Mathematics-I (25SMT-121) UNIT-2

Mathematics-II (25SMT-125)

Compiled by : Subhayu

Unit-2
Chapter 1 : Vector Space - Vector Space, Basis and dimension of a
vector space.

ify
Chapter 2 : Linear Transformations - Linear transformations (maps),range
and kernel of a linear map, rank and nullity, Inverse of a linear transforma-
tion, rank- nullity theorem (Without Proof) , composition of linear maps,
Matrix associated with a linear map.
te
No

1
Chapter 1

Vector Space

1.1 Vector Space

ify
Aim: To understand the concepts of Vector Space, Basis, and Dimension in
a rigorous yet intuitive way.

Imagine you are at a playground. You can move forward/backward or

te
left/right. These two directions are enough to reach any point on the ground.
This playground is like a vector space.

Definition
No

A Vector Space V over a field F (for example R or C) is a set together

with two operations:
1. Vector addition: If u, v ∈ V , then u + v ∈ V . 2. Scalar
multiplication: If c ∈ F and v ∈ V , then cv ∈ V .
These operations must satisfy: - Commutativity: u + v = v + u -
Associativity: (u + v) + w = u + (v + w) - Existence of zero vector:
0 + v = v - Existence of additive inverse: For v, there exists −v such
that v +(−v) = 0 - Distributivity of scalar multiplication over addition
- Compatibility of scalar multiplication

Examples

1. The set R2 (all ordered pairs of real numbers) with usual addition
and multiplication is a vector space. 2. The set of polynomials of
degree ≤ n with real coefficients forms a vector space. 3. The set of
continuous functions f : R → R forms a vector space.

2
Mathematics-I (25SMT-121) UNIT-2

1.2 Basis of a Vector Space

Think of building with Lego blocks. You do not need every block in the
store, just a minimal set of blocks that can create everything. These minimal
building blocks in a vector space are called the basis.

Definition
A basis of a vector space V is a set of vectors {v1 , v2 , . . . , vk } such
that:
1. They are linearly independent. 2. They span V , i.e. every v ∈ V

Example
ify
can be written as a linear combination of them.

In R2 , the set {(1, 0), (0, 1)} is a basis. For any (x, y) ∈ R2 , we can
write:
te
(x, y) = x(1, 0) + y(0, 1)

1.3 Dimension of a Vector Space

No

The number of Lego blocks in your minimal set is like the number of basis
vectors. This number is called the dimension.

Definition
The dimension of a vector space V is the number of vectors in any
basis of V .

Examples

1. R2 has dimension 2. 2. The space of polynomials of degree ≤ n has

dimension n + 1. 3. The space of all m × n matrices with real entries
has dimension mn.

3
Mathematics-I (25SMT-121) UNIT-2

1.4 Visual Flowchart

Vector Space: A set with vector
addition + scalar multiplication

Basis: A minimal set of linearly inde-

pendent vectors that span the space

Dimension: The num-

ber of vectors in a basis

1.5
—

Solved Example
Example
ify
te
Problem: Find the dimension of the space of polynomials of degree
≤ 2.
Solution: A general polynomial is a0 + a1 x + a2 x2 . The set {1, x, x2 }
spans the space and is linearly independent. Thus, dim = 3.
No

4
Mathematics-I (25SMT-121) UNIT-2

1.6 Solved Problems on Vector Spaces, Basis

and Dimension
Problem 1
Verify whether the set {(1, 2), (2, 4)} is linearly independent in R2 .
Solution: Let a(1, 2) + b(2, 4) = (0, 0). This gives the equations:
a + 2b = 0, 2a + 4b = 0.
From the first equation: a = −2b. Substituting into the second: 2(−2b) +
4b = 0 ⇒ 0 = 0. Thus, there exists a non-trivial solution. Hence, the
vectors are linearly dependent.
—

