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

Concept:

Vector Space and Dual Space:

The vector space consists of all polynomials of degree ≤ 3 over .

These polynomials are of the form where .

Therefore, . A basis of this space is .

The set A consists of all -linear maps from to . This is called the dual space of .

Dual space of any vector space has the same dimension, so .

The set B consists of those functionals such that .

This imposes one linear condition, reducing the dimension by 1. So, .

By the Rank-Nullity Theorem: .

If is non-zero, then is one-dimensional as it maps into . So, .

If is the zero map (which belongs to B), then is 0-dimensional and .

Therefore, not every has kernel dimension 3. It can be 4 if .

Calculation:

Given,

Dimension of = 4

⇒ Dual space has dimension 4.

⇒ Subset defined by , so .

⇒ By rank-nullity, for non-zero :

⇒  

⇒ But for zero map ,  

Option-wise Explanation:

Option 1: If , then

This is false.

If is the zero functional, then the entire space maps to 0. So, .

The correct statement should be: "If and , then ".

Option 2:  

This is true.

The condition imposes one linear restriction on (dimension 4), reducing it to dimension 3.

Option 3: 

This is true.

Since is the dual space of , they have the same dimension: 4.

Option 4: If , then the image of is a one-dimensional -vector space.

This is false.

If is the zero functional, then the image is  , whose dimension is 0.

Only for non-zero functionals, the image is one-dimensional.

So, the correct version would be: "If and , then the image is one-dimensional."

∴ The correct statements are Option 2 and Option 3.

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:

• Vector Space V : Set of real-valued continuous functions defined on the interval .
• Inner Product: For functions , the inner product is defined as .
• Set S : This is a set of functions in .
• Subspace W : Defined as the span of set S , i.e., .
• Basis: A set is a basis of a subspace if it is linearly independent and spans the space.
• Orthogonality: Two functions are orthogonal if their inner product is zero: .
• Orthonormal Set: A set where all vectors are orthogonal and of unit norm.

Calculation:

Given:

Step 1: Is S a basis of W ?

⇒ By definition, W = {span}(S)

⇒ S spans W trivially

⇒ Now check linear independence of S

⇒ 

⇒ One function is a linear combination of others

⇒ So, S is linearly dependent

⇒ But basis of a space can be the set itself if we define the space as its span

⇒ Hence, S is a basis of W (though not minimal)

Statement 1 is TRUE

Step 2: Is S an orthonormal basis?

⇒ Check  

⇒ But   , not orthogonal 

⇒ Also norms are not 1 

⇒ So, not orthogonal, not normalized

⇒ Statement 2 is FALSE

Step 3: Do there exist orthogonal functions in S ?

⇒   

⇒ Statement 3 is TRUE

Step 4: Does S contain an orthonormal basis of W ?

⇒ Since S is linearly dependent, and orthonormal sets must be linearly independent

⇒ No subset of S is orthonormal and spans W

⇒ Statement 4 is FALSE

∴ Final Answer: Only Statements 1 and 3 are TRUE.

Concept -

(1) If a is the eigen value of M then the eigen value of Mⁿ is aⁿ.

(2) If all the eigen value of the matrix is different then the matrix is diagonalizable.

Explanation -

Given - 

Now characteristic equation for the given matrix is -

⇒ | M - λ I | = 0

⇒ 

⇒ -λ (4 - λ ) + 3 = 0 ⇒ λ² - 4λ + 3 = 0

Now solve this equation we get the eigen values of the matrix -

 ⇒ λ2 - 3λ - λ  + 3 = 0  ⇒ (λ -3)(λ -1) = 0  ⇒ λ = 1, 3 

So the eigen value of M are 1 and 3.

Now solve the given statements -

(P) 𝑀8 + 𝑀12 is diagonalizable.

Now the eigen value of 𝑀8 + 𝑀12  are  

Hence both the eigen value of 𝑀8 + 𝑀12  are different So 𝑀8 + 𝑀12  is diagonalizable.

(𝑄) 𝑀7 + 𝑀9 is diagonalizable.

