Unit - 3

Vector Space

3.1 Vector addition and scalar multiplication

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

Vector Sub Space

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

Key takeaways:

1. If V is a vector space over the field F, then we will denote vector space as V(F).
2. If then represents addition in vectors(V).
3. For V to be an abelian group with respect to addition vectors, the following conditions must be followed-

1.

2.

3.

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

5. 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.

6. 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.

3.2 Linear dependence of vectors

In the theory of vector spaces, a set of vectors is said to be linearly dependent if at least one of the vectors in the set can be defined as a linear combination of the others; if no vector in the set can be written in this way, then the vectors are said to be linearly independent. These concepts are central to the definition of dimension.

Linear Dependence and Independence: The set of vectors in an vector space V is said to be Linearly dependent if there exist scalars not all zero such that

Example 1: Are linearly independent?

Independent if 

Has only the trivial solution 

Example 2: Check whether the vectors are independent.

Solution: Calculate the coefficients in which a linear combination of these vectors vector.

This vector equation can be written as a system of linear equations

Solve this system using the gauss method

From 2 row we subtract the 1-th row; from 3row we subtract the 1-th row:

From 1 row we subtract the 2row; for 3 row add 2 row:

This solution shows that the system has many solutions, i.e exist nonzero combination c such that the linear combination of is equal to the zero vector,

For Example, 

Means vectors are linearly dependent.

Answer: Vectors are linearly dependent.

Key takeaways:

1. 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-

2.     The set of non-zero vectors of V(F) is linearly dependent if some Is a linear combination of the preceding ones.

3.3 Linear independence of vectors

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.

3.4 Basis and dimension

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

Example 1: Is the set S = {(1,1),(1,-1)} a basis for

1. Does S span

Solve

This system is consistent for every x and y, therefore S spans

2.IS S linearly independent?

Solve

The system has a unique solution (trivial solution)

Therefore, S is linearly independent.

Consequently, S is a basis for

Example 2:  

Is a basis for the vector space

1.+

Is equivalent to:

Which is consistent for every a,b,c and  d.

Therefore, S spans

2.+  is equivalent to:

The system has only the trivial solution is linearly independent.

Consequently, S is a basis for

Key takeaways:

1. 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).
2. If V(F) is a finite dimensional vector spaces, then any two bases of V have the same number of elements.
3. Every linearly independent subset of a finitely generated vector space V(F) forms of a part of a basis of V.

3.5 Linear transformations, range and kernel of a linear map

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.

Theorem: Let L: V. Then ker L is a subspace of V.

Proof: Notice that if L(v)=0 and L(u)=0, then for any constants c,d, L(cu+dv)=0.

Then by the subspace theorem, the kernel of L is a subspace of V.

This theorem has a nice interpretation in terms of the eigenspace of L. Suppose L has a zero eigenvalue. Then the associated eigenspace consists of all vectors v such that Lv=0v=0; in other words , the 0-eigenspace of L is exactly the kernel of L

Returning to the previous example, let L(x,y) =(x+y,x+2y,y). L is clearly not surjective, Since L sends to a plane in .

Notice that if x = L(v) and y = L(v), then for any constants c,d then cx+dy=L(cx+du). Then the subspace theorem strikes again, and we have the following theorem.

Example 1: defined by is linear. Describe its kernel and range and give the dimension of each.

It should be clear that  if, and only if, b= -a and d= -c. The kernel of T is therefore all matrices of the form

The two matrices and are not scalar multiples of each other, so they must be linearly independent. Therefore, the dimension of ker(T) is two.

Now suppose that we have any vector Clearly . So, the range of T of Thus the dimension of ran(T)is two.

Example 2: defined by T(ax+b) = 2bx-a is linear. Describe its kernel and range and give the dimension of each.

T(ax+b) =2bx-a=0 if, and only if, both a and b are zero. Therefore, the kernel of T is only the zero polynomial. By definition, the dimension of the subspace consisting of only the zero vector is zero, so ker(T) has dimension zero.

Suppose that we take a random polynomial cx+d in the codomain. If we consider the polynomial -dx+ in the domain we see that

This shows that the range of T is all of it has dimension two.

Key takeaways:

1. 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.

