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

Explanation:

Let A =  [0, 2] , B =  [2, 3]  and A UB=[0,3]

Let f ∈ V

\(f_A(x) = a_0 +a_1 x +a_2 x^2 + a_3 x^3 \) where \(a_0 ,a_1,a_2,a_3 ∈ \mathbb{R} \)

\(f_B(x) = b_0 +b_1 x +b_2 x^2 + b_3 x^3 + b_4 x^4 \) where \(b_0 ,b_1,b_2,b_3, b_4 ∈ \mathbb{R} \)

Given: f(0) = 3 (this point will lie in A)

\(3= a_0 +0+0+0 \implies a_0=3 \)

Since V be the real vector space of all continuous functions

\(f_{A}(2) = f_{B}(2) \)  

⇒ \(a_0 +a_1 2 +a_2 2^2 + a_3 2^3 = b_0 +b_1 2 +b_2 2^2 + b_3 2^3 + b_4 2^4 \)

⇒ \(a_0 +2a_1 + 4a_2 + 8a_3 = b_0 + 2b_1 +4 b_2 +8 b_3 +16 b_4 \)

Now we can define 

\(f(x) = \begin{cases} 3 +a_1 x +a_2 x^2 + a_3 x^3, & \text{when } x ∈ A\\ b_0 +b_1 x +b_2 x^2 + b_3 x^3 + b_4 x^4, & \text{when } x ∈ B\\ \end{cases} \)

With restriction \(a_0 +2a_1 + 4a_2 + 8a_3 = b_0 + 2b_1 +4 b_2 +8 b_3 +16 b_4 \)

Then the dimension of V is equal to = Number of scalar - No of Restrictions = 8-1 = 7

Hence 7 is the answer.

Concept:

An invertible linear transformation (or nonsingular transformation) is a linear transformation \(T: V \to V \) on

a vector space V that has an inverse transformation \(T^{-1} \) such that

\(T(T^{-1}(v)) = v \quad \text{and} \quad T^{-1}(T(v)) = v\)

for all \(v \in V\) .

Explanation:

T, S : \(\mathbb{R}^4 \rightarrow \mathbb{R}^4\) are non-zero, non-identity linear transformations.

\(T^2 = T \), meaning T is an idempotent transformation.

Option 1: T is necessarily invertible:

Since \(T^2 = T \) , T is idempotent. An idempotent matrix (transformation) cannot be invertible unless

it is the identity matrix because if T were invertible, we would have \(T^2 = T \Rightarrow T = I\) , which contradicts

the given condition that T is non-identity. Therefore, Option 1 is false.

Option 2: T and S are similar if \(S^2 = S\) and \(\text{Rank}(T) = \text{Rank}(S) :\)

Two idempotent transformations T and S are similar if they have the same rank, as similar transformations

have the same Jordan form, and for idempotent matrices, the form is characterized by the rank. So, if \(S^2 = S\) and

\(\text{Rank}(T) = \text{Rank}(S)\) , then T and S can be similar. Therefore, Option 2 is true.

Option 3: T and S are similar if S has only 0 and 1 as eigenvalues:

While both T and S would have eigenvalues 0 and 1 (since they are idempotent), similarity requires

not only the same eigenvalues but also the same rank, as well as a similarity transformation that relates their

Jordan forms. This condition alone (only 0 and 1 as eigenvalues) is not sufficient to guarantee similarity. Therefore, Option 3 is false.

Option 4: T is necessarily diagonalizable:

 An idempotent matrix (with eigenvalues only 0 and 1) is always diagonalizable, as it can be expressed in a form

where it is similar to a diagonal matrix with 0s and 1s on the diagonal, corresponding to the projection on the

eigenspaces associated with each eigenvalue. Therefore, Option 4 is true.

Hence option 2) and 4) are correct.

Concept Use: 

Basis of the Vector space : 

Basis: let V(F) be a vector space and S be a non-empty subset of V then S is said to be basis of the vector space V if (a) S is linearly independent and (b) L(S) = V i.e., S spans V.

Characteristic Polynomial of the Transformation have the degree same to that of Dimension of the Vector space

A is the matrix of the form \(\rm A=\left(\begin{array}{ll} 0 & 2 \\ 2 & 0 \end{array}\right) \) then A has eigen value +-(√a × b) = +- 2

Explanation:

Since, T Contains the Matrix of 2 × 2 order where element are belongs to Complex Number 

So, option 2 and 3 Discarded 
 

Take the Basis for T that is \(\left(\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right) \),\(\left(\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right) \), \(\left(\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right) \), \(\left(\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right) \)

So, T(\(\left(\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right) \)) = \(\left(\begin{array}{ll} 0 & 2 \\ 2 & 0 \end{array}\right) \)\(\left(\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right) \)

, T(\(\left(\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right) \)) = \(\left(\begin{array}{ll} 0 & 2 \\ 2 & 0 \end{array}\right) \)\(\left(\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right) \)  

Concept:

Linear Transformation and Matrix Representation:

T is a linear transformation that maps any \(2 \times 2\) matrix \(B \in V\) to \(AB\), where \(A\) is a given \(2 \times 2\) matrix.