Problem 2 ify
Show that {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is a basis of R3 .
Solution: Any (x, y, z) ∈ R3 can be written as
te
(x, y, z) = x(1, 0, 0) + y(0, 1, 0) + z(0, 0, 1).
Thus, the given vectors span R3 . Also, no vector is a linear combination
of the others, so they are linearly independent. Therefore, this set forms a
basis.
No

Problem 3
Find the dimension of the space of all polynomials of degree ≤ 4.
Solution: A general polynomial is
a0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 .
The basis is {1, x, x2 , x3 , x4 }. Therefore, the dimension is 5.
—

Problem 4
Find the dimension of the space of 2 × 3 real matrices.
Solution: Each 2 × 3 matrix has 6 entries, each of which can take inde-
pendent real values. Hence, the dimension of the space is 2 · 3 = 6.
—

5
Mathematics-I (25SMT-121) UNIT-2

Problem 5
Decide if {1 + x, x + x2 , 1 + x2 } forms a basis of the space of polynomials of
degree ≤ 2.
Solution: Consider a linear combination:

a(1 + x) + b(x + x2 ) + c(1 + x2 ) = 0.

Expanding:
(a + c) + (a + b)x + (b + c)x2 = 0.
Equating coefficients:

a + c = 0, a + b = 0, b + c = 0.

vectors, this is a basis.

—

Problem 6
ify
The only solution is a = b = c = 0. Hence, the vectors are linearly inde-
pendent. Since the dimension of the space is 3 and we have 3 independent
te
Find a basis of the solution space of x1 + x2 + x3 = 0.
Solution: Let x1 = −x2 − x3 . General solution: (−x2 − x3 , x2 , x3 ). This
can be written as
x2 (−1, 1, 0) + x3 (−1, 0, 1).
No

Hence, a basis is {(−1, 1, 0), (−1, 0, 1)} and the dimension is 2.

—

Problem 7
Prove that {(1, 1, 0), (1, 0, 1), (0, 1, 1)} is linearly independent.
Solution: Consider

a(1, 1, 0) + b(1, 0, 1) + c(0, 1, 1) = (0, 0, 0).

This gives equations:

a + b = 0, a + c = 0, b + c = 0.

From b + c = 0 ⇒ b = −c. Substitute in a + b = 0 ⇒ a = −b. Then

a + c = 0 ⇒ −b + c = 0 ⇒ c = b. Thus b = c = 0, hence a = 0. Therefore,
the vectors are linearly independent.
—

6
Mathematics-I (25SMT-121) UNIT-2

Problem 8
Find the dimension of the space of symmetric 2 × 2 matrices.
Solution: A general symmetric matrix is
" #
a b
.
b c

There are three free parameters (a, b, c). Hence, the dimension is 3. A basis
is
1 0 0 1 0 0
(" # " # " #)
, , .
0 0 1 0 0 1
—

Problem 9


0
ify
Find the dimension of the space of skew-symmetric 3 × 3 matrices.
Solution: A skew-symmetric matrix has the form

a b

te
−a 0 c  .
 

−b −c 0

The free parameters are a, b, c. Thus, the dimension is 3.

—
No

Problem 10
Find the dimension of the null space of the matrix

1 2 3
" #
A= .
4 5 6

Solution: The rank of A is 2 (rows are independent). By the rank-nullity

theorem:
nullity = n − rank = 3 − 2 = 1.
Hence, the null space has dimension 1.
—

7
Mathematics-I (25SMT-121) UNIT-2

Problem 11
Show that {(1, 0, −1), (0, 1, 1), (1, 1, 0)} forms a basis of R3 .
Solution: Consider

a(1, 0, −1) + b(0, 1, 1) + c(1, 1, 0) = (0, 0, 0).

This gives equations:

a + c = 0, b + c = 0, −a + b = 0.

