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

Concept:

Matrix Commutator and Diagonalizability:

• Let A and B be 2×2 real matrices.
• Commutator: For matrices A and B, the expression M = AB − BA is called the commutator of A and B.
• Diagonalizable Matrix: A matrix is said to be diagonalizable over ℝ if it is similar to a diagonal matrix via a real invertible matrix. That is, there exists an invertible P such that P⁻¹MP is diagonal.
• Nilpotent Matrix: A matrix M is nilpotent if M^k = 0 for some positive integer k. All nilpotent matrices are diagonalizable only if they are the zero matrix.
• Properties of the commutator:
  • Tr(M) = Tr(AB − BA) = 0 for any A, B ∈ M_n(ℝ).
  • If A and B are both diagonalizable, M = AB − BA is not necessarily diagonalizable.
  • If A and B are both upper triangular, then AB and BA are also upper triangular. Their difference M is upper triangular with zero diagonal, hence nilpotent.
  • Every nilpotent 2×2 matrix is similar over ℝ to or the zero matrix. So such matrices are diagonalizable only if zero.
  • However, for 2×2 nilpotent matrices (like above), M² = 0, so M² = λI₂ for λ = 0. So, Option 4 always holds.

Calculation:

Given:

A, B ∈ M₂(ℝ), M = AB − BA

Let’s analyze Option 1:

If A and B are upper triangular:

⇒ AB and BA are upper triangular

⇒ Diagonal of AB and BA = same (as only diagonal entries contribute to Tr)

⇒ M is upper triangular with zero diagonal ⇒ Tr(M) = 0

⇒ Such M is strictly upper triangular ⇒ nilpotent

⇒ In 2×2 case: any nilpotent matrix M satisfies M² = 0 ⇒ diagonalizable only if M = 0

⇒ So M is not necessarily diagonalizable unless it is 0

⇒ Option 1 is False

Option 2: A and B diagonalizable

⇒ M need not be diagonalizable

⇒ Counterexamples exist (e.g., A = diag(1,2), B = off-diagonal)

⇒ Option 2 is False

Option 3: A and B diagonalizable

⇒ M = λI₂?

⇒ Not always true. Commutator is traceless

⇒ can't be scalar matrix unless 0

⇒ I₂ has nonzero trace

⇒ M ≠ λI₂ unless λ = 0 and M = 0

⇒ Option 3 is False

Option 4:

⇒ M is always traceless 2×2 matrix.

So minimal polynomial is x² + ax + b or x² = 0 if nilpotent

⇒ ∃ λ ∈ ℝ such that M² = λI₂ is always true for 2×2 real matrices where eigenvalues are purely imaginary or zero

⇒ Option 4 is always true

∴ Correct answer: Option 4 only

Concept:

Linear Transformation on Matrix Spaces:

• Let M₂(ℝ) denote the set of all 2×2 matrices with real entries. This forms a 4-dimensional vector space over ℝ.
• Any linear transformation T on M₂(ℝ) can be represented as a 4×4 matrix when expressed in terms of the standard basis: {E₁₁, E₁₂, E₂₁, E₂₂} where E_ij has 1 in the (i,j) position and 0 elsewhere.
• Given transformation T is defined by T(X) = A × X × B^T, where A and B are fixed 2×2 matrices, and B^T is the transpose of B.
• Matrix multiplication preserves linearity, so T is a linear transformation.
• To compute the matrix representation of T, we apply T to each of the 4 basis matrices and express the result in terms of the same basis, forming a 4×4 matrix.
• Determinant of T: This is the determinant of the 4×4 matrix representation of T. It represents how T scales 4-dimensional volume.
• Trace of T: This is the sum of the diagonal entries in the 4×4 matrix. It equals the sum of eigenvalues of T (counted with multiplicity).

Given:

• A =
• B =
• ⇒ B^T =

Calculation:

We compute T(X) = A × X × B^T on each basis matrix of M₂(ℝ):

Let the standard basis matrices be:

• E₁₁ =
• E₁₂ =
• E₂₁ =
• E₂₂ =

⇒ Compute T(E_ij) = A × E_ij × B^T for each i,j.

⇒ Write each result as a linear combination of basis matrices to form the 4×4 matrix representation of T.

⇒ From computation (shown internally):

⇒ Matrix representation of T =  

⇒ Determinant = (−1)×5×(−3)×15 = 225

⇒ Trace = −1 + 5 + (−3) + 15 = 16

∴ Correct statements: Option 1 and Option 3

Concept:

Bilinear Form:

• A bilinear form B: ℝ⁴ × ℝ⁴ → ℝ is a function linear in each variable.
• It can be written in the form , where A is a matrix representing the bilinear form with respect to the standard basis.
• Given:
• This expression is linear in both x and y, hence it is a valid bilinear form.
• We find the matrix A such that

Calculation:

Given,

⇒ Identify terms to build matrix A using

⇒ Matrix A is:

⇒ Check properties:

⇒ A is skew-symmetric:

⇒ For skew-symmetric matrices of even order (like 4 × 4),  

⇒ But here, |A| = 1 

⇒ Check statements:

Option 1: False det A = 0 is incorrect.

Option 2: False det A = –1 is incorrect.

Option 3: False   = 0 for all x, since is skew-symmetric ⇒ B(x, x) = 0 

Option 4: True  If x ≠ 0, then since A is not the zero matrix, there exists some y such that B(x ,y ) ≠ 0

∴ The correct answer is Option 4:

If x ∈ ℝ⁴ is nonzero, there exists y ∈ ℝ⁴ such that B(x, y) ≠ 0.

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:

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

Explanation:

T : ℝ3 → ℝ3 is a linear transformation such that T(1, 2, 3) = (1, 2, 3), T(1, 5, 0) = (2, 10, 0) and T(-1, 2, -1) =(-3, 6, -3).

let a = (1, 2, 3), b = (1, 5, 0), c = (-1, 2, -1)

Then Linear transformation is

T(1, 2, 3) = (1, 2, 3) i.e., T(a) = a = 1a + 0b + 0c

T(1, 5, 0) = (2, 10, 0) = 2(1, 5, 0) i.e., T(b) = 2b = 0a + 2b + 0c

T(-1, 2, -1) =(-3, 6, -3) = 3(-1, 2, -1) i.e., T(c) = 3c = 0a + 0b + 3c

So matrix representation is

Rank of the matrix is 3

So, the dimension of the vector space spanned by all the eigenvectors of T is 3

Option (4) is true