Characteristic Polynomial:

The characteristic polynomial of a matrix A is given by the determinant of \(A - \lambda I\), where \(\lambda\) is

the eigenvalue and \(I \) is the identity matrix. The characteristic polynomial is \( \text{det}(A - \lambda I)\) where \(I \) is

the identity matrix of the same dimension as A, and \(\lambda\) represents the eigenvalues.

Explanation:
A =\( \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}\)
 

Here, T  is acting \(2 \times 2 \)  matrices. So, we need to understand how  A acts on any \(B \in V \), where B = \(\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}\) .

The action of A  on B is

T(B) = AB = \(\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}\) = \(\begin{pmatrix} 2b_{11} & 2b_{12} \\ b_{21} & b_{22} \end{pmatrix} \)

This shows how the transformation T  scales the first row of the matrix B by 2 and leaves the second row unchanged.

Now, we represent T as a matrix that acts on the vectorization of the \(2 \times 2 \)  matrix B. If we write the entries

of B as a vector \(\mathbf{b} \in \mathbb{R}^4\) (by stacking the columns of  B), i.e.,

\(\mathbf{b} = \begin{pmatrix} b_{11} \\ b_{21} \\ b_{12} \\ b_{22} \end{pmatrix}\)

Then the action of T on \(\mathbf{b}\)  can be represented as a \(4 \times 4 \) matrix. The effect of  T  is:

\(T(\mathbf{b}) = \begin{pmatrix} 2b_{11} \\ b_{21} \\ 2b_{12} \\ b_{22} \end{pmatrix}\)

This can be written as the matrix multiplication

\(T(\mathbf{b}) = \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} b_{11} \\ b_{21} \\ b_{12} \\ b_{22} \end{pmatrix}\)

Thus, the matrix representation of  T is

\(T = \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\)
 

The characteristic polynomial of a matrix  T is given by:

\(p(\lambda) = \det(T - \lambda I)\)

where  \(I\) is the \(4 \times 4 \) identity matrix. Substituting  T into this expression:

\(T - \lambda I = \begin{pmatrix} 2 - \lambda & 0 & 0 & 0 \\ 0 & 1 - \lambda & 0 & 0 \\ 0 & 0 & 2 - \lambda & 0 \\ 0 & 0 & 0 & 1 - \lambda \end{pmatrix}\)

Now, we compute the determinant

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\det(T - \lambda I) = (2 - \lambda)(1 - \lambda)(2 - \lambda)(1 - \lambda)\)

Simplifying,

\(p(\lambda) = (2 - \lambda)^2 (1 - \lambda)^2\)

Thus, the characteristic polynomial of  T is

\(p(\lambda) = (2 - \lambda)^2 (1 - \lambda)^2\)

Hence the correct option is 3).

Explanation:

The linear transformation associated with the matrix A is given by:

⇒ \(T(x, y) = A \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2x + 3y \\ 4x - y \end{bmatrix}\)

Step 1: Compute T(1,3) and express it in terms of the basis S

⇒ \(T(1,3) = \begin{bmatrix} 2(1) + 3(3) \\ 4(1) - 3 \end{bmatrix} = \begin{bmatrix} 2 + 9 \\ 4 - 3 \end{bmatrix} = \begin{bmatrix} 11 \\ 1 \end{bmatrix}\)

We express (11,1) as a linear combination of the basis vectors 

⇒ \( \begin{bmatrix} 11 \\ 1 \end{bmatrix} = a \begin{bmatrix} 1 \\ 3 \end{bmatrix} + b \begin{bmatrix} 2 \\ 5 \end{bmatrix}\)

​​This gives the system: ​​​​

 ⇒ a + 2b = 11

⇒ 3a + 5b = 1

Solving for a and b :

Multiplying the first equation by 3

⇒ 3a + 6b = 33

Subtracting from the second equation

⇒ (3a + 5b) - (3a + 6b) = 1 - 33

\(-b = -32 \Rightarrow b = 32\)

Substituting b = 32 into a + 2b = 11 : 