From −a + b = 0 ⇒ b = a. From a + c = 0 ⇒ c = −a. Substituting:

b + c = a − a = 0. Thus a = b = c = 0. Therefore, the set is linearly
independent. Since there are 3 vectors in R3 , they form a basis.
—

Problem 12 ify
Find the dimension of the subspace of R4 spanned by {(1, 1, 0, 0), (0, 1, 1, 0), (0, 0, 1, 1), (1, 0, 0, 1)}.
Solution: Arrange vectors as rows:
te
1 1 0 0
 
0 1 1 0
.
 
0 0 1 1


1 0 0 1
No

Row-reducing, we find rank = 3. Hence, the dimension of the span is 3.

Therefore, these 4 vectors span a 3-dimensional subspace of R4 .

8
Chapter 2

Linear Transformations

2.1 Introduction
ify
Aim: To understand the concepts of Linear Transformations (maps), range,
kernel, rank and nullity, and to develop skills in solving related problems.

In many real-life situations, we want to study functions that preserve struc-

te
ture. For example: - Stretching or rotating a rubber sheet keeps straight
lines straight. - Scaling an image enlarges or shrinks it but does not distort
proportionality.
Such structure-preserving functions in vector spaces are called linear
No

transformations.

2.2 Definition
A function T : V → W between vector spaces over the same field F is called
a linear transformation if for all u, v ∈ V and scalars c ∈ F :

T (u + v) = T (u) + T (v), T (cu) = cT (u).

Key idea: A linear transformation respects both vector addition and
scalar multiplication.

2.3 Examples
1. T : R2 → R2 defined by T (x, y) = (x + y, x − y).
2. Differentiation operator D : Pn → Pn−1 where D(f (x)) = f ′ (x).

9
Mathematics-I (25SMT-121) UNIT-2

3. T : R3 → R defined by T (x, y, z) = 2x − y + 3z.

—

2.4 Kernel and Range

Kernel (Null Space):

ker(T ) = {v ∈ V | T (v) = 0}.

It represents all vectors that are sent to the zero vector.

Range (Image):

Range(T ) = {T (v) | v ∈ V } ⊆ W.

2.5
—

Rank and Nullity

ify
It is the set of all possible outputs.
te
Rank: Dimension of the range of T .

rank(T ) = dim(Range(T ))

Nullity: Dimension of the kernel of T .

No

nullity(T ) = dim(ker(T ))

Rank–Nullity Theorem: For a linear map T : V → W ,

dim(V ) = rank(T ) + nullity(T ).

2.6 Solved Problems

Problem 1: Let T : R2 → R2 be defined by T (x, y) = (x + y, x − y). Find
ker(T ) and Range(T ).
Solution: Let (x, y) ∈ R2 . Then

T (x, y) = (x + y, x − y).

10
Mathematics-I (25SMT-121) UNIT-2

For kernel: Solve T (x, y) = (0, 0). That gives equations:

x + y = 0, x − y = 0.

Adding: 2x = 0 =⇒ x = 0. Then y = 0. So ker(T ) = {(0, 0)}.

For range: Output is (x + y, x − y). Let u = x + y, v = x − y. For any
(u, v) ∈ R2 , we can solve for x, y. So Range(T ) = R2 .
Hence rank(T ) = 2, nullity(T ) = 0.
—
Problem 2: Check whether T : R3 → R3 defined by T (x, y, z) = (x, 0, z)
is linear.
Solution: Take (x1 , y1 , z1 ) and (x2 , y2 , z2 ).

T ((x1 , y1 , z1 ) + (x2 , y2 , z2 )) = T (x1 + x2 , y1 + y2 , z1 + z2 ).

ify
= (x1 + x2 , 0, z1 + z2 ).
= (x1 , 0, z1 ) + (x2 , 0, z2 ).
= T (x1 , y1 , z1 ) + T (x2 , y2 , z2 ).
Also scalar property holds:
te
T (c(x, y, z)) = (cx, 0, cz) = c(x, 0, z) = cT (x, y, z).

Thus T is linear.
—
No

Problem 3: Find ker(T ) and rank of T (x, y, z) = (x + z, y + z, z).

