Explore Exams
Login Signup
Sign in Open App
All Exams  >   Mathematics  >   Mathematics for Competitive Exams  >   All Questions

All questions of Vector Space & Linear Transformation for Mathematics Exam

Clear all your doubts with EduRev

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 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 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?

Chirag Verma 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.,

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 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.

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?

Chirag Verma 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 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.

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 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 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: 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.
 

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 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 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 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'.

A is any nxn matrix with all entries equal to 1 then 0 is an eigenvalue of A and
  • a)
    multiplicity of 0 is n - 1
  • b)
    multiplicity of 0 is 1
  • c)
    multiplicity of 0 is n
  • d)
    multiplicity of 0 is 0
Correct answer is option 'A'. Can you explain this answer?

Qiana Sharma answered
Explanation:

To prove that 0 is an eigenvalue of matrix A, we need to show that there exists a nonzero vector x such that Ax = 0.

Let's consider the matrix A with all entries equal to 1.

Matrix A:
[1 1 1 ... 1]
[1 1 1 ... 1]
[1 1 1 ... 1]
...
[1 1 1 ... 1]

Step 1:
To find the eigenvalues of matrix A, we need to solve the equation (A - λI)x = 0, where λ is the eigenvalue and I is the identity matrix.

Step 2:
Substituting A and I into the equation, we get:
[1-λ 1 1 ... 1]
[1 1-λ 1 ... 1]
[1 1 1-λ ... 1]
...
[1 1 1 ... 1-λ] * [x1, x2, x3, ..., xn] = 0

Step 3:
Expanding this equation, we get n equations:
(1-λ)x1 + x2 + x3 + ... + xn = 0
x1 + (1-λ)x2 + x3 + ... + xn = 0
x1 + x2 + (1-λ)x3 + ... + xn = 0
...
x1 + x2 + x3 + ... + (1-λ)xn = 0

Step 4:
Simplifying these equations, we get:
(1-λ)(x1 + x2 + x3 + ... + xn) = 0

Step 5:
Since all entries of matrix A are equal to 1, the sum of all elements in each row is n.

Step 6:
Therefore, the equation becomes:
(1-λ)(n) = 0

Step 7:
Solving this equation, we get:
1-λ = 0
λ = 1

Conclusion:
From the above steps, we can see that the only eigenvalue of matrix A is 1. Therefore, 0 is an eigenvalue of A with multiplicity n-1.

Hence, the correct answer is option A) multiplicity of 0 is n - 1.

Which of the following statement is correct?
  • a)
    The matrix A and its transpose AT have the same characteristic polynomial.
  • b)
    Similar matrices have not the same characteristic polynomial.
  • c)
    The trace of two similar matrices are not same.
  • d)
    The determinant of two similar matrices is not same.
Correct answer is option 'A'. Can you explain this answer?

Abhishek Nambiar answered
Explanation:

Characteristic polynomial:
The characteristic polynomial of a square matrix A is defined as the polynomial det(A - λI) where det denotes the determinant, I is the identity matrix, and λ is a scalar variable.

Statement (a):
The matrix A and its transpose AT have the same characteristic polynomial.
This statement is true. The proof is given below:
- Let A be a square matrix of order n.
- Then, the characteristic polynomial of A is det(A - λI) where I is the n × n identity matrix.
- Now, consider the transpose of A, denoted by AT.
- The characteristic polynomial of AT is det(AT - λI).
- We know that (AT - λI) = (A - λI)T.
- Therefore, det(AT - λI) = det((A - λI)T) = det(A - λI).
- Hence, the matrix A and its transpose AT have the same characteristic polynomial.

Statement (b):
Similar matrices have not the same characteristic polynomial.
This statement is false. Similar matrices have the same characteristic polynomial.
- Two matrices A and B are said to be similar if there exists an invertible matrix P such that B = P-1AP.
- If A and B are similar matrices, then they have the same eigenvalues.
- The eigenvalues of a matrix are the roots of its characteristic polynomial.
- Hence, similar matrices have the same characteristic polynomial.

Statement (c):
The trace of two similar matrices are not same.
This statement is false. Similar matrices have the same trace.
- The trace of a matrix is the sum of its diagonal elements.
- If A and B are similar matrices, then they have the same characteristic polynomial and hence the same eigenvalues.
- The sum of the eigenvalues of a matrix is equal to its trace.
- Therefore, similar matrices have the same trace.

Statement (d):
The determinant of two similar matrices is not same.
This statement is false. Similar matrices have the same determinant.
- If A and B are similar matrices, then they have the same characteristic polynomial and hence the same eigenvalues.
- The determinant of a matrix is equal to the product of its eigenvalues.
- Therefore, similar matrices have the same determinant.

where T be the reflection of the points through the line y = -x then the matrix of T with respect to standard basis is
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Shubha Madevi answered
The reflection of the point (x,y) across the line y= -x is (-y,-x).
The transformation is defined by T(x,y)=(-y,-x).
Find the basis for this transformation w.r.t standard basis you will get the solution

If T be a linear operator on a vector space V such that T2 - T + 1 = 0 then
  • a)
    T is one-one but may not be onto.
  • b)
    T is onto but may not be one-one.
  • c)
    T is invertible.
  • d)
    No such T exists.
Correct answer is option 'C'. Can you explain this answer?

Eesha Mehta answered
To prove that the correct answer is option 'C', we need to show that if T is a linear operator on a vector space V such that T^2 - T + 1 = 0, then T is invertible.

Given: T^2 - T + 1 = 0

To prove: T is invertible

Proof:
1. Assume that T is not invertible. This means that there exists a nonzero vector v in V such that T(v) = 0.
2. Applying T to both sides of the given equation, we get:
T(T^2 - T + 1) = T(0)
T^3 - T^2 + T = 0
3. Rearranging the terms, we have:
T(T^2 - T) = T^2
4. Since T(v) = 0, we can substitute T(v) with 0 in the above equation:
T(0) = T^2
0 = T^2
5. Multiplying both sides by T, we get:
T(0) = T(T^2)
0 = T^3
6. Substituting T^3 with T^2 - T in the above equation (from step 2), we have:
0 = T^2 - T
T^2 = T
7. Substituting T^2 with T in the given equation, we get:
T - T + 1 = 0
1 = 0
8. This is a contradiction, as 1 cannot be equal to 0.
9. Therefore, our assumption that T is not invertible is incorrect, and T must be invertible.

Hence, the correct answer is option 'C': T is invertible.

Chapter doubts & questions for Vector Space & Linear Transformation - Mathematics for Competitive Exams 2025 is part of Mathematics exam preparation. The chapters have been prepared according to the Mathematics exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Mathematics 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

Chapter doubts & questions of Vector Space & Linear Transformation - Mathematics for Competitive Exams in English & Hindi are available as part of Mathematics exam. Download more important topics, notes, lectures and mock test series for Mathematics Exam by signing up for free.

Mathematics for Competitive Exams

98 videos|37 docs|32 tests

Top Courses Mathematics

© EduRev
Education Revolution
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup