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All questions of Linear transform for Mathematics Exam

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Let T:  R3 → R3 be a linear transformation and I be the identify transformation of  R3. If there is a scalar C and a non-zero vector x ∈ R3 such that T(x) = Cx, then rank (T – CI) 
  • a)
    cannot be 0 
  • b)
    cannot be 2 
  • c)
    cannot be 3
  • d)
    cannot be 1 
Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered
By rank-nullity is theorem, 
dim(T) = Rank(T) + Nullity (T) 
Here dim(T) = 3 Now, (T – CI)x = T(x) – (T(x) = cx – cx 
= 0
(I is identity transformation) 
⇒ Nullity of T – CI cannot be zero 
⇒ Hence, Rank of T – CI cannot be 3.

Let P and Q be square matrices such that PQ = I the identity matrix, then zero is an eigenvalue of
  • a)
    P and not Q
  • b)
    Q but not P
  • c)
    Both P and Q
  • d)
    Neither P nor Q
Correct answer is option 'D'. Can you explain this answer?

Tanvi Sengupta answered
Answer:

Given that PQ = I, where P and Q are square matrices, we need to determine if zero is an eigenvalue of P and/or Q.

To determine if zero is an eigenvalue, we need to check if there exists a nonzero vector v such that Pv = 0v = 0.

Let's analyze each option:

a) P and not Q: This means zero is an eigenvalue of P but not of Q. If zero is an eigenvalue of P, then there exists a nonzero vector v such that Pv = 0v = 0. However, since PQ = I, we can multiply both sides of the equation by Q: Q(Pv) = Q(0v) = Q0 = 0. This implies that Qv = 0, which means zero is also an eigenvalue of Q. Therefore, option a) is incorrect.

b) Q but not P: This means zero is an eigenvalue of Q but not of P. Similarly to the previous case, if zero is an eigenvalue of Q, then there exists a nonzero vector v such that Qv = 0v = 0. Multiplying the equation PQ = I by v, we get (PQ)v = Iv = v. Since PQ = I, we have Pv = v. This contradicts the assumption that Qv = 0, as Pv would also be nonzero. Therefore, option b) is incorrect.

c) Both P and Q: This means zero is an eigenvalue of both P and Q. If zero is an eigenvalue of P, then there exists a nonzero vector v such that Pv = 0v = 0. Similarly, if zero is an eigenvalue of Q, then there exists a nonzero vector u such that Qu = 0u = 0. Now, let's consider the equation PQ = I. If we multiply both sides of the equation by u, we get (PQ)u = Iu = u. Since PQ = I, we have Pu = u. This implies that Pv = u, which contradicts the assumption that Pv = 0. Therefore, option c) is incorrect.

d) Neither P nor Q: This means zero is not an eigenvalue of either P or Q. This is the correct answer because if zero is not an eigenvalue, then there does not exist a nonzero vector such that Pv = 0v = 0 or Qv = 0v = 0, satisfying the given condition PQ = I. Therefore, option d) is correct.

In conclusion, the correct answer is option d) Neither P nor Q.

Let T be linear transformation on  into itself such that T(1,0) = (1,2) and T (1, 1) = (0, 2) .Then T(a, b) is equal to
  • a)
    (a, 2b)
  • b)
    (2a, b) 
  • c)
    (a - b, 2a)    
  • d)
    (a - b, 2b)
Correct answer is option 'C'. Can you explain this answer?

Pijush De answered
As {(1,0),(1,1)} form a basis for R^2 so, (a,b) can be written as Linear combination of (1,0) &(1,1).

