector space V(F) is said to be finite dimensional if there exists a finite subset S of V such that V = L(S).

Dimension theorem for vector spaces-

Statement- If V(F) is a finite dimensional vector spaces, then any two bases of V have the same number of elements.

Proof: Let V(F) is a finite dimensional vector space. Then V possesses a basis.

Let and be two bases of V.

We shall prove that m = n

Since V = L( and therefore can be expressed as a linear combination of

Consequently the set which also generates V(F) is linearly dependent. So that there exists a member of this set. Such that is the linear combination of the preceding vectors

If we omit the vector from then V is also generated by the remaining set.

Since V = L( and therefore can be expressed as a linear combination of the vectors belonging to

Consequently the set-

Is linearly dependent.

Therefore there exists a member of this set such that is a linear combination of the preceding vectors. Obviously will be different from

Since {} is a linearly independent set

If we exclude the vector from then the remaining set will generate V(F).

We may continue to proceed in this manner. Here each step consists in the exclusion of an and the inclusion of in the set

Obviously the set of all can not be exhausted before the set and

Otherwise V(F) will be a linear span of a proper subset of thus become linearly dependent. Therefore we must have-

Now interchanging the roles of we shall get that

Hence,

Extension theorem-

Every linearly independent subset of a finitely generated vector space V(F) forms of a part of a basis of V.

Proof: Suppose be a linearly independent subset of a finite dimensional vector space V(F) if dim V = n, then V has a finite basis, say

Let us consider a set-

…………. (1)

Obviously L(, since there can be expressed as linear combination of therefore the set is linearly dependent.

So that there is some vector of which is linear combination of its preceding vectors. This vector can not be any of the since the are linearly independent.

Therefore this vector must be some say

Now omit the vector from (1) and consider the set-

Obviously L(. If is linearly independent, then will be a basis of V and it is the required extended set which is a basis of V.

If is not linearly independent, then repeating the above process a finite number of times. We shall get a linearly independent set containing and spanning V. This set will be a basis of V contains the same number of elements, so that exactly n-m elements of the set of will be adjoined to S so as to form a basis of V.

Example-1: Show that the vectors (1, 2, 1), (2, 1, 0) and (1, -1, 2) form a basis of

Sol. We know that set {(1,0,0) , (0,1,0) , (0,0,1)} forms a basis of

So that dim .

Now if we show that the set S = {(1, 2, 1), (2, 1, 0) , (1, -1, 2)} is linearly independent, then this set will form a basis of

We have,

(1, 2, 1)+ (2, 1, 0)+ (1, -1, 2) = (0, 0, 0)

(

Which gives-

On solving these equations we get,

So we can say that the set S is linearly independent.

Therefore it forms a basis of

Definition-

Suppose U(F) and V(F) are two vector spaces, then a mapping is called a linear transformation of U into V, if:

1.

2.

It is also called homomorphism.

Kernel of a linear transformation-

Let f be a linear transformation of a vector space U(F) into a vector space V(F).

The kernel W of f is defined as-

The kernel W of f is a subset of U consisting of those elements of U which are mapped under f onto the zero vector V.

Since f(0) = 0, therefore atleast 0 belong to W. So that W is not empty.

Example: The mapping defined by-

Is a linear transformation .

What is the kernel of this linear transformation?

Sol. Let be any two elements of

Let a, b be any two elements of F.

We have

(

=

So that f is a linear transformation.

To show that f is onto . Let be any elements .

Then and we have

So that f is onto

Therefore f is homomorphism of onto .

If W is the kernel of this homomorphism then

We have

∀

Also if then

Implies

Therefore

Hence W is the kernel of f.

References

1. G.B. Thomas and R.L. Finney, Calculus and Analytic geometry, 9th Edition,Pearson, Reprint, 2002.

2. Erwin kreyszig, Advanced Engineering Mathematics, 9th Edition, John Wiley & Sons, 2006.

3. Veerarajan T., Engineering Mathematics for first year, Tata McGraw-Hill, New Delhi, 2008.

4. Ramana B.V., Higher Engineering Mathematics, Tata McGraw Hill New Delhi, 11thReprint, 2010.

5. D. Poole, Linear Algebra: A Modern Introduction, 2nd Edition, Brooks/Cole, 2005.

6. N.P. Bali and Manish Goyal, A text book of Engineering Mathematics, Laxmi Publications, Reprint, 2008.

7. B.S. Grewal, Higher Engineering Mathematics, Khanna Publishers, 36th Edition, 2010.