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All questions of Vector Space & Linear Transformation 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 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 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?

Mahira Patel answered
Rank of Matrix A in terms of its Nonzero Eigenvalues
Rank of a matrix A is defined as the dimension of the column space of A. Let the set of all nonzero eigenvalues of A be denoted by E = {λ1, λ2, ..., λr}.
  • Statement: Rank A ≥ r

    • Since the nonzero eigenvalues of A are λ1, λ2, ..., λr, the rank of A must be at least r because each nonzero eigenvalue corresponds to a linearly independent eigenvector. Therefore, the rank of A is greater than or equal to the number of nonzero eigenvalues, which is r.

Hence, the correct statement is rank A ≥ r.

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: 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.

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 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.,

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

Let be a linear transformation defined by T(x, y, z) = (x + y - z, x + z, y - z) then the matrix of the linear transformation T with respect to ordered basis β = {(0,1,0), (0,0,1), (1,0,0)} of is 
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Avik Das answered
T(0,1,0)=(1,0,1)=0(0,1,0)+1(0,0,1)+1(1,0,0)
T(0,0,1)=(-1,1,-1)=1(0,1,0)-1(0,0,1)-1(1,0,0)
T(1,0,0)=(1,1,0)=1(0,1,0)+0(0,0,1)+1(1,0,0)
so, the matrix should be
column wise (0,1,1),(1,-1,-1),(1,0,1)

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