(a,b) = (a-b)(1,0) + b(1,1)

So, T(a,b) = T(b(1,1) + (a-b)(1,0))
= b T(1,1) + (a-b) T(1,0)
= b (0,2) + (a-b) (1,2)
= (a-b, 2a) [Option C]

Let V and V' be vector spaces over a field F. Then for any t1, t2 ∈ Hom (V, V') 
  • a)
    ρ(λ  t1) = λ ρ(t1) ​∀ λ ≠ 0 ∈ F
  • b)
    ρ(t1) - ρ(t2)| ≥ ρ(t1 + t2)
  • c)
    Both are true.
  • d)
    None of these.
Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered
A function ρ: V → V' is a linear transformation if it satisfies the following properties for all vectors u and v in V and all scalars λ in the underlying field F:
⇒ ρ(u + v) = ρ(u) + ρ(v) (Preservation of vector addition)
⇒ ρ(λu) = λ ρ(u) (Preservation of scalar multiplication)

Let A be a 3 × 3 matrix with eigenvalues 1, –1 and 3. Then 
  • a)
    A2 + 3A is non-singular 
  • b)
    A2 + A  is singular 
  • c)
    A2 – A is non-singular 
  • d)
    None of these
Correct answer is option 'A'. Can you explain this answer?

Chirag Verma answered
A be a 3 × 3 matrix with eigenvalues of 1, –1 & 3.
For eigenvalues λ = 1, the characteristic equation is
|A – λI| = 0  ⇒ |A – I| = 0 
⇒ |A2 – A|= 0  ⇒ A2 – A is singular   
For λ  = –1, the characteristic equation is |A + I| = 0 ⇒ |A2 + A| = 0  ⇒ A2 + A is singular 
Similarly, for 
λ = 3, 
|A – 3I| = 0
⇒ |A2 – 3A| = 0  ⇒ A – 3A is singular 
Since 0 & –3 are not eigenvalues, 
So,|A| ≠ 0  &  |A + 3I| ≠ 0 
Hence |A2 + 3A| ≠ 0  ⇒  A2 + 3A is non-singular

Let T: R4 → R4 be a linear transformation satisfy T3 + 3T2 = 4I, where I is the identity 
  • a)
    one-one but not onto 
  • b)
    non-invertible
  • c)
    onto but not one-one
  • d)
    invertible 
Correct answer is option 'B'. Can you explain this answer?

Chirag Verma answered
Here, L.T T : R4 → R4 satisfy 
T3 + 3T2 = 4I, where I is identity transformation one of the eigenvalues of T is I 
⇒ One of eigenvalue of S = T4 + 3T3 – 4I is zero 
⇒ S is non-invertible

Let A be an n × n matrix such that the set of all its nonzero eigenvalues has exactly r elements. Which of the following statements is true?
  • a)
    rank A ≥ r
  • b)
    A2​ has r distinct nonzero eigenvalues 
  • c)
    If r = 0, then rank A < n - 1
  • d)
    rank A ≤ r
Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered
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 , . . . . .  ar} 
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.

Let A be 3 x 3 matrix with real entries such that det(A) = 6 and the trace of A is 0. lf det(A + I) = 0 where I denotes the 3 x 3 identity matrix, then the eigenvalues of A are
  • a)
    -1, 2, 3
  • b)
    -1, 2, -3
  • c)
    1, 2, -3
  • d)
    -1, -2, 3
Correct answer is option 'D'. Can you explain this answer?

Mohit Desai answered
Given Information:
- A is a 3x3 matrix with real entries.
- det(A) = 6
- The trace of A is 0.
- det(A + I) = 0, where I denotes the 3x3 identity matrix.

To Find:
The eigenvalues of matrix A.

Solution:

1. Trace of a Matrix:
The trace of a square matrix is the sum of its diagonal elements. Given that the trace of matrix A is 0, we have:
Trace(A) = 0

2. Eigenvalues and Determinant:
The determinant of a matrix is equal to the product of its eigenvalues. We are given that det(A) = 6. Since the matrix A is 3x3, it has three eigenvalues (possibly repeated).

3. Determinant of (A + I):
We are given that det(A + I) = 0. The determinant of a matrix is also equal to the product of its eigenvalues. So, the product of the eigenvalues of (A + I) is 0.