Now the eigen value of 𝑀7 + 𝑀9  are  

Hence both the eigen value of 𝑀7 + 𝑀9 are different So 𝑀7 + 𝑀9 is diagonalizable.

Hence the option (4) is true.

Concept:

Odd degree polynomial must have at least one real root

Explanation:

A is a a 3 × 3 matrix with real entries.

So characteristic polynomial of A will be of degree 3.

(1): Since we know that, odd degree polynomial must have at least one real root so A must have a real eigenvalue.

(1) is true

(2): As we know that determinant of a matrix is equal to the product of eigenvalues. So if the determinant of A is 0, then 0 is an eigenvalue of A.

(2) is true

(3): The determinant of A is negative and 3 is an eigenvalue of A.

If possible let the other two eigenvalues of A are not real and they are α + iβ, α - iβ

So determinant = 3(α + iβ)(α - iβ) = 3(α² + β²) > 0 for all α, β which is a contradiction.

So A must have three real eigenvalues.

(3) is true and (4) is false statement    

Explanation:

T be a linear operator on ℝ3. So T is a 3 × 3 matrix.

(a): T is non-zero and 0 is an eigen value of T. Let other two eigenvalues are α, β, so f(x) = x(x - α)(x - β)

As f(X) = X g(X) hence g(x) = (x - α)(x - β)

If α = 2, β = -2 then g(x) = (x - 2)(x + 2) = x² - 4

So g(T) = T² - 4I 

Now, eigenvalues of T are 0, 2, -2 so eigenvalues of g(T) are 4, 0, 0

So g(T) ≠ 0 

(a) is false.

(b): 0 is an eigenvalue of T with at least two linearly independent eigen vectors.

So GM = 2 for 0

We know that AM ≥ GM

So AM ≥ 2 ⇒ AM = 2 or AM = 3 for eigenvalue 0

For the case AM = 2

Let T =  λ ≠ 0

So characteristic polynomial f(x) = x²(x - λ)

Therefore g(x) = x(x - λ) = x² - λx

so g(T) = T² - λT

Now, eigenvalue of T is 0, 0, λ

eigenvalue of T2 - λT is 0 - 0λ, 0 - 0λ, λ² - λ² = 0, 0, 0

so g(T) = 0

If λ = 0 then f(x) = x³ so g(x) = x² Hence g(T) = 0

(b) is correct

Option (4) is correct

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 quadratic form is degenerate if at least one eigenvalue is 0 

Explanation:

B =  and S = .

So quadratic form is Q(x, y, z) = x² + y² -2z² + 2xy

Now, Q(a, b, c) = 0

⇒ a2 + b2 -2c2 + 2ab = 0....(i)

(1): On xy-plane, z = 0 so c = 0

Therefore (i) implies

a 2 + b2 + 2ab = 0 ⇒ (a + b)² = 0 ⇒ a + b = 0

i.e., x + y = 0, which is a line

So option (1) is TRUE

(2) On xz-plane, y = 0 so b = 0

Therefore (i) implies

a2 - 2c2 = 0 ⇒ x2 - 2z2 = 0 which is not an equation of ellipse

So option (2) is FALSE

(3): (i) ⇒  a2 + b2 -2c2 + 2ab = 0

  ⇒ (a + b)² = 2c²

  ⇒ a + b = ± √2c

So S is the union of two planes.

Option (3) is TRUE

(4): B =  

eigenvalues of B are -2, 2, 0

So Q is a degenerate quadratic form.

Option (4) is TRUE

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

Concept:

(a) Let V be a vector space. A function β : V × V → ℝ, usually denoted β(x, y) = , is called an inner product on V if it is positive, symmetric, and bilinear. That is, if

(i) ≥ 0,  = 0 only for x = x

Explanation:

x = (x1, …, xn) and y = (y1, …, yn) ∈ ℝn 

For n = 2

(1) ⟨x, y⟩ =  xiyj = x1y1 + x1y2 + x2y1 + x2y2 

So, A =  

Here a > 0, d > 0 but ad - bc = 1 - 1 = 0

So it does not define inner product space.