⇒ a + 2(32) = 11

⇒ a + 64 = 11

⇒ a = -53

Thus, the coordinate vector of T(1,3) in the basis S is}

⇒ \(\begin{bmatrix} -53 \\ 32 \end{bmatrix}\)

Compute T(2,5) and express it in terms of the basis S 

⇒ \(T(2,5) = \begin{bmatrix} 2(2) + 3(5) \\ 4(2) - 5 \end{bmatrix} = \begin{bmatrix} 4 + 15 \\ 8 - 5 \end{bmatrix} = \begin{bmatrix} 19 \\ 3 \end{bmatrix}\)

We express  (19,3)  as: 

⇒ \( \begin{bmatrix} 19 \\ 3 \end{bmatrix} = a \begin{bmatrix} 1 \\ 3 \end{bmatrix} + b \begin{bmatrix} 2 \\ 5 \end{bmatrix}\)

This gives the system

⇒ a + 2b = 19

⇒ 3a + 5b = 3

Solving for a and b: 

Multiplying the first equation by 3

⇒ 3a + 6b = 57

Subtracting from the second equation

⇒ (3a + 5b) - (3a + 6b) = 3 - 57

\(-b = -54 \Rightarrow b = 54\)

Substituting b = 54 into a + 2b = 19 

⇒ a + 2(54) = 19

⇒ a + 108 = 19

⇒ a = -89 

Thus, the coordinate vector of T(2,5) in the basis S is

⇒ \(\begin{bmatrix} -89 \\ 54 \end{bmatrix}\)

The transformation matrix in the basis S is:

⇒ \(B=[T]_S = \begin{bmatrix} -53 & -89 \\ 32 & 54 \end{bmatrix}\)

Hence option 2 is correct

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 = \(\begin{bmatrix}0&0&0\\0&0&0\\0&0&λ\end{bmatrix}\) 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

Concept:

Linear Transformation and Matrix Representation:

T is a linear transformation that maps any \(2 \times 2\) matrix \(B \in V\) to \(AB\), where \(A\) is a given \(2 \times 2\) matrix.

Characteristic Polynomial:

The characteristic polynomial of a matrix A is given by the determinant of \(A - \lambda I\), where \(\lambda\) is

the eigenvalue and \(I \) is the identity matrix. The characteristic polynomial is \( \text{det}(A - \lambda I)\) where \(I \) is

the identity matrix of the same dimension as A, and \(\lambda\) represents the eigenvalues.

Explanation:
A =\( \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}\)
 

Here, T  is acting \(2 \times 2 \)  matrices. So, we need to understand how  A acts on any \(B \in V \), where B = \(\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}\) .

The action of A  on B is

T(B) = AB = \(\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}\) = \(\begin{pmatrix} 2b_{11} & 2b_{12} \\ b_{21} & b_{22} \end{pmatrix} \)

This shows how the transformation T  scales the first row of the matrix B by 2 and leaves the second row unchanged.

Now, we represent T as a matrix that acts on the vectorization of the \(2 \times 2 \)  matrix B. If we write the entries

of B as a vector \(\mathbf{b} \in \mathbb{R}^4\) (by stacking the columns of  B), i.e.,

\(\mathbf{b} = \begin{pmatrix} b_{11} \\ b_{21} \\ b_{12} \\ b_{22} \end{pmatrix}\)

Then the action of T on \(\mathbf{b}\)  can be represented as a \(4 \times 4 \) matrix. The effect of  T  is:

\(T(\mathbf{b}) = \begin{pmatrix} 2b_{11} \\ b_{21} \\ 2b_{12} \\ b_{22} \end{pmatrix}\)

This can be written as the matrix multiplication

\(T(\mathbf{b}) = \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} b_{11} \\ b_{21} \\ b_{12} \\ b_{22} \end{pmatrix}\)

Thus, the matrix representation of  T is

\(T = \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\)
 

The characteristic polynomial of a matrix  T is given by:

\(p(\lambda) = \det(T - \lambda I)\)

where  \(I\) is the \(4 \times 4 \) identity matrix. Substituting  T into this expression:

\(T - \lambda I = \begin{pmatrix} 2 - \lambda & 0 & 0 & 0 \\ 0 & 1 - \lambda & 0 & 0 \\ 0 & 0 & 2 - \lambda & 0 \\ 0 & 0 & 0 & 1 - \lambda \end{pmatrix}\)

Now, we compute the determinant

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\det(T - \lambda I) = (2 - \lambda)(1 - \lambda)(2 - \lambda)(1 - \lambda)\)

Simplifying,

\(p(\lambda) = (2 - \lambda)^2 (1 - \lambda)^2\)

Thus, the characteristic polynomial of  T is

\(p(\lambda) = (2 - \lambda)^2 (1 - \lambda)^2\)

Hence the correct option is 3).