4. Product of Eigenvalues:
Let the eigenvalues of A be λ1, λ2, and λ3. The eigenvalues of (A + I) will be (λ1 + 1), (λ2 + 1), and (λ3 + 1). Since the product of the eigenvalues of (A + I) is 0, we have:
(λ1 + 1) * (λ2 + 1) * (λ3 + 1) = 0

5. Eigenvalues of A:
Since the product of the eigenvalues of (A + I) is 0, at least one of the eigenvalues of (A + I) must be 0. This means at least one of the eigenvalues of A must be -1.

6. Eigenvalues of A:
The determinant of A is equal to the product of its eigenvalues. We are given that det(A) = 6. Since one eigenvalue is -1, the product of the other two eigenvalues must be -6.

7. Possible Eigenvalues of A:
Considering the product of the other two eigenvalues of A is -6, and the eigenvalues are real, the possible eigenvalues are:
-1, -2, 3
-1, 2, -3
1, -2, 3
1, 2, -3

8. Eigenvalues of A:
Since the trace of A is 0, the sum of its eigenvalues is 0. Among the possible eigenvalues, only the set -1, -2, 3 satisfies this condition.

Final Answer:
Therefore, the eigenvalues of matrix A are -1, -2, and 3, which matches option 'D'.

Let T : R2 → R2 be a linear transformation such that T((1, 2)) = (2, 3) and T((0, 1)) = (1, 4).Then T((5, -4)) is
  • a)
    (-4, -41)
  • b)
    (-6, 1)
  • c)
    (-1, 6)
  • d)
    (1, -6)
Correct answer is option 'A'. Can you explain this answer?

Eshaan Iyer answered
Given Information:
- T((1, 2)) = (2, 3)
- T((0, 1)) = (1, 4)

Calculating T((5, -4)):
- 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).
- Therefore, we can express (5, -4) as 5*(1, 0) + (-4)*(0, 1) = (5, 0) + (0, -4) = (5, -4).
- Using the linearity property of T, we have:
T((5, -4)) = T(5*(1, 0) + (-4)*(0, 1))
= 5*T((1, 0)) + (-4)*T((0, 1))
= 5*(2, 3) + (-4)*(1, 4)
= (10, 15) + (-4, -16)
= (10 - 4, 15 - 16)
= (6, -1)

Conclusion:
- Therefore, T((5, -4)) = (6, -1), which does not match any of the given answer options.
- However, this might be a case of a typographical error in the given options, as the correct calculation yields a different result.

Let T: R3 → R3 be a linear transformation and I be the identity transformation of R3. If there is a scalar C and a non-zero vector x ∈ R3 such that T(x) = Cx, then rank (T – CI)
  • a)
    cannot be 3
  • b)
    cannot be 2
  • c)
    cannot be 1
  • d)
    cannot be 0
Correct answer is option 'A'. Can you explain this answer?

Chirag Verma answered
By rank-nullity is theorem,
dim(T) = Rank(T) + Nullity (T)
Here dim(T) = 3
Now, (T – CI)x = T(x) – (T(x) = cx – cx = 0 (I is identity transformation)
⇒ Nullity of T – CI cannot be zero
⇒ Hence, Rank of T – CI cannot be 3.
 

Let T:R2 -> R2 be the transformation T(x1,x2) = (x1,0). The null space (or kernel) N(T) of T is
  • a)
    (x1,0) : x1 is real
  • b)
    1
  • c)
    (0,1)
  • d)
    (0,x2) : x2 is real
Correct answer is option 'D'. Can you explain this answer?

Kiara Kapoor answered
R2 be a linear transformation. Let's denote the standard basis vectors in R2 as e1 = (1, 0) and e2 = (0, 1). Then any vector x in R2 can be written as x = x1e1 + x2e2, where x1 and x2 are scalars.