Solution: We write in matrix form:
1 0 1 x
  

T (x, y, z) = 0 1 1 y  .
  

0 0 1 z

Call this matrix

 A. Rank of T = rank of A.
1 0 1
A= 0 1 1.


0 0 1
Row-reduction: Already upper triangular with 3 pivots. So rank(T ) = 3.
For kernel: Solve A[x, y, z]T = 0. That gives x + z = 0, y + z = 0, z = 0.
So x = 0, y = 0, z = 0. Thus ker(T ) = {(0, 0, 0)}. Nullity=0.
Hence dim(V ) = 3 = rank + nullity = 3 + 0.
—
Problem 4: Let T : R2 → R3 defined by T (x, y) = (x, y, x + y). Find
basis of Range(T ) and ker(T ).

11
Mathematics-I (25SMT-121) UNIT-2

Solution: Matrix form:

1 0 " #
 
 x
T (x, y) = 0 1 .

y
1 1

So columns are (1, 0, 1)T and (0, 1, 1)T . Range is span of these two vectors.
Check independence: a(1, 0, 1) + b(0, 1, 1) = (0, 0, 0) gives a = 0, b = 0. So
they are independent.
Therefore Range(T ) = span{(1, 0, 1), (0, 1, 1)}. Basis of range given. Rank=2.
For kernel: Solve T (x, y) = (0, 0, 0). That means (x, y, x + y) = (0, 0, 0).
So x = 0, y = 0. Kernel = {(0, 0)}. Nullity=0.
—

ify
Problem 5: Let T : P2 → P2 be differentiation: T (f (x)) = f ′ (x). Find
rank and nullity.
Solution: Any f (x) = a0 + a1 x + a2 x2 .

T (f (x)) = a1 + 2a2 x.

Range: all polynomials of degree ≤ 1. So rank(T ) = 2.

te
Kernel: Solve f ′ (x) = 0 =⇒ a1 = 0, a2 = 0. So f (x) = a0 . Kernel =
constant polynomials, which form 1-dimensional space. Nullity=1.
Check theorem: dim(P2 ) = 3 = 2 + 1.
—
Problem 6: Let T : R3 → R2 given by T (x, y, z) = (x + y, y + z). Find
No

rank and nullity.

Solution: Matrix:
 
x
1 1 0  
" #
T (x, y, z) = y  .
0 1 1
z

1 1 0
" #
Matrix A = .
0 1 1
Rank: Row-reduction → independent rows. So rank=2.
Nullity = dim(V ) − rank = 3 − 2 = 1.
Kernel: Solve x + y = 0, y + z = 0. So x = −y, z = −y. Kernel =
span{(−1, 1, −1)}.
—

12
Mathematics-I (25SMT-121) UNIT-2

2.7 Recap
- A linear transformation preserves vector addition and scalar multipli-
cation. - The kernel is the set of inputs mapped to zero. - The range is
the set of outputs. - Rank = dimension of range. - Nullity = dimension of
kernel. - Rank–Nullity theorem: dim(V ) = rank(T ) + nullity(T ).

2.8 Inverse of a Linear Transformation

Definition: A linear transformation T : V → W is said to be invertible if
there exists another linear transformation S : W → V such that

ify
S(T (v)) = v
T (S(w)) = w
∀v ∈ V,
∀w ∈ W.
In that case, S is called the inverse of T and denoted by T −1 .
Condition for existence: - T is invertible if and only if it is bijective
(one–one and onto). - For finite-dimensional spaces, T is invertible if and
te
only if rank(T ) = dim(V ) = dim(W ).
—
Problem 7: Check if T : R2 → R2 given by T (x, y) = (2x + y, x + 3y) is
invertible.
Solution: Matrix representation:
No

2 1
" #
A= .
1 3

Determinant:

det(A) = 2 · 3 − 1 · 1 = 6 − 1 = 5 ̸= 0.