Concept:

Eigen Values:

1. Let A be a square matrix of order ‘n’ and ‘λ’ be a scalar.

|A−λI| = 0 is called the characteristic equation of matrix A.

2 .The roots of the characteristic equation are called Eigenvalues.

3. Corresponding to each eigenvalue ‘λ’, there exists a non-zero vector ‘v’ such that Av = λv or (A -λI)v = 0

Trace of a Matrix: 

Let A be a matrix of order n × n. Let \(λ_1,λ_2,λ_3,...,λ_n\) be the eigenvalues of M. Then:

1. Trace of a matrix is equal to the sum of its diagonal elements. It is represented by tr(A).

2. tr(A + B) = tr(A) + tr(B) 

3. tr(A) = \(\sum_{i=1}^{n}λ_i\)

4. tr(A^k) = \(\sum_{i=1}^{n}λ_i^k\), where \(λ_i\) is the i^th eigen value

Calculation:

We have, A is a 4 × 4 matrix such that -1, 1, 1, -2 are its eigenvalues.

∴ tr(A) = (-1) + 1 + 1 + (-2) 

⇒ tr(A) = -1

Given, B = A4 - 5A2 + 5I

⇒ tr(B) = tr(A⁴) - 5tr(A²) + 5tr(I)

tr(A⁴) = (- 1)⁴ +  1⁴ + 1⁴ + (-2)⁴ = 1 + 1 + 1 + 16 = 19

tr(A2) = (- 1)² + 1² + 1² + (- 2)² = 1 + 1 + 1 + 4 = 7

tr(I)  = 1 + 1 + 1 + 1 = 4

⇒ tr(B) = 19 - 5 × 7 + 5 × 4

⇒ tr(B) = 19 - 35 + 20 = 39 - 35

⇒ tr(B) = 4

tr(A + B) = tr(A) + tr(B)

= (-1) + 4

∴ tr(A + B) = 3

Alternate Method 
Eigenvalues of A are -1, 1, 1, -2

So tr(A) = -1 + 1 + 1 - 2 = -1

We know that if λ is an eigenvalue of A then λ^m is an eigenvalue of A^m. 

B =  A4 - 5A2 + 5I 
So eigenvalues of B are 

1 - 5 + 5 = 1, 1 - 5 + 5 = 1, 1 - 5 + = 1 and 16 - 20 + 5 = 1

So tr(B) = 1 + 1 + 1 + 1 = 4

Hence tr(A + B) = -1 + 4 = 3
 (3) correct

Concept:

Eigenvalues:

1. Let A be a square matrix of order ‘n’ and ‘λ’ be a scalar.

|A−λI|=0 is called the characteristic equation of matrix A.

2. The roots of the characteristic equation are called Eigenvalues.

3. Corresponding to each eigenvalue ‘λ’, there exists a non-zero vector ‘v’ such that Av = λv or (A -λI)v = 0

Let A be a matrix of order n × n. Let \(λ_1,λ_2,λ_3,...,λ_n\) be the eigenvalues of M. Then:

1. det(A) = \(λ_1λ_2λ_3\cdot...\cdotλ_n\) i.e., the determinant of A is equal to the product of the eigenvalues 

2. tr(A) = \(λ_1+λ_2+λ_3+...+λ_n\) i.e., the trace of A is equal to the sum of the eigenvalues. [ Trace of A is the sum of the diagonal elements of A ]

Calculation:

\(M = \left[ {\begin{array}{*{20}{c}} 0&{ - 1}&0\\ 1&2&{ - 1}\\ 1&1&3 \end{array}} \right]\)

⇒ |M| = 0 + 1(3 + 1) + 0

⇒ |M| = 4

Let \(λ_1, λ_2,λ_3\) be the eigenvalues.

∴ \(λ_1λ_2λ_3\) = 4...(i)

and, \(λ_1+λ_2+λ_3\) = 5...(ii)

According to the question, 1 is an eigenvalue of M.

Let \(λ_1\) = 1

∴ (i) ⇒ \(λ_2λ_3\) = 4...(iii)

and, (ii) ⇒ \(1+λ_2+λ_3\) = 5

⇒ \(λ_2+λ_3\) = 4...(iv)

Solving (iii) and (iv), we get:

\(λ_2\) = 2 and \(λ_2\) = 2

∴ The eigenvalues of M are 1, 2, and 2. 

Options 1 and 2 are false.

