Unit – 6 | unit 6 vector spaces

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Unit – 6

Vector space

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) 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.

A2)

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-

4. Where 1 is the unity element of F.

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

Q3) The vector space of all polynomials over a field F. Prove.

A3)

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.

Q4) Suppose V(F) is a vector space and 0 be the zero vector of V. Then-

A4)

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

Q5) Define vector sub space.

A5)

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.

Q6) 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-

A6)

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.

Q7) The intersection of any two subspaces and of a vector space V(F) is also a subspace of V(F). Prove.

A7)

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).

Q8) 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 Prove.

A8)

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

Q9) Let R be a field of real numbers. Now check whether which one of the following are subspaces of

1. {

2. {

3. {

A9)

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

Q10) Define symmetric and skew-symmetric matrix.

A10)

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.

Q11) Any intersection of subspaces of a vector space V is a subspace of V. Prove.

A11)

Let C be a collection of subspaces of V, and let W denote the intersection of the subspaces in C. Since every subspace contains the zero vector, 0 ∈ W. Let a ∈ F and x, y ∈ W. Then x and y are contained in each subspace in C. Because each subspace in C is closed under addition and scalar multiplication, it follows that x+y and ax are contained in each subspace in C. Hence x+y and ax are also contained in W, so that W is a subspace of V

Q12) Define homomorphism.

A12)

Definition-

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

It is also called homomorphism.

Q13) What is kernel of a linear transformation.

A13)

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.

Q14) The mapping defined by-

Is a linear transformation .

What is the kernel of this linear transformation.

A14)

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.

Q15) Let V and W be vector spaces and T: V → W be linear. Then N(T) and R(T) are subspaces of V and W, respectively. Prove.

A15)

Here we will use the symbols 0 V and 0W to denote the zero vectors of V and W, respectively.

Since T(0 V) = 0W, we have that 0 V ∈ N(T). Let x, y ∈ N(T) and c ∈ F.

Then T(x + y) = T(x)+T(y) = 0W+0W = 0W, and T(c x) = c T(x) = c0W = 0W. Hence x + y ∈ N(T) and c x ∈ N(T), so that N(T) is a subspace of V. Because T(0 V) = 0W, we have that 0W ∈ R(T). Now let x, y ∈ R(T) and c ∈ F. Then there exist v and w in V such that T(v) = x and T(w) = y. So T(v + w) = T(v)+T(w) = x + y, and T(c v) = c T(v) = c x. Thus x+ y ∈ R(T) and c x ∈ R(T), so R(T) is a subspace of W.

Q16) Let A is a matrix of order m by n, then-

Prove.

A16)

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.

Q17) What is linear span?

A17)

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) =

Q18) The linear span L(S) of any subset S of a vector space V(F) is s subspace of V generated by S. Prove.

A18)

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).

Q19) What is linear dependence and linear independence?

A19)

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-

Q20) 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

A20)

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.

Q21) If are linearly independent vectors of V(F).where F is the field of complex numbers, then so also are . Prove.

A21)

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.

Q22) State and prove extension theorem.

A22)

Statement-

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.

Q23) Show that the vectors (1, 2, 1), (2, 1, 0) and (1, -1, 2) form a basis of

A23)

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

Unit – 6

Vector space

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).

A2)

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-

4. Where 1 is the unity element of F.

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

Q3) The vector space of all polynomials over a field F. Prove.

A3)

Q4) Suppose V(F) is a vector space and 0 be the zero vector of V. Then-

A4)

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

Q5) Define vector sub space.

A5)

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.

Q6) 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-

A6)

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.

Q7) The intersection of any two subspaces and of a vector space V(F) is also a subspace of V(F). Prove.

A7)

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).

Q8) 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 Prove.

A8)

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

Q9) Let R be a field of real numbers. Now check whether which one of the following are subspaces of

1. {

2. {

3. {

A9)

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

Q10) Define symmetric and skew-symmetric matrix.

A10)

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.

Q11) Any intersection of subspaces of a vector space V is a subspace of V. Prove.

A11)

Q12) Define homomorphism.

A12)

Definition-

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

It is also called homomorphism.

Q13) What is kernel of a linear transformation.

A13)

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.

Q14) The mapping defined by-

Is a linear transformation .

What is the kernel of this linear transformation.

A14)

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.

Q15) Let V and W be vector spaces and T: V → W be linear. Then N(T) and R(T) are subspaces of V and W, respectively. Prove.

A15)

Here we will use the symbols 0 V and 0W to denote the zero vectors of V and W, respectively.

Since T(0 V) = 0W, we have that 0 V ∈ N(T). Let x, y ∈ N(T) and c ∈ F.

Q16) Let A is a matrix of order m by n, then-

Prove.

A16)

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.

Q17) What is linear span?

A17)

L(S) =

Q18) The linear span L(S) of any subset S of a vector space V(F) is s subspace of V generated by S. Prove.

A18)

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).

Q19) What is linear dependence and linear independence?

A19)

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-

Q20) 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

A20)

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.

Q21) If are linearly independent vectors of V(F).where F is the field of complex numbers, then so also are . Prove.

A21)

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.

Q22) State and prove extension theorem.

A22)

Statement-

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.

Q23) Show that the vectors (1, 2, 1), (2, 1, 0) and (1, -1, 2) form a basis of

A23)

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

Unit – 6

Vector space

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).

A2)

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-

4. Where 1 is the unity element of F.

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

Q3) The vector space of all polynomials over a field F. Prove.

A3)

Q4) Suppose V(F) is a vector space and 0 be the zero vector of V. Then-

A4)

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

Q5) Define vector sub space.

A5)

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.

Q6) 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-

A6)

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.

Q7) The intersection of any two subspaces and of a vector space V(F) is also a subspace of V(F). Prove.

A7)

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).

Q8) 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 Prove.

A8)

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

Q9) Let R be a field of real numbers. Now check whether which one of the following are subspaces of

1. {

2. {

3. {

A9)

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

Q10) Define symmetric and skew-symmetric matrix.

A10)

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.

Q11) Any intersection of subspaces of a vector space V is a subspace of V. Prove.

A11)

Q12) Define homomorphism.

A12)

Definition-

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

It is also called homomorphism.

Q13) What is kernel of a linear transformation.

A13)

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.

Q14) The mapping defined by-

Is a linear transformation .

What is the kernel of this linear transformation.

A14)

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.

Q15) Let V and W be vector spaces and T: V → W be linear. Then N(T) and R(T) are subspaces of V and W, respectively. Prove.

A15)

Here we will use the symbols 0 V and 0W to denote the zero vectors of V and W, respectively.

Since T(0 V) = 0W, we have that 0 V ∈ N(T). Let x, y ∈ N(T) and c ∈ F.

Q16) Let A is a matrix of order m by n, then-

Prove.

A16)

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.

Q17) What is linear span?

A17)

L(S) =

Q18) The linear span L(S) of any subset S of a vector space V(F) is s subspace of V generated by S. Prove.

A18)

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).

Q19) What is linear dependence and linear independence?

A19)

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-

Q20) 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

A20)

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.

Q21) If are linearly independent vectors of V(F).where F is the field of complex numbers, then so also are . Prove.

A21)

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.

Q22) State and prove extension theorem.

A22)

Statement-

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.