Vector Spaces MCQ Quiz in తెలుగు - Objective Question with Answer for Vector Spaces - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept: 
To determine if each set is a subspace of  \(\mathbb{R}^3 \), it must satisfy the following conditions:
1. Contain the zero vector: The zero vector (0, 0, 0) must be in W .
2. Closed under addition: If \( \mathbf{u}, \mathbf{v} \in W \)  , then  \(\mathbf{u} + \mathbf{v} \in W \) .
3. Closed under scalar multiplication: If  \(\mathbf{u} \in W \) and c is a scalar, then \(c \mathbf{u} \in W \) .

Let's examine each option:

Option 1: W =  \(\{ (x, y, z) \in \mathbb{R}^3 : x + 4y - 10z = -2 \} \)
This is not a subspace because it does not contain the zero vector.

For any vector to satisfy x + 4y - 10z = -2 , it cannot equal zero (the equation is not homogeneous).

Hence, it fails the first condition.

Option 2:   W = \(\{ (x, y, z) \in : x y = 0 \} \)
This set includes vectors where either x = 0 or y = 0 .

However, it is not closed under addition.

For example, (1, 0, 0) \( \in \) W and (0, 1, 0)  \(\in \) W , but (1, 1, 0)  \(\notin \) W since xy  \(\neq \) 0 .

Thus, this set is not a subspace.

Option 3: W =\( \{ (x, y, z) \in \mathbb{R}^3 : 2x + 3y - 4z = 0 \} \) 

This is a subspace because it satisfies all three conditions:
It contains the zero vector (0, 0, 0) since 2(0) + 3(0) - 4(0) = 0 .
It is closed under addition: if  \(2x_1 + 3y_1 - 4z_1 = 0 \)  and  \(2x_2 + 3y_2 - 4z_2 = 0 \), then

\(2(x_1 + x_2) + 3(y_1 + y_2) - 4(z_1 + z_2) = 0 \) .

It is closed under scalar multiplication: if 2x + 3y - 4z = 0 ,

then 2(cx) + 3(cy) - 4(cz) = 0 for any scalar c .

Therefore, W defined by 2x + 3y - 4z = 0 is a subspace.

Option 4: W =  \(\{ (x, y, z) \in \mathbb{R}^3 : x \in \mathbb{Q} \} \)
This set includes points where the x -coordinate is rational.

However, it is not closed under scalar multiplication.

For example, if (x, y, z) \( \in \) W with x = 1 (a rational number), multiplying this vector by an irrational scalar would result in an irrational x -component,

which would no longer belong to W .

Thus, this is not a subspace.

The correct answer is Option 3.

W = \( \{ (x, y, z) \in \mathbb{R}^3 : 2x + 3y - 4z = 0 \} \)  is a subspace of R3.

Concept: 

Dimension of a vector space:

The dimension of a vector space is the number of vectors in a basis for the space.

A vector space is finite-dimensional if it is spanned by a finite set,

and infinite-dimensional if it is not.

Explanation:

1. Understanding the Condition of Invertibility:

The space \(M_7(\mathbb{C})\) consists of all \(7 \times 7\) matrices over \(\mathbb{C}\).

For a matrix to be invertible, its determinant must be non-zero.

Therefore, if every nonzero matrix in V is invertible,

then V cannot contain any singular (non-invertible) matrices.

2. Implications of Invertibility on Subspace V :

The only way for a subspace V to have the property that

every nonzero matrix in V is invertible

if V consists of scalar multiples of a single invertible matrix.

This is because:

If V contained more than one linearly independent matrix,

it would also contain linear combinations of these matrices.

Linear combinations of invertible matrices may result in a singular matrix (non-invertible),

depending on the coefficients.

Therefore, to ensure that all nonzero elements in V are invertible,

V must be spanned by a single invertible matrix.

3. Determining the Dimension of V :

Since V is spanned by a single invertible matrix,

it is a one-dimensional subspace of \(M_7(\mathbb{C})\) .

Therefore, the dimension of V over \(\mathbb{C}\)  is 1 .

Hence Option(1) is the correct.

Concept: 

Two Step test for subspace 

Let V be a vector space over a field ℝ 

Let U be a non-empty subset of V such that 

(i) zero vector belongs to U 

(ii) ∀ u ∈ U, λ ∈ ℝ then, λu ∈ U 

(iii) ∀ u, v ∈ U, u + v ∈ U 

Explanation-  

Option (1): Let U = {(x, y, z) ∈ ℝ3 : (y + z)2 + (2x − 3y)2 = 0}

(i) (0, 0, 0) ∈ ℝ3, \((0+0)^2+(0-0)^2 = 0 \)

So, zero vector belongs to vector space

(ii) Let a ∈ ℝ, a.(x, y, z) = (ax, ay, az) ∈ ℝ3 

\((ay+az)^2+(2ax-3ay)^2\) = a²((y + z)2 + (2x − 3y)2) = a2.0 = 0

So a.(x, y, z) ∈ U

(iii) Let u = \((x_1,y_1,z_1) \) and v \(=(x_2,y_2,z_2)\) 

u + v = \((x_1+x_2, y_1+y_2, z_1+z_2)\) 

Now, Check u + v belongs to vector space 

\((y_1+y_2+z_1+z_2)^2 + (2(x_1+x_2)- 3(y_1+y_2))^2= 0 \)

\((y_1+y_2)^2+(z_1+z_2)^2 +2(y_1+y_2)(z_1+z_2)\) + \(4(x_1+x_2)^2 + 9 (y_1+y_2)^2 - 12 (x_1+x_2)(y_1+y_2) \) = 0

\((y_1+z_1)^2 + (2x_1-3y_1)^2 + (y_2+z_2)^2 +(2x_2-3y_2)^2 = 0\)

Therefore, \(u+v ∈ U\)

Hence it is a vector space. 

(1) is correct

Option (2): Let U = {(x, y, z) ∈ ℝ3 : y ∈ ℚ}

Scalar multiplication does not hold 

\(\sqrt2(x,y,z)\) ∈ ℝ3 but \(\sqrt 2y\notin \mathbb Q\)

Hence it is not a vector space 

(2) is false

Option (3): U = {(x, y, z) ∈ ℝ3 : yz = 0} 

Let u = (0,1,0) then y.z = 0 

v = ( 0,0,1) then y.z = 0 

So u, v ∈ U

but u + v = ( 0,1,1) \(\notin\) U as y.z = 1.1 = 1 \(\neq 0 \) 

Hence, it is not a vector space. 

(3) is false 

Option (4): U = {(x, y, z) ∈ ℝ3 : x + 2y − 3z + 1 = 0}

 0 + 0 + 0 + 1 \(\neq 0\) 

So zero vector does not belongs to U

Hence, It is not a vector space 

(4) is false