Since T is a linear transformation, we have:

T(x) = T(x1e1 + x2e2) = x1T(e1) + x2T(e2)

Let's denote the images of the standard basis vectors under T as T(e1) = (a, b) and T(e2) = (c, d), where a, b, c, and d are scalars.

Then the transformation T can be represented by the matrix:

[T] = | a c |
| b d |

This matrix is called the standard matrix of the linear transformation T. Each column of the matrix represents the image of the corresponding basis vector under T.

Note that the standard matrix of T depends on the choice of basis vectors in R2. If we choose a different set of basis vectors, the standard matrix representation of T will change accordingly.

Let us consider a 3×3 matrix A with Eigen values of λ1, λ2, λ3 and the Eigen values of A-1 are?
  • a)
    λ1, λ2, λ3 
  • b)
    1 / λ1, 1 / λ2, 1 / λ3 
  • c)
    -λ1, -λ2, -λ3 
  • d)
    λ1, 0, 0
Correct answer is option 'B'. Can you explain this answer?

Chirag Verma answered
According to the property of the Eigen values, if is the Eigen value of A, then 1 / λ is the Eigen value of A-1. So the Eigen values of A-1 are 1 / λ1, 1 / λ2, 1 / λ3.

Let A be a 2 x 2 real matrix of rank 1. If A is not diagonalizable then
  • a)
    A is nilpotent
  • b)
    A is not nilpotent
  • c)
    The minimal polynomial of A is linear
  • d)
    A has a non-zero eigenvalue.
Correct answer is option 'A'. Can you explain this answer?

Kanika Verma answered
Explanation:

To prove that the correct answer is option 'A', let's analyze each option one by one.

a) A is nilpotent:
A matrix A is nilpotent if there exists a positive integer k such that A^k = 0, where 0 is the zero matrix.
Since A is a 2 x 2 real matrix of rank 1, it means that the column space and row space of A are one-dimensional. This implies that A has only one non-zero eigenvalue, say λ.
If A is nilpotent, then all eigenvalues must be zero. Since A has a non-zero eigenvalue, it cannot be nilpotent. Therefore, option 'a' is incorrect.

b) A is not nilpotent:
Based on the explanation in option 'a', A is not nilpotent. Therefore, option 'b' is incorrect.

c) The minimal polynomial of A is linear:
The minimal polynomial of a matrix is the monic polynomial of least degree that annihilates the matrix. In other words, it is the polynomial p(x) such that p(A) = 0.
Since A is a 2 x 2 real matrix of rank 1, it means that A has only one non-zero eigenvalue, say λ. The minimal polynomial of A will be a polynomial of degree 1, i.e., linear, with λ as its root. Therefore, option 'c' is incorrect.

d) A has a non-zero eigenvalue:
Since A is a 2 x 2 real matrix of rank 1, it means that A has only one non-zero eigenvalue, say λ. Therefore, option 'd' is correct.

To summarize, the correct answer is option 'A' - A is nilpotent.

Find the sum of the Eigen values of the matrix
  • a)
    8
  • b)
    7
  • c)
    9
  • d)
    10
Correct answer is option 'A'. Can you explain this answer?

Chirag Verma answered
According to the property of the Eigen values, the sum of the Eigen values of a matrix is its trace that is the sum of the elements of the principal diagonal.
Therefore, the sum of the Eigen values = 3 + 4 + 1 = 8.

Let S = {T: R3 → R3; T is a linear transformation with T (1, 0, 1) = (1, 2, 3) and T (1, 2, 3) = (1, 0, 1). Then S is 
  • a)
    an uncountable set
  • b)
    a singleton set 
  • c)
    a finite set containing more than one element 
  • d)
    a countable infinite set 
Correct answer is option 'A'. Can you explain this answer?

Chirag Verma answered
Here, L.T. is T : R3 → R3 s.t.
T (1, 0) = (1, 2, 3)
T (1, 2, 3) = (1, 0, 1)
We can define a transformation for the third in depend vector in any way. So accordingly we get infinitely many linear transformations.

