Algebra of Linear Transformations MCQ Quiz - Objective Question with Answer for Algebra of Linear Transformations - Download Free PDF

Explanation:   

Let X ∈ N_A(T) 

T(A) = 0 

⇒ AX - XA = 0 

⇒ AX = XA 

⇒  = 

⇒     

after comparing   

⇒ Nullity = 8

Hence 8 is the correct answer.

Explanation: 

Statement P: Given linear transformation

 defined by

T(x, y, z, u) = (3x, 2y, 0, 0)    (x, y, z, u)  ∈

To Find the Kernel of T:

Let (x, y, z, u) ∈ ker T

Then T(x, y, z, u) = (0, 0, 0, 0)

Using the definition of linear transformation, we get (3x, 2y, 0, 0) = (0, 0, 0, 0)

⇒ x = 0, y = 0, and z and u are arbitrary

Therefore, ker T = {(0, 0, z, u) : z, u ∈

since the kernel has two free variables, z and u

⇒ Nullity of T = 2 

Find Rank of T using Rank-Nullity Theorem:

Rank of T = dim - Nullity T = 4 - 2 = 2

⇒ Rank of T = 2 

Hence Rank of T = Nullity of T = 2

⇒ Statement P is False .

Statement Q: 

 Let  be defined by  

Ker T  = {(x, y, z) such that T(x, y ,z) = 0}

Ker T  = {(x, y, z) : (3x, 2y ,0) = 0}

Ker T  = {(0, 0, z) : z ∈ \mathbb{R}} 

Therefore, Ker T ≠ {0, 0, 0}

⇒ T is Singular

and    be defined by S(x, y)= (2x, 3y) 

Ker S  = {(x, y) such that S(x, y) = 0}

Ker S  = {(x, y) : (2x, 3y) = 0} = {(0, 0)} 

⇒ S is non-singular

⇒  T is Singular and S is non-singular

⇒ Statement Q is also False.

Hence Option(2) is the correct answer.

Explanation:

Let's consider the standard basis of P₂(R):

{1, x, x²}

Applying T to each basis element, we get:

here p(x) = 1 so p(2x²) = 1

T(1) = 1 = 1 + 0.x + 0.x²

here p(x)=x so p(2x²) = 2x2

T(x) = 2x² = 0 + 0.x + 2.x²

here p(x) = x2 so p(2x2) = 4x4

T(x²) = 4x4 = 4x (mod x³)

T(x²) = 4x⁴ = 4x (mod x³) = 0 + 4.x + 0.x²

However, since we are working in P₂(R), the polynomial x⁴ is not in the space. We can reduce it modulo x³ to get:

So, the image of T is spanned by the polynomials {1, x, x²}.

This set is linearly independent and spans P₂(R).

Matrix Representation of T will be:

Rank of T = 3

Hence 3 is the correct answer.

Explanation:

Let's go through each question one by one.

Let T :  be a linear transformation defined by

T(x, y) = (x, x + y, y) .

To find the rank of T ,

let's write the transformation matrix for T .

Since T(x, y) = (x, x + y, y) ,

we can represent this in matrix form as:

T(x, y) = 

The matrix of T is:

A =  

To determine the rank of T , we find the rank of matrix A .

Observing A ,

the columns are linearly independent (no column is a linear combination of the others).

Thus, the rank of A is 2 .

Hence Option(4) is correct. 

Explanation:

Since T is a linear transformation,

we can express any vector in R² as a linear combination of the basis vectors (1, 0) and (0, 1)

Expressing (1, 0) as a linear combination:

(1, 0) = a(1, 2) + b(3, 4)

We need to find scalars a and b such that:

a + 3b = 1 

2a + 4b = 0

Solving this system, we get:

a = -2 

b = -1

Since T is linear, we have:

T(1, 0) = T(-2(1, 2) - 1 (3, 4))

Using the linearity property (  ), we get:

T(1, 0) = -2T(1, 2) + 1 T(3, 4)

Substituting the given values:

T(1, 0) = -2(3, 2, 1) + 1 (6, 5, 4)

T(1, 0) = (-6, -4, -2) + (6, 5, 4)

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

Therefore, the correct answer is (0 , 1 , 2).