(i) Dimension of eigenspace of (λ) = G.M.(λ)

(ii) G.M.(λ) = n - rank(A - λI), A is n × n matrix.

Now, M - 2I =  \( \left[ {\begin{array}{*{20}{c}} 0&{ - 1}&0\\ 1&2&{ - 1}\\ 1&1&3 \end{array}} \right]- 2 \left[ {\begin{array}{*{20}{c}} 1&{ 0}&0\\ 0&1&{ 0}\\ 0&0&1 \end{array}} \right] \)

M - 2I  = \( \left[ {\begin{array}{*{20}{c}} -2&{ - 1}&0\\ 1&0&{ - 1}\\ 1&1&1 \end{array}} \right]\)

Applying row operations, R₁ → R₁ + 2R₂

and R₃ → R₃ - R₂

= \( \left[ {\begin{array}{*{20}{c}} 0&{ - 1}&-2\\ 1&-2&{ - 1}\\ 0&1&2 \end{array}} \right]\)

R₃ → R₃ + R₁

= \( \left[ {\begin{array}{*{20}{c}} 0&{ - 1}&-2\\ 1&0&{ - 1}\\ 0&0&0\end{array}} \right]\)

∴ Rank (M - 2I) = 2

⇒ G.M.(2) = n - rank (A - 2I) = 3 - 2 = 1 = Eigenspace(2)

∵ A.M. ≥ G.M. and A.M.( λ = 1) = 1

⇒ G.M. (1) = 1 = Eigenspace (1)

Also, A.M.(2) ≠ G.M.(2) ⇒ M is not diagonalizable.

∴ Option 4 is false.

Since all the eigenvalues are having G. M. 1 so the eigenspace of each eigenvalue has dimension 1 option 3 is true.

Explanation:

M be a 5 × 5 matrix with real entries such that Rank(M) = 3

Then echelon form of M is of the form

\(\left[\begin{array}{ccccc} 1 & 0 & 0 & * & * \\ 0 & 1 &0& * & * \\ 0 & 0 &1& * & * \\ 0 & 0 &0& 0& 0 & \\ 0 & 0 &0 & 0 & 0 \end{array}\right] \)

Linear system Mx = b has a solution and R = [M|b]

So Rank(M) = Rank(R) = 3

∴ \(\left[\begin{array}{ccccc} 1 & 0 & 0 & * & * &*\\ 0 & 1 &0& * & *&* \\ 0 & 0 &1& * & *&* \\ 0 & 0 &0& 0& 0 &0\\ 0 & 0 &0 & 0 & 0 &0\end{array}\right] \)

Hence R[4, ∶] = [0 0 0 0 0 0]

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 = \(\begin{bmatrix}0&0&0\\0&0&0\\0&0&λ\end{bmatrix}\) 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

Concept:

The row space of two matrices are same if the row echelon form of both matrices are same.

Explanation:

\(\left(\begin{array}{lll} 4 & 8 & 4 \\ 3 & 6 & 1 \\ 2 & 4 & 0 \end{array}\right)\)

 ∼\(\left(\begin{array}{lll} 2 & 4 & 0 \\ 3 & 6 & 1 \\ 4 & 8 & 4 \end{array}\right)\) (\(R_1\leftrightarrow R_3\))

 ∼\(\left(\begin{array}{lll} 1 & 2 & 0 \\ 3 & 6 & 1 \\ 4 & 8 & 4 \end{array}\right)\) (\(R_1\rightarrow \frac12R_1\))

∼\(\left(\begin{array}{lll} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 4 \end{array}\right)\) (\(R_2\rightarrow R_2-3R_1\), \(R_3\rightarrow R_3-4R_1\))

∼\(\left(\begin{array}{lll} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right)\) (\(R_3\rightarrow R_3-4R_2\),

So \(\left(\begin{array}{lll} 1 & 2 & 0 \\ 0 & 0 & 1 \end{array}\right)\) has the same row space as the given matrix.

(1) is correct

Explanation:

we know that if A = \(\begin{bmatrix}a&1\\0&a\end{bmatrix}\) then Aⁿ = \(\begin{bmatrix}a^n&na^{n-1}\\0&a^n\end{bmatrix}\)

Given \(\rm A=\left[\begin{array}{ll}2 & 1 \\ 0 & 2\end{array}\right] \)

using above result we have

A10 = \(\left[\begin{array}{ll}2^{10} & 10\cdot2^9 \\ 0 & 2^{10}\end{array}\right] \) = \(\left[\begin{array}{cc}4^5 & 20 \cdot 4^4 \\ 0 & 4^5\end{array}\right]\)

(3) correct