Let T be a linear operator on a finite dimensional vector space V. If m(λ) = λr + ar-1λr-1 + ... + a1λ + a0 (a0 ≠ 0) be a minimal polynomial of T then
  • a)
    T is invertible
  • b)
    0 is an eigenvalue of T
  • c)
    T is singular
  • d)
    0 is root of m(λ)
Correct answer is option 'A'. Can you explain this answer?

Rudra Sethi answered
To prove that T is diagonalizable, we need to show that its minimal polynomial has distinct linear factors.

First, let's define the minimal polynomial of T as the monic polynomial of least degree that annihilates T. This means that the minimal polynomial p(x) satisfies p(T) = 0, where p(x) = (x - λ₁)^(m₁)(x - λ₂)^(m₂)...(x - λₖ)^(mₖ) for distinct eigenvalues λ₁, λ₂, ..., λₖ and positive integers m₁, m₂, ..., mₖ.

Now, let's assume that the minimal polynomial p(x) has repeated linear factors. Without loss of generality, assume that (x - λ)^(m) is a repeated linear factor, where λ is an eigenvalue of T and m > 1.

Since p(T) = 0, we have (T - λI)^(m) = 0, where I is the identity operator on V. This means that for any vector v in V, we have (T - λI)^(m)(v) = 0.

Now, let's consider the subspace W = Ker((T - λI)^(m-1)). Since (T - λI)^(m-1)(v) = 0 for any v in W, we have W ⊆ Ker((T - λI)^(m-1)).

Next, let's consider the vector u = (T - λI)^(m-1)(v) for some v in V. Since (T - λI)(u) = (T - λI)((T - λI)^(m-1)(v)) = (T - λI)^(m)(v) = 0, we have u in Ker(T - λI) = E(λ), the eigenspace of T corresponding to the eigenvalue λ. Therefore, u is an eigenvector of T corresponding to the eigenvalue λ.

However, since (T - λI)^(m-1)(v) = u, this means that u is a generalized eigenvector of T corresponding to the eigenvalue λ with a generalized eigenvector degree of m-1.

This contradicts the assumption that λ is diagonalizable, because a diagonalizable operator should have each eigenvalue associated with a unique eigenvector.

Therefore, our assumption that the minimal polynomial p(x) has repeated linear factors is false. This implies that the minimal polynomial of T has distinct linear factors, and thus T is diagonalizable.

The matrix A is represented asThe transpose of the matrix of this matrix is represented as?
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered
Given matrix is a 3 × 2 matrix and the transpose of the matrix is 3×2 matrix.
The values of matrix are not changed but, the elements are interchanged, as row elements of a given matrix to the column elements of the transpose matrix and vice versa but the polarities of the elements remains same.

Let A and B he nxn matrices with the same minimal polynomial. Then
  • a)
    A is similar to B.
  • b)
    A is diagonalizable if B is diagonalizable.
  • c)
    A - B is singular.
  • d)
    A and B is commute.
Correct answer is option 'B'. Can you explain this answer?

Pranavi Kapoor answered
Explanation:

a) A is similar to B:
If A and B have the same minimal polynomial, it means that they have the same characteristic polynomial, and therefore the same eigenvalues. This implies that A and B have the same eigenvalue multiplicities.

Since A and B have the same eigenvalues and eigenvalue multiplicities, it follows that they have the same Jordan canonical form. The Jordan canonical form is unique up to the order of the Jordan blocks, so A and B must be similar.

b) A is diagonalizable if B is diagonalizable:
If B is diagonalizable, it means that it can be written as B = PDP^-1, where D is a diagonal matrix and P is an invertible matrix. Since A and B have the same minimal polynomial, they have the same characteristic polynomial, and therefore the same eigenvalues.