So A is invertible. Thus T is invertible.

Inverse matrix:
1 3 −1
" #
A =
−1
.
5 −1 2
 
Hence T −1 (u, v) = 3u−v
5
, −u+2v
5
.
—
Problem 8: Let T : R2 → R2 be given by T (x, y) = (x + y, x + y). Is T
invertible?

13
Mathematics-I (25SMT-121) UNIT-2

Solution: Matrix:
1 1
" #
A= .
1 1
det(A) = 1 · 1 − 1 · 1 = 0. So matrix is not invertible. Therefore T is not
invertible.
In fact, Range(T ) = span{(1, 1)}, so rank= 1 < 2.
—
Problem 9: Check whether differentiation operator D : P1 → P1 defined
by D(f (x)) = f ′ (x) is invertible.
Solution: Basis of P1 : {1, x}. For f (x) = a + bx, D(f (x)) = b.
So range = span{1} (only constants). dim(P1 ) = 2, but rank= 1. So not
invertible.
—

2.9 Rank–Nullity Theorem ify

Statement (without proof): If T : V → W is a linear transformation
from a finite-dimensional vector space V , then
te
dim(V ) = rank(T ) + nullity(T ).

This result connects the input space dimension with the dimensions of
the range and kernel.
No

—
Problem 10: Let T : R3 → R2 be given by T (x, y, z) = (x + y, y + z).
Verify Rank–Nullity theorem.
Solution: Matrix form:
1 1 0
" #
A= .
0 1 1

Rank: Rows independent =⇒ rank=2. dim(R3 ) = 3. So nullity =

3 − 2 = 1.
Kernel: Solve (x + y, y + z) = (0, 0). That gives x = −y, z = −y. Kernel
= span{(−1, 1, −1)}, dimension=1.
Hence rank+nullity = 2 + 1 = 3 = dim(V ). Verified.
—
Problem 11: Let T : P2 → P2 be T (f (x)) = f ′ (x). Verify Rank–Nullity
theorem.
Solution: dim(P2 ) = 3.

14
Mathematics-I (25SMT-121) UNIT-2

Kernel: f ′ (x) = 0 =⇒ f (x) = a (constant). So ker(T ) has dimension 1.

Nullity=1.
Range: f ′ (x) = a1 + 2a2 x, i.e. degree ≤ 1. So rank=2.
Rank + Nullity = 2 + 1 = 3 = dim(P2 ). Verified.
—
Problem 12: Consider T : R2 → R3 defined by T (x, y) = (x, y, 0).
Verify Rank–Nullity theorem.
Solution: Matrix:
1 0
 

A = 0 1 .
 

0 0
Columns are independent =⇒ rank=2.
dim(R2 ) = 2. So nullity = 2 − 2 = 0.

2.10 Recap
ify
Kernel = {(0, 0)}, indeed dimension 0.
Thus rank+nullity = 2 + 0 = 2 = dim(V ). Verified.
te
- A linear map is invertible if it is bijective. - For finite-dimensional spaces,
invertibility ⇐⇒ determinant ̸= 0 (matrix form). - The Rank–Nullity
Theorem relates input dimension, rank, and nullity:
No

dim(V ) = rank(T ) + nullity(T ).

- Every problem must check consistency by verifying this identity.

Composition of Linear Maps

Definition: Let U, V, and W be vector spaces over the same field F. Suppose
T1 : U → V and T2 : V → W are linear maps. Then the composition
T2 ◦ T1 : U → W is defined as

(T2 ◦ T1 )(u) = T2 (T1 (u)) ∀u ∈ U

Properties:

• Composition of linear maps is again a linear map.

• If T1 and T2 are both one-one (injective), then T2 ◦ T1 is also one-one.

15
Mathematics-I (25SMT-121) UNIT-2

• If T1 and T2 are both onto (surjective), then T2 ◦ T1 is also onto.