Hence Option(1) is the correct answer. 

Explanation:

Characteristic polynomial p(T) is divisible by T2.

p(x)/x2

So p(x) = x2(x + a) where a can be zero also

Option (1): Let A =  Here 0 and a are the eigenvalues and eigenspace of A for 0 is 1

So option (1) is false

Here A is 3 × 3 matrix and two eigenvalues are 0, 0. Since complex eigenvalue are always complex conjugate and they are in pairwise. So here third eigenvalue must be real.

Option (2) is correct

 For A = , A3 ≠ 0

Option (3) is false

also AM of eigenvalue 0 is 2 and GM of eigenvalue 0 is 1 

Since AM ≠ GM so not diagonalizable.

Option (4) is false

Concept:

The generalized coordinate transformation from (p, q) to (P, Q) is canonical if  = 1

Explanation:

Given Q = pq(a+1), P = qb is canonical if  = 1

Now,  = q(a+1),  = (a+1)pqa,  = 0,  = bqb-1 

 = 1

⇒  = 1

⇒  = 1

⇒ - bq^a+b = 1

Only option (4) is satisfying the above relation.

Hence option (4) is corret

Calculation: 

Let A be an n × n matrix such that the set of all its nonzero eigenvalues has exactly r elements.

let E = { a₁ , a₂ , . . . . .  a_r} 

for each non zero eigen values there is at least one eigen vector .

for r non zero distinct eigenvector .

range space is at least r .

Hence option 3 is correct .

Option (1): 

Let A =  then eigenvalues are 0, 0 ⇒ r = 0

rank(A) = 1 = 2 - 1  2 - 1

Option (2) is false

Rank(A) = 1  r = 0

Option (1) is false

Option (4):

A has r non-zero eigenvalues

⇒ A² has r non-zero eigenvalues

But if A has r distinct eigenvalues does not imply A2 has r distinct eigenvalues.

Let A =  then eigenvalues of A are i, -1

but A2 has eigenvalues -1, -1 which are not distinct.

Option (4) is false

Calculation: 

Given 

and 

A ²⁰ = 

Hence option (2) is correct 

Calculation: 

Let 

and 

now 

det (A + B ) = (ad - bc) + (xw - yz) + aw + xd - bz - yc  . . ..(i) 

Now 

det (A - B) = ( ad - bc) + (xw - yz) - xd - aw + bz + yc   . . . .  (ii) 

Now from equations (i) and (ii), we get 

det (A + B ) + det(A - B) = 2 det(A) + 2 det (B) 

Hence option (3) is correct

Alternate MethodLet  and  

, 

then (1) ⇒ 1 + 1 = 1 + 0 (→ ←) - option (1) is false 

(4) ⇒ 1 - 1 = 2 - 0 (→ ←) option (4) is false

Let A = I_2×2, B = -I_2×2 then A + B = O_2×2, A - B = 2l_2×2

(2) ⇒ 0 + 2 = 2 - 2 (→ ←) option (2) is false.

So, option (3) is true.

Concept:

Linear Transformation: A function   between two vector spaces that preserves the operations of

vector addition and scalar multiplication. In this case,  is a linear transformation from  

(an  m -dimensional space) to  (an n -dimensional space).

One-to-One (Injective): A linear transformation is injective if distinct vectors in  are mapped to

distinct vectors in  . In terms of matrices,  is injective if its null space only contains the zero vector.

Onto (Surjective): A linear transformation is surjective if for every vector in , there is at least one vector

in  that maps to it. In terms of matrices,  is surjective if its image spans the entire space  (i.e., if ).

Bijective: A linear transformation is bijective if it is both injective (one-to-one) and surjective (onto).

A bijection implies that the linear transformation has an inverse, meaning  can map  to  perfectly

without losing or repeating information.

Explanation:

Option 1: 

Consider the set X = {1, 2, 3} so m = 3 and set Y = {a, b, c, d} so n = 4.

Define the function  Y by A(1) = a, A(2) = b and A(3) = c

This function is one-to-one (no two elements in X map to the same element in Y)

 but not onto (the element d \in Y is not mapped by any element of X).