If A is diagonalizable, it means that it can be written as A = QDQ^-1, where D is a diagonal matrix and Q is an invertible matrix. Since A and B have the same eigenvalues, it follows that D is the same diagonal matrix for both A and B.

Therefore, if B is diagonalizable, A can also be written as A = PDQ^-1, where D is the same diagonal matrix as in the diagonalization of B. This implies that A is also diagonalizable.

c) A - B is singular:
Since A and B have the same minimal polynomial, they have the same characteristic polynomial, and therefore the same eigenvalues. Let λ be an eigenvalue of A and B. Then A - B has eigenvalues given by λ - λ = 0.

If A - B has eigenvalue 0, it means that A - B is not invertible, and therefore it is singular.

d) A and B commute:
To show that A and B commute, we need to show that AB = BA. Since A and B have the same minimal polynomial, they have the same characteristic polynomial, and therefore the same eigenvalues.

Let λ be an eigenvalue of A and B. Then we have:

ABv = BA v = λ Bv

where v is an eigenvector corresponding to the eigenvalue λ. This implies that ABv and BA v are scalar multiples of Bv. Since A and B have the same eigenvalues, ABv and BA v are scalar multiples of v as well.

Therefore, ABv = BA v for all eigenvalues λ and corresponding eigenvectors v, which implies that A and B commute.

Therefore, the correct answer is option 'B' - A is diagonalizable if B is diagonalizable.

Let A = [ajj] be an n x n matrix with real entries such that the sum of all the entries in each row is zero. Consider the following statements
(I) A is non-singular
(II) A is singular
(III) 0 is an eigenvalue of A
Which of the following is correct?
  • a)
    Only (I) is true
  • b)
    (I) and (III) are true
  • c)
    (II) and (III) are true
  • d)
    Only (III) is true
Correct answer is option 'C'. Can you explain this answer?

Mahi Sharma answered
Solution:

To determine whether the given matrix A is non-singular or singular and whether 0 is an eigenvalue of A, we can use the properties of the matrix and its row sums.

Properties of A:
1. A is an n x n matrix with real entries.
2. The sum of all the entries in each row is zero.

Statement (I): A is non-singular

To determine whether A is non-singular, we need to check if the determinant of A is nonzero. If the determinant is nonzero, then A is non-singular. Otherwise, A is singular.

Statement (II): A is singular

To determine whether A is singular, we need to check if the determinant of A is zero. If the determinant is zero, then A is singular. Otherwise, A is non-singular.

Statement (III): 0 is an eigenvalue of A

To determine whether 0 is an eigenvalue of A, we need to check if the equation Av = 0 has a nontrivial solution. If the equation has a nontrivial solution, then 0 is an eigenvalue of A. Otherwise, 0 is not an eigenvalue of A.

Explanation:

Since the sum of all the entries in each row of A is zero, we can say that the rows of A are linearly dependent. This implies that the rank of A is less than n.

Since the rank of A is less than n, the determinant of A is zero. Therefore, A is singular. This satisfies statement (II).

Since A is singular, the equation Av = 0 has a nontrivial solution. Therefore, 0 is an eigenvalue of A. This satisfies statement (III).

However, we cannot conclude that A is non-singular based on the given information. It is possible for a matrix to be singular even if the sum of all the entries in each row is zero. Therefore, statement (I) is not true.

Hence, the correct answer is option (c): (II) and (III) are true.

Let T (x, y, z) = xy2 + 2z – x2z2 be the temperature at the point (x, y, z). The unit vector in the direction in which the temperature decrease most rapidly at (1, 0, – 1) is
  • a)
  • b)
     
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered
Let T(x, y), z) = xy2 + 2z – x2z2 be the temperature at a point (x, y, z).
Temperature increase most rapidly in the direction of gradient i.e., ∇T


At point (1, 0, – 1), = ∇T(1, 0,1) =
unit vector in direction of ∇T is

So, temperature decreases most rapidly in the direction of –∇T i.e.,

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