2. 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-

3.6 Rank and nullity-theorem and inverse of a linear transformation

Rank-nullity theorem:

Statement: Let {\displaystyle V}V,{\displaystyle W}W be vector spaces, where {\displaystyle V} V is finite dimensional. Let {\displaystyle T\colon V\to W} T: V be a linear transformation. Then

{\displaystyle \operatorname {Rank} (T)+\operatorname {Nullity} (T)=\dim V}Rank(T)+Nullity(T) = dim

Where

Rank(T) = dim(Image(T)) and Nullity(T) = dim(Ker(T)){\displaystyle \operatorname {Rank} (T):=\dim(\operatorname {Image} (T))}Rank(TRR{\displaystyle \operatorname {Nullity} (T):=\dim(\operatorname {Ker} (T)).}

Proof: Let {\displaystyle V,W} V, W  be vector spaces over some field {\displaystyle \mathbb {F} } F and {\displaystyle T} T defined as in the statement of the theorem with {\displaystyle \dim V=n}dim V = n

As {\displaystyle \operatorname {Ker} T\subset V} Ker (T)  is a subspace, there exists a basis for it. Suppose{\displaystyle \dim \operatorname {Ker} T=k}

Dim Ker (T) = k and let

{\displaystyle {\mathcal {K}}:=\{v_{1},\ldots ,v_{k}\}\subset \operatorname {Ker} (T)}

Be such a basis.

We may now, by the Steinitz exchange lemma, extend {\displaystyle {\mathcal {K}}} with {\displaystyle n-k}n-k linearly independent vectors {\displaystyle w_{1},\ldots ,w_{n-k}} to form a full basis of {\displaystyle V} V.

Let

S = {\displaystyle {\mathcal {S}}:=\{w_{1},\ldots ,w_{n-k}\}\subset V\setminus \operatorname {Ker} (T)}

Such that

{\displaystyle {\mathcal {B}}:={\mathcal {K}}\cup {\mathcal {S}}=\{v_{1},\ldots ,v_{k},w_{1},\ldots ,w_{n-k}\}\subset V}B= =

Is a basis for {\displaystyle V}V.

From this, we know that

{\displaystyle \operatorname {Im} T=\operatorname {Span} T({\mathcal {B}})=\operatorname {Span} \{T(v_{1}),\ldots ,T(v_{k}),T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} \{T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} T({\mathcal {S}})}Im T = SpanT(B) = Span = Span(T)

We now claim that {\displaystyle T({\mathcal {S}})} T(S) is a basis for {\displaystyle \operatorname {Im} T}ImT. By definition, {\displaystyle T({\mathcal {S}})}T(S) is generating; it remains to be shown that it is also linearly independent to conclude that it is a basis.

Suppose {\displaystyle T({\mathcal {S}})}T(s) is not linearly independent, and let

{\displaystyle \sum _{j=1}^{n-k}\alpha _{j}T(w_{j})=0_{W}}

For some {\displaystyle \alpha _{j}\in \mathbb {F} }.

Thus, owing to the linearity of {\displaystyle T}T, it follows that

T{\displaystyle T\left(\sum _{j=1}^{n-k}\alpha _{j}w_{j}\right)=0_{W}\implies \left(\sum _{j=1}^{n-k}\alpha _{j}w_{j}\right)\in \operatorname {Ker} T=\operatorname {Span} {\mathcal {K}}\subset V}

This is a contradiction to B being a basis, unless all {\displaystyle {\mathcal {B}}}{\displaystyle \alpha _{j}}are equal to zero. This shows that T(S){\displaystyle T({\mathcal {S}})}0  is linearly independent, and more specifically that it is a basis for ImT .{\displaystyle \operatorname {Im} T}To summarise, we have {\displaystyle {\mathcal {K}}}KKKKnakhxkjwhx,jw              hxK, a basis for Ker T,  {\displaystyle \operatorname {Ker} T}and T(S), {\displaystyle T({\mathcal {S}})} a basis for {\displaystyle \operatorname {Im} T}ImT.

Finally, we may state that

Rank(T)+Nullity(T)=dim ImT+ dimKer T=

This concludes our proof.

Inverse of a Matrix:

To understand this concept better let us take a look at the following example.

The Inverse of Linear Transformation:

Definition: A function T from X to Y is called invertible if the equation T(x)=y has a unique solution x in X for each y in Y.