Real-life analogy: Think of T1 as a machine that converts wheat into
flour, and T2 as another machine that converts flour into bread. The com-
position T2 ◦ T1 is like a single machine that directly converts wheat into
bread.

Solved Examples
Example 1: Let T1 : R2 → R2 be defined by T1 (x, y) = (x + 1, y), and
T2 : R2 → R2 be defined by T2 (x, y) = (2x, 3y). Find (T2 ◦ T1 )(x, y).
Solution:
T1 (x, y) = (x + 1, y)

Hence,
ify
T2 (T1 (x, y)) = T2 (x + 1, y) = (2(x + 1), 3y) = (2x + 2, 3y)

(T2 ◦ T1 )(x, y) = (2x + 2, 3y)

Example 2: Check whether T2 ◦ T1 = T1 ◦ T2 for the above maps.
Solution: We already have (T2 ◦ T1 )(x, y) = (2x + 2, 3y).
Now,
te
T2 (x, y) = (2x, 3y), T1 (2x, 3y) = (2x + 1, 3y)
Thus,
(T1 ◦ T2 )(x, y) = (2x + 1, 3y)
Clearly,
No

(T2 ◦ T1 )(x, y) ̸= (T1 ◦ T2 )(x, y)

So, composition is not commutative in general.

Matrix Associated with a Linear Map

Definition: Every linear map T : Rn → Rm can be represented by a matrix
A such that:
T (x) = Ax, ∀x ∈ Rn
Procedure to find the matrix of T :
1. Choose a basis {e1 , e2 , . . . , en } for Rn .
2. Compute T (ei ) for each basis vector.
3. Express each T (ei ) as a column in the matrix.
4. The resulting m × n matrix is the matrix representation of T .

16
Mathematics-I (25SMT-121) UNIT-2

Solved Examples
Example 3: Let T : R2 → R2 be defined by

T (x, y) = (x + 2y, 3x + y)

Find the matrix representation of T with respect to the standard basis {e1 =
(1, 0), e2 = (0, 1)}.
Solution: Step 1: Compute T (e1 ) and T (e2 ).

T (1, 0) = (1 + 0, 3 · 1 + 0) = (1, 3)
T (0, 1) = (0 + 2, 0 + 1) = (2, 1)
Step 2: Arrange them as columns.

Thus, the matrix of T is

ify
A=
1 2
3 1
" #
te
1 2
" #
A=
3 1

Example 4: Let T : R3 → R2 be defined by

No

T (x, y, z) = (x + 2y + z, 3x − y)

Find the matrix associated with T .

Solution: Step 1: Apply T to basis vectors.

T (1, 0, 0) = (1, 3), T (0, 1, 0) = (2, −1), T (0, 0, 1) = (1, 0)

Step 2: Arrange as columns in a 2 × 3 matrix.

1 2 1
" #
A=
3 −1 0

Hence, A is the matrix representation of T .

17
Mathematics-I (25SMT-121) UNIT-2

Solved Problems
Problem 1: Let T : R2 → R2 be defined by T (x, y) = (x + 2y, 3x − y).
Check whether T is linear.
Solution: Check addition:

T ((x1 , y1 )+(x2 , y2 )) = T (x1 +x2 , y1 +y2 ) = ((x1 +x2 )+2(y1 +y2 ), 3(x1 +x2 )−(y1 +y2 ))

= (x1 + 2y1 , 3x1 − y1 ) + (x2 + 2y2 , 3x2 − y2 ) = T (x1 , y1 ) + T (x2 , y2 )

Check scalar multiplication:

T (c(x, y)) = T (cx, cy) = (cx + 2cy, 3cx − cy) = c(x + 2y, 3x − y) = cT (x, y)

Hence T is linear.
—
ify
Problem 2: Find ker(T ) and Range(T ) for T (x, y) = (x + 2y, 3x − y).
Solution: Kernel: Solve T (x, y) = (0, 0)

x + 2y = 0, 3x − y = 0
te
Solve simultaneously: y = 3x, substitute: x + 2(3x) = 7x = 0 =⇒ x =
0, y = 0
ker(T ) = {(0, 0)}, nullity = 0
Range: Matrix form
No

