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 $a_0,a_1,a_2,a_3\in \mathbb{R}$

$f_B(x)=b_0+b_{1}x+b_2{x}^2+b_3{x}^3+b_4{x}^4$ where $b_0,b_1,b_2,b_3,b_4\in \mathbb{R}$

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

$3=a_0+0+0+0⟹a_0=3$

Since V be the real vector space of all continuous functions

$f_A(2)=f_B(2)$  

⇒ $a_0+a_{1}2+a_2{2}^2+a_3{2}^3=b_0+b_{1}2+b_2{2}^2+b_3{2}^3+b_4{2}^4$

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

Now we can define 

$f(x)=\{\begin{array}{ll}3+a_{1}x+a_2{x}^2+a_3{x}^3,& \text{when }x\in A\\ b_0+b_{1}x+b_2{x}^2+b_3{x}^3+b_4{x}^4,& \text{when }x\in B\end{array}$

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

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

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

for all $v\in V$ .

Explanation:

T, S : ${\mathbb{R}}^4\to {\mathbb{R}}^4$ 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 $\mathrm{A}=\left(\begin{array}{ll}0& 2\\ 2& 0\end{array}\right)$ 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 =$\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$
 

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

The action of A  on B is

T(B) = AB = $\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right)$ = $\left(\begin{array}{cc}2b_{11}& 2b_{12}\\ b_{21}& b_{22}\end{array}\right)$

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.,

$\mathbf{b}=\left(\begin{array}{c}{b}_{11}\\ b_{21}\\ b_{12}\\ b_{22}\end{array}\right)$

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

$T(\mathbf{b})=\left(\begin{array}{c}2b_{11}\\ b_{21}\\ 2b_{12}\\ b_{22}\end{array}\right)$

This can be written as the matrix multiplication

$T(\mathbf{b})=\left(\begin{array}{cccc}2& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{c}{b}_{11}\\ b_{21}\\ b_{12}\\ b_{22}\end{array}\right)$

Thus, the matrix representation of  T is

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

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:

$T-\lambda I=\left(\begin{array}{cccc}2-\lambda & 0& 0& 0\\ 0& 1-\lambda & 0& 0\\ 0& 0& 2-\lambda & 0\\ 0& 0& 0& 1-\lambda \end{array}\right)$

Now, we compute the determinant

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

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:

⇒ $T(x,y)=A\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}2x+3y\\ 4x-y\end{array}\right]$

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

⇒ $T(1,3)=\left[\begin{array}{c}2(1)+3(3)\\ 4(1)-3\end{array}\right]=\left[\begin{array}{c}2+9\\ 4-3\end{array}\right]=\left[\begin{array}{c}11\\ 1\end{array}\right]$

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

⇒ $\left[\begin{array}{c}11\\ 1\end{array}\right]=a\left[\begin{array}{c}1\\ 3\end{array}\right]+b\left[\begin{array}{c}2\\ 5\end{array}\right]$

​​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}

⇒ $\left[\begin{array}{c}-53\\ 32\end{array}\right]$

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

⇒ $T(2,5)=\left[\begin{array}{c}2(2)+3(5)\\ 4(2)-5\end{array}\right]=\left[\begin{array}{c}4+15\\ 8-5\end{array}\right]=\left[\begin{array}{c}19\\ 3\end{array}\right]$

We express  (19,3)  as: 

⇒ $\left[\begin{array}{c}19\\ 3\end{array}\right]=a\left[\begin{array}{c}1\\ 3\end{array}\right]+b\left[\begin{array}{c}2\\ 5\end{array}\right]$

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

⇒ $\left[\begin{array}{c}-89\\ 54\end{array}\right]$

The transformation matrix in the basis S is:

⇒ $B=[T{]}_S=\left[\begin{array}{cc}-53& -89\\ 32& 54\end{array}\right]$

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 = $\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& \lambda \end{array}\right]$ 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 =$\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$
 

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

The action of A  on B is

T(B) = AB = $\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right)$ = $\left(\begin{array}{cc}2b_{11}& 2b_{12}\\ b_{21}& b_{22}\end{array}\right)$

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.,

$\mathbf{b}=\left(\begin{array}{c}{b}_{11}\\ b_{21}\\ b_{12}\\ b_{22}\end{array}\right)$