Denote the inverse of T as from Y to X, and write

Note

If a function T is invertible, then so is

Example 1: Is the given, matrix Invertible

Solution:

A fail to be invertible, since rref(A)

Example 2:

Given 33 Rectangular Matrix 

The augumented matrix is as follows

After applying the Gauss-Jordan elimination method:

The inverse of a matrix is as follows,

Example 3: Find the inverse of

A=    

Solution:

Step 1: Adjoin the identity matrix to the right side of A:

Step 2: Apply row operations to this matrix until the left side is reduced to I. The computations are:

R2  R2-R1  , R3R3-R1

R3 R3 + 2R2

R1R1 -3R3  , R2R2+3R3

R1R1-2R2

Step 3: Conclusion: The inverse matrix is:

Example 4: Find the inverse of matrix A given below:

Solution:   Let A = IA

Or

Thus the inverse of matrix A is given by I =

Therefore,

3.7 Composition of linear maps

The composition of linear maps is linear: if f : V → W and g : W → Z are linear, then so is their composition g ∘ f : V → Z. It follows from this that the class of all vector spaces over a given field K, together with K -linear maps as morphisms, forms a category. The inverse of a linear map, when defined, is again a linear map.

Example 1: Theorem: Let () be a basis of V and () an arbitrary list of vectors in W. Then there exists a unique linear map

Proof:

First we verify that there is at most one linear map T with . Take any Since () is a basis of V there are unique scalars such that . By linearity we must have

And hence T(v) is completely determined. To show existence , use (3) to define T. It remains to show that this T is linear and that. These two conditions are not hard to show and are left to the reader.

The set of linear maps is itself a vector space. For S,T addition is defined as

(S+T)v = S v + Tv

For a F an T scalar multiplication is defined as

(aT)(v) = a(Tv) 

We should verify that S+T and aT are indeed linear maps again and that all properties of a vector space is satisfied.

I.e., associativity, identity and distributive property. Hence by the above properties we conclude that the given transformation is linear transformation.

Example 2: consider a linear map Tv: which rotates vectors be an angle  0 . From a known statement we have the corresponding matrix for this linear transformation is given below, solve for linear mapping

Solution: the composition of two rotations b angles are given by v and v’ that is

TV’TV is clearly the rotation TV+V’ hence we must have

Av+v’ = Av’Av

The left hand side of this equation is given by

And the right hand-side is,

Comparing corresponding entries we find the claim that Tv’Tv = Tv+v’ is equivalent to well-known addition formulas

Cos(v+v’) = cosv. Cosv’ – sinv. Sinv’

Sin(v+v’) = cosv’. Sinv +cosv. Sinv’.

3.8 Matrix associated with a linear mapping

Let V and W be a finite dimensional F- vector spaces such that dimV = n and dimW = m and let BV = be a basis of  V and BW = be a basis of W. Let T   L(V, W)such that for each k = 1,2,…,n we have T (vk) = .then the matrix of linear map T w.r.to the bases BV and BW is

=

Example 1: Let T L be defined by T(p(x)) = 2xp’(x), and let B1 = be a basis of (R). Determine

Solution: First we will calculate the images of the basis vectors in B1 under the linear transformation T.

T(1) = 2x(1)’ = 2x(0) = 0(1) + 0(x) + 0(x2) + 0(x3) + 0(x4)

T(x) = 2x(x)’ = 2x(1) = 0(1) +0(x) + 0(x2) + 0(x3) + 0(x4)

T(x2) = 2x(x2)’ = 2x(2x) = 0(1) +0(x) + 0(x2) + 0(x3) + 0(x4)

T(x3) = 2x(x3)’ = 2x(3x3) = 0(1) +0(x) + 0(x2) + 0(x3) + 0(x4)

T(x4) = 2x(x4)’ = 2x (4x3) = 0(1) +0(x) + 0(x2) + 0(x3) + 0(x4)

Therefore, we can now construct our matrix as follows.

The first column of corresponds to the co-efficients determinend by T(1).

The second column corresponds to the co-efficients determined by T(x) and so fourth,

  =

Example 2:Let T L(M22, M22) be defined by T = for a,b,c,d R and let B1 = be a basis of M22.Determine

Solution: we will first calculate the images of our basic vectors under T.

T = = 0 + 0 + 0 + 1

T = = 0 -1 + 0 + 0

T = = 0 + 0 -1 + 0

T = = 1 + 0 + 0 +0

Therefore, we construct

=

References:

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

2. V. Krishnamurthy, V.P. Mainra And J.L. Arora, An Introduction to Linear Algebra, Affiliated East–West Press, Reprint 2005.

3. Erwin Kreyszig, Advanced Engineering Mathematics, 9th Edition, John Wiley & Sons, 2006.

4. Veerarajan T., Engineering Mathematics for First Year, Tata McGraw-Hill,

New Delhi, 2008.

5. N.P. Bali and Manish Goyal, A Text Book of Engineering Mathematics, Laxmi Publications, Reprint, 2010.