Option (1) is false

 xj yn−j+1= x1y2 + x2y1  

So, A =  

Here a = 0, d = 0 but ad - bc = 0 - 1 = -1

So it does not define inner product space.

Option (4) is false

(2) 

⟨x, y⟩ = 

= 

= 

Consider  x = (1,1)   and y = (1,1)

then = 2(2)² + 2 (2)² + 2 (1)² + 2(1)² = 20

But 2 = 2( 2 (1)2 + 2(1)2 +  2 (1)2 + 2(1)2 ) = 16

Thus  ≠ 2 and so  is not an inner product. 

Option (2) is false

Hence option (3) is true.

Concept:

(i) Q ∶ R2 → R is said to be positive definite if Q(x, y) > 0 ∀ (x, y) ≠ (0, 0)

(ii) A symmetric matrix is positive definite ⇔ It's all eigenvalues are positive

Explanation:

(1) Q(1, -1) = -1  0, so no positive definite.

(3) Q(x, -x) = x² - 2x² + x² = 0  0, so no positive definite.

(4) Q(x, -x) = x² - x² = 0  0, so no positive definite. 

So, option (1), (3), (4) are false and option (2) is true.

Alternate Method

(1). Matrix Representation for Q(x, y) = xy is,

 then det (A) = -¼

So, It's all eigenvalues can not be positive.

∵ det (A) = λ₁ λ2)

option (1) is false

(2)  then chA(x) = x2 - 2x + 

So, Eigen values of A are, x2 - 2x +  = 0

⇒ all eigen values of A are positive.

⇒ quadratic form is positive definite. 

option (2) is true

(3)  then det (A) = 0

⇒ one eigenvalue of A is zero and other is 2.

⇒ all eigenvalues are not positive.

⇒ Q(x, y) is not positive definite.

option (3) is false.

(4)  then det (A) = -1/4

⇒ All eigenvalues can not be positive.

⇒ Q(x, y) is not positive definite 

option (4) is false.

Concept:

Rank of a Matrix:

The rank of a matrix is the dimension of the image of the matrix, which represents the number

of linearly independent rows or columns. In this case, the matrix A has a rank of 7, meaning there are

7 linearly independent rows and columns, while 3 dimensions of the space are in the null space.

Null Space:

Since the matrix A is  and has rank 7, the null space (kernel) of A has dimension 10 - 7 = 3. This means there exist 3 linearly independent vectors  v  such that Av = 0 .

Nilpotent Matrix Consideration:
 

A nilpotent matrix is one where  for some positive integer k . In this case, nothing explicitly

indicates that A is nilpotent, but we do need to consider the powers of  A  when analyzing the vector space transformations.

Explanation:

Option 1: 
This suggests that the matrix  A  might act in a way where  v  is mapped by  A  to a non-zero value, but

applying  A  again results in zero. This is possible in the case of a matrix with a non-trivial null space.

This statement could be true for a matrix with certain properties, but it is not necessarily true in every case.

Counter example:

Consider the following   matrix, which can easily be expanded to larger dimensions:

This is a simple matrix that does not satisfy the given condition for vectors. Now, let’s check if we can

find a vector  v  that meets the conditions   and  .
 

Thus,  , which means   for all vectors  v .

Then 

Option 2: 

Since the matrix has rank 7, it implies that A  maps some vectors to non-zero values. If

 v is not in the null space of  A , then applying A  twice to  v  might not result in zero.

This statement is necessarily true, as there are vectors in  (those not in the null space) for which  .

Option 3: 
While the matrix has rank 7, this doesn't guarantee that the matrix has non-zero eigenvalues.

A matrix with non-trivial null space can still have zero eigenvalues. This is not necessarily true.

Option 4: " ."
This would imply that  A  is nilpotent and becomes zero after applying it 7 times.

There is no information in the problem that suggests  A  is nilpotent. This is not necessarily true.

The correct answer is Option 2).
 

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