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

$T(\mathbf{b})=\left(\begin{array}{c}2b_{11}\\ b_{21}\\ 2b_{12}\\ b_{22}\end{array}\right)$

This can be written as the matrix multiplication

$T(\mathbf{b})=\left(\begin{array}{cccc}2& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{c}{b}_{11}\\ b_{21}\\ b_{12}\\ b_{22}\end{array}\right)$

Thus, the matrix representation of  T is

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

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:

$T-\lambda I=\left(\begin{array}{cccc}2-\lambda & 0& 0& 0\\ 0& 1-\lambda & 0& 0\\ 0& 0& 2-\lambda & 0\\ 0& 0& 0& 1-\lambda \end{array}\right)$

Now, we compute the determinant

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

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 ${\lambda }_1,{\lambda }_2,{\lambda }_3,...,{\lambda }_n$ 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) = $\sum _{i=1}^n{\lambda }_i$

4. tr(A^k) = $\sum _{i=1}^n{\lambda }_i^k$, where ${\lambda }_i$ 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 ${\lambda }_1,{\lambda }_2,{\lambda }_3,...,{\lambda }_n$ be the eigenvalues of M. Then:

1. det(A) = ${\lambda }_1{\lambda }_2{\lambda }_3\cdot ...\cdot {\lambda }_n$ i.e., the determinant of A is equal to the product of the eigenvalues 

2. tr(A) = ${\lambda }_1+{\lambda }_2+{\lambda }_3+...+{\lambda }_n$ 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:

$M=\left[\begin{array}{ccc}0& -1& 0\\ 1& 2& -1\\ 1& 1& 3\end{array}\right]$

⇒ |M| = 0 + 1(3 + 1) + 0

⇒ |M| = 4

Let ${\lambda }_1,{\lambda }_2,{\lambda }_3$ be the eigenvalues.

∴ ${\lambda }_1{\lambda }_2{\lambda }_3$ = 4...(i)

and, ${\lambda }_1+{\lambda }_2+{\lambda }_3$ = 5...(ii)

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

Let ${\lambda }_1$ = 1

∴ (i) ⇒ ${\lambda }_2{\lambda }_3$ = 4...(iii)

and, (ii) ⇒ $1+{\lambda }_2+{\lambda }_3$ = 5

⇒ ${\lambda }_2+{\lambda }_3$ = 4...(iv)

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

${\lambda }_2$ = 2 and ${\lambda }_2$ = 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 =  $\left[\begin{array}{ccc}0& -1& 0\\ 1& 2& -1\\ 1& 1& 3\end{array}\right]-2\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

M - 2I  = $\left[\begin{array}{ccc}-2& -1& 0\\ 1& 0& -1\\ 1& 1& 1\end{array}\right]$

Applying row operations, R₁ → R₁ + 2R₂

and R₃ → R₃ - R₂

= $\left[\begin{array}{ccc}0& -1& -2\\ 1& -2& -1\\ 0& 1& 2\end{array}\right]$

R₃ → R₃ + R₁

= $\left[\begin{array}{ccc}0& -1& -2\\ 1& 0& -1\\ 0& 0& 0\end{array}\right]$

∴ 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

$\left[\begin{array}{ccccc}1& 0& 0& \ast & \ast \\ 0& 1& 0& \ast & \ast \\ 0& 0& 1& \ast & \ast \\ 0& 0& 0& 0& 0& \\ 0& 0& 0& 0& 0\end{array}\right]$

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

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

∴ $\left[\begin{array}{cccccc}1& 0& 0& \ast & \ast & \ast \\ 0& 1& 0& \ast & \ast & \ast \\ 0& 0& 1& \ast & \ast & \ast \\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right]$

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 = $\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& \lambda \end{array}\right]$ 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:

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

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

 ∼$\left(\begin{array}{lll}1& 2& 0\\ 3& 6& 1\\ 4& 8& 4\end{array}\right)$ ($R_1\to \frac{1}{2}{R}_1$)

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

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

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 = $\left[\begin{array}{cc}a& 1\\ 0& a\end{array}\right]$ then Aⁿ = $\left[\begin{array}{cc}{a}^n& n{a}^{n-1}\\ 0& a^n\end{array}\right]$

Given $\mathrm{A}=\left[\begin{array}{ll}2& 1\\ 0& 2\end{array}\right]$

using above result we have

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

(3) correct