In this case, m = 3 and n = 4, but m

providing a counterexample to the condition m > n.

Option 2:

Consider the set X = {1, 2, 3, 4} (so m = 4) and set Y = {a, b, c} (so n = 3).

Define the function  Y by
A(1) = a, A(2) = b, A(3) = c and A(4) = c

This function is onto because every element in Y is mapped by some element in X.

However, it is not one-to-one because two elements in X (3 and 4) map to the same element c in Y.

In this case, m = 4 and n = 3, but m > n. Therefore, the function is surjective but not injective,

providing a counterexample to the condition m

Option 3: 

A transformation is bijective if it is both one-to-one and onto, meaning every element of  has

a unique preimage in . This can only happen if m = n, so this is a correct statement.

Option 4:

Consider the set X = {1, 2, 3} (so m = 3) and the set Y = {a, b, c, d} (so n = 4).

Define the function  Y by A(1) = a, A(2) = b and A(3) = c

This function is one-to-one (injective) because no two elements of X map to the same element of Y.

However, m  n, as m = 3 and n = 4.

Thus, the function is injective but m  n, providing a valid counter example.

Thus, the correct statements is option 3).

Calculation: 

Let a and b are two eigen values of A .

a + b = 3   . . . . .  . 1

ab = 1  . . . . .  . . 2

(a - b )² = (a +b)² - 4ab 

(a-b ) ² = 5 

a - b =    . . . .  3 

now from equation 1 and 3 , we get 

a =   and  b =  

trace ( A² ) = a² + b² = 7 

Hence option (4) is correct

Alternate Method

Method -I Let λ₁ and λ₂ be two eigen values of A. 

Then det A = λ₁ λ₂ = 1

& Trace A = λ₁ + λ₂ = 3

And, Eigen values of A² are λ₁², λ₂².

⇒ Trace (A²) = λ12 + λ22

Now, (λ₁ + λ₂)² = λ12 + λ22 + 2λ₁λ₂

⇒ (3)² = λ₁² + λ₂² + 2

⇒ λ12 + λ22 = 9 - 2 = 7

Hence option (4) is correct

Method - II

Ch_A (x) = x² - Trace (A) x + Det (A) = 0 

⇒ x² - 3x + 1 = 0 ⇒ 

 λ₂ which are eigenvalues of A .

then eigenvalues of A² are,

Then trace  = 7

Hence option (4) is correct

Explanation:

T: ℝ2 →ℝ2 be defined by 

 be a basis of ℝ2 

So,  = 

  = 

So, matrix representation is

Option (3) is true and others are false

Explanation:

Characteristic polynomial p(T) is divisible by T2.

p(x)/x2

So p(x) = x2(x + a) where a can be zero also

Option (1): Let A =  Here 0 and a are the eigenvalues and eigenspace of A for 0 is 1

So option (1) is false

Here A is 3 × 3 matrix and two eigenvalues are 0, 0. Since complex eigenvalue are always complex conjugate and they are in pairwise. So here third eigenvalue must be real.

Option (2) is correct

 For A = , A3 ≠ 0

Option (3) is false

also AM of eigenvalue 0 is 2 and GM of eigenvalue 0 is 1 

Since AM ≠ GM so not diagonalizable.

Option (4) is false

Concept:

A linear transformation: The Linear transformation T : V → W is that for any vectors v₁ and v₂ in V and scalars a and b of the underlying field, it satisfies following condition:

T(av₁ + bv₂) = a T(v₁) + b T(v₂).

Calculation:

Here, Linear transformation T : R4 → R4 satisfy T3 + 3T2 = 4I, where I is the identity transformation

i.e. T is satisfied by annihilating polynomial λ³ + 3λ² = 4.

which is satisfied by λ = 1

Now, S = T4 + 3T3 – 4I = T(T³ + 3T²) - 4I = 4(T - I) = 0

i.e. λ = 1 satisfy | T - λI | = 0

i.e. S = 0i.e. λ = 0 satisfy | S - λI | = 0

or λ = 0 is a root of eigen value of S.

Hence, S is non-invertible.

Hence, the correct answer is option 4)