Unit - 1

Vector spaces

Q1) Define vector space.

A1)

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).

Q2) Suppose V(F) is a vector space and 0 be the zero vector of V. Then prove that

A2)

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-

Q3) What are the necessary and sufficient conditions for subspace?

A3)

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.

Q4) The intersection of any two subspaces and of a vector space V(F) is also a subspace of V(F).

A4)

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).

Q5) Define a subspace.

A5)

A subset W of a vector space V over a field F is called a subspace of V if W is a vector space over F with the operations of addition and scalar multiplication defined on V.

In any vector space V, note that V and {0 } are subspaces. The latter is called the zero subspace of V.

Q6) What are symmetric and skew symmetric matrices?

A6)

The transpose of an m × n matrix A is the n × m matrix obtained from A by interchanging the rows with the columns

Suppose,

Then

Transpose of this matrix,

A symmetric matrix is a matrix A such that = A.

A skew- symmetric matrix is a matrix A such that = A.

Q7) Define sum and direct sum.

A7)

Sum- If S1 and S2 are nonempty subsets of a vector space V, then the sum of S1 and S2, denoted S1+S2, is the set {x + y : x ∈ S1 and y ∈ S2}.

Direct sum- A vector space V is called the direct sum of W1 and W2 if W1 and W2 are subspaces of V such that W1 ∩W2 = {0} and W1 +W2 = V. We denote that V is the direct sum of W1 and W2 by writing V = W1 ⊕ W2.

Q8) What do you understand by linear combinations of vectors?

A8)

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

Where

Is called a linear combination of the vectors

Q9) The linear span L(S) of any subset S of a vector space V(F) is s subspace of V generated by S.

A9)

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).

Q10) What is linear dependence and linear independence?

A10)

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-

Q11) The set of non-zero vectors of V(F) is linearly dependent if some

Is a linear combination of the preceding ones.

A11)

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.

Q12) 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.

A12)

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.

Q13) If are linearly independent vectors of V(F) where F is the field of complex numbers, then so also are .

A13)

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.

Q14) State and prove dimension theorem for vector spaces.

A14)

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,

Q15) State and prove extension theorem.

A15)

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.

Q16) What is null space?

A16)

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.

Q17) The mapping defined by-

Is a linear transformation .

What is the kernel of this linear transformation.

A17)

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.

Q18) What is rank nullity theorem?

A18)

Statement:

Let A is a matrix of order m by n, then-

Proof:

If rank (A) = n, then the only solution to Ax = 0 is the trivial solution x = 0by using invertible matrix.

So that in this case null-space (A) = {0}, so nullity (A) = 0.

Now suppose rank (A) = r < n, in this case there are n – r > 0 free variable in the solution to Ax = 0.

Let represent these free variables and let denote the solution obtained by sequentially setting each free variable to 1 and the remaining free variables to zero.

Here is linearly independent.

Moreover every solution is to Ax = 0 is a linear combination of

Which shows that spans null-space (A).

Thus is a basis for null-space(A) and nullity (A) = n – r.

Q19) Let V and W be vector spaces, and let T: V → W be linear. Then T is one-to-one if and only if N(T) = {0}.

A19)

Suppose that T is one-to-one and x ∈ N(T). Then T(x) = 0 = T(0 ). Since T is one-to-one, we have x = 0 . Hence N(T) = {0 }.

Now assume that N(T) = {0 }, and suppose that T(x) = T(y). Then 0 = T(x) − T(y) = T(x − y).

Therefore x − y ∈ N(T) = {0 }. So x − y = 0, or x = y. This means that T is one-to-one.