1 2
" #
A=
3 −1
Determinant det(A) = 1·(−1)−2·3 = −1−6 = −7 ̸= 0 Rank= 2, Range= R2
—
Problem 3: Let T : R3 → R3 , T (x, y, z) = (x + y, y + z, x + z). Find
rank and nullity.
Solution: Matrix form:
1 1 0
 

A = 0 1 1


1 0 1

Row-reduce or check linear independence: 3 rows independent =⇒

rank=3 dim(R3 ) = 3, nullity= 3 − 3 = 0
—
Problem 4: Check if T : P2 → P2 , T (f (x)) = f ′ (x) is invertible.
Solution: P2 basis: {1, x, x2 } D(1) = 0, D(x) = 1, D(x2 ) = 2x

18
Mathematics-I (25SMT-121) UNIT-2

Kernel: f ′ (x) = 0 =⇒ f (x) = a, dim=1 Range: all degree ≤ 1, dim=2

Since dim(P2 ) = 3 ̸= rank = 2, T is not invertible.
—
Problem 5: Find the inverse of# T : R2 → R2 , T (x, y) = (x + y, x − y).
1 1
"
Solution: Matrix: A = Determinant det(A) = −1−1 = −2 ̸= 0
1 −1
→ invertible
Inverse:
1 −1 −1 1 1 1
" # " #
A =
−1
=
−2 −1 1 2 1 −1
 
Hence T −1 (u, v) = u+v
2
, u−v
2
—
Problem 6: Verify rank-nullity
Solution: Matrix: A =
"

ify
theorem
1 0 1
0 1 1
# for T (x, y, z) = (x + z, y + z)
Rank=2, dim(R3 ) = 3 Nullity =
3 − 2 = 1 Kernel: Solve x + z = 0, y + z = 0 =⇒ x = −z, y = −z →
dimension 1 Rank+Nullity=2+1=3.
—
Problem 7: Composition: T1 (x, y) = (x + y, x), T2 (x, y) = (2x, y − x).
te
Compute T2 ◦ T1 .
Solution:

T2 (T1 (x, y)) = T2 (x + y, x) = (2(x + y), x − (x + y)) = (2x + 2y, −y)

No

—
Problem 8: Check if T2 ◦ T1 = T1 ◦ T2 for above maps.
Solution:

T1 (T2 (x, y)) = T1 (2x, y − x) = (2x + (y − x), 2x) = (x + y, 2x)

Clearly T1 ◦ T2 ̸= T2 ◦ T1
—
Problem 9: Matrix associated with T (x, y) = (3x − 2y, x + y)
Solution: Basis e1 = (1, 0), e2 = (0, 1) T (e1 ) = (3, 1), T (e2 ) = (−2, 1)
Matrix:
3 −2
" #
A=
1 1
—
Problem 10: Matrix associated with T (x, y, z) = (x + z, y − z)
Solution: e1 = (1, 0, 0) → T (e1 ) = (1, 0) e2 = (0, 1, 0) → T (e2 ) = (0, 1)
e3 = (0, 0, 1) → T (e3 ) = (1, −1)

19
Mathematics-I (25SMT-121) UNIT-2

Matrix:
1 0 1
" #
A=
0 1 −1
—
Problem 11: Find ker(T ) if T (x, y, z) = (x + y + z, x − y + 2z)
Solution: Solve:

x + y + z = 0, x − y + 2z = 0

Add equations: 2x + 3z = 0 =⇒ x = −3z/2 From first: −3z/2 + y + z =

0 =⇒ y = z/2 Kernel: span{(−3/2, 1/2, 1)}, dim=1 → nullity=1
—
Problem 12: Verify rank-nullity theorem for T (x, y, z) = (x − y, y − z)
Solution: Matrix: A =
"

ify
1 −1 0
0 1 −1
#

Rank: Rows independent → rank=2 dim(V ) = 3 → nullity = 3 − 2 = 1

Kernel: Solve x − y = 0, y − z = 0 =⇒ x = y = z → span{(1, 1, 1)}
Rank+Nullity=2+1=3
te
No

20
Mathematics-I (25SMT-121) UNIT-2

MST 2 - Practice Sheet

General Instructions: Attempt all questions
Total Time: 1 hour
Total Marks: 20
Q.No Statement
Section A
1 Define a vector space over a field. Give two examples of
vector spaces other than Rn .
2 Explain the concepts of basis and dimension of a vector
space with examples.
3 Let T : R2 → R2 , T (x, y) = (x + 2y, 3x − y). Determine

5
ify
ker(T ) and Range(T ).
Let T : P2 → P2 be defined by T (f (x)) = f ′ (x). Find
the kernel and rank of T .
State the Rank-Nullity theorem and explain it with a
simple example.
Section B
te
6 Let T : R → R be defined by T (x, y) = (x + y, x − y).
2 2

Determine if T is invertible. If yes, find T −1 .

7 Let T1 (x, y) = (x + y, x) and T2 (x, y) = (2x, y − x).
Compute (T2 ◦ T1 )(x, y) and (T1 ◦ T2 )(x, y). Are they
equal?
No

21
Mathematics-I (25SMT-121) UNIT-2

Answers
Section A

1. Vector Space:
A vector space V over a field F is a set of objects (vectors) with oper-
ations of vector addition and scalar multiplication satisfying: closure,
associativity, commutativity, existence of zero vector, existence of ad-
ditive inverse, distributivity, and compatibility of scalars.

Examples: 1. The set of all 2 × 2 matrices over R. 2. The set of

continuous functions f : R → R.

2. Basis and Dimension:

ify
Basis: A set of vectors in V that is linearly independent and spans V .
Dimension: Number of vectors in any basis.
Example: R2 , basis = {(1, 0), (0, 1)}, dimension = 2.

3. Kernel and Range: T (x, y) = (x + 2y, 3x − y)

te
Kernel: Solve T (x, y) = (0, 0)

x + 2y = 0, 3x − y = 0

Solve: y = 3x =⇒ x + 2(3x) = 7x = 0 =⇒ x = 0, y = 0
No

ker(T ) = {(0, 0)}, nullity = 0

1 2
" #
Range: Matrix A = , det(A) = −7 ̸= 0 Rank=2 → Range= R2
3 −1

4. Derivative Map T (f ) = f ′ : Basis of P2 = {1, x, x2 }

Kernel: f ′ (x) = 0 =⇒ f (x) = c, dim=1 Range: Span{1, x}, dim=2
→ rank=2

5. Rank-Nullity Theorem: For T : V → W , dim(V ) = rank(T ) +

nullity(T )
Example: T (x, y, z) = (x + y, y + z), rank=2, nullity=1, dim(R3 ) = 3,
satisfies theorem.

Section B

22
Mathematics-I (25SMT-121) UNIT-2

6. T (x, y) = (x + y, x)
1 1
" #
Matrix A = , det(A) = −2 ̸= 0 =⇒ T invertible
1 −1
1 1
" #
A −1
= 1
2 1 −1

u+v u−v
 
T −1
(u, v) = ,
2 2

7. Composition: T1 (x, y) = (x + y, x), T2 (x, y) = (2x, y − x)

T2 ◦ T1 (x, y) = T2 (x + y, x) = (2(x + y), x − (x + y)) = (2x + 2y, −y)
T1 ◦ T2 (x, y) = T1 (2x, y − x) = (2x + y − x, 2x) = (x + y, 2x)
Not equal → composition not commutative.

ify
te
For more resources, visit https://subhayu18.github.io/Noteify/
No

23