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

Concept:

• Vector Space V : Set of real-valued continuous functions defined on the interval .
• Inner Product: For functions , the inner product is defined as .
• Set S : This is a set of functions in .
• Subspace W : Defined as the span of set S , i.e., .
• Basis: A set is a basis of a subspace if it is linearly independent and spans the space.
• Orthogonality: Two functions are orthogonal if their inner product is zero: .
• Orthonormal Set: A set where all vectors are orthogonal and of unit norm.

Calculation:

Given:

Step 1: Is S a basis of W ?

⇒ By definition, W = {span}(S)

⇒ S spans W trivially

⇒ Now check linear independence of S

⇒ 

⇒ One function is a linear combination of others

⇒ So, S is linearly dependent

⇒ But basis of a space can be the set itself if we define the space as its span

⇒ Hence, S is a basis of W (though not minimal)

Statement 1 is TRUE

Step 2: Is S an orthonormal basis?

⇒ Check  

⇒ But   , not orthogonal 

⇒ Also norms are not 1 

⇒ So, not orthogonal, not normalized

⇒ Statement 2 is FALSE

Step 3: Do there exist orthogonal functions in S ?

⇒   

⇒ Statement 3 is TRUE

Step 4: Does S contain an orthonormal basis of W ?

⇒ Since S is linearly dependent, and orthonormal sets must be linearly independent

⇒ No subset of S is orthonormal and spans W

⇒ Statement 4 is FALSE

∴ Final Answer: Only Statements 1 and 3 are TRUE.

Concept:

A Vector space is called subspace if it satisfies the following conditions

1. Zero Vector: The subset must contain the zero vector.

2. Closure under Addition: The sum of any two vectors in the subset must also be in the subset.

3. Closure under Scalar Multiplication: The product of any scalar and a vector in the subset must also be in the subset.

Explanation:

Option (1):

The only solution to (3y + 2z)² + (2x - 3y)² = 0 is (0, 0, 0).

This set contains only the zero vector and satisfies all three conditions for a subspace.

Option (2):

This set does not contain all rational numbers for the y-coordinate.

For example, (1, √2, 0) is not in the set, even though it is a vector in ℝ³.

Therefore, it fails the closure under addition and scalar multiplication conditions.

Option (3):

This set contains the zero vector (0, 0, 0).

However, it fails the closure under addition condition.

For example, (1, 0, 0) and (0, 1, 0) are in the set, but their sum (1, 1, 0) is not.

Option (4):

This set does not contain the zero vector (0, 0, 0) because it does not satisfy the equation 5x + 7y - 3z + 1 = 0.

Therefore, it fails the first condition.

Therefore, the only correct option is (1): {(x, y, z) ∈ ℝ³ : (3y + 2z)² + (2x - 3y)² = 0}.

Hence Option(1) is the correct answer.

Explanation:

S and T are subspaces of a vector space V(F).

From the results we know that

(i) S ⊂ L(T) ⇔ L(S) ⊂ L(T)

(ii) L (S ∪ T) = L(S) + L(T)

(iii) L[L(T)] = L[T]

Hence option (1) is not correct.

Concept: 
To determine if each set is a subspace of  , 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   , then   .
3. Closed under scalar multiplication: If   and c is a scalar, then  .

Let's examine each option:

Option 1: W =  
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 =
This set includes vectors where either x = 0 or y = 0 .

However, it is not closed under addition.

For example, (1, 0, 0)  W and (0, 1, 0)   W , but (1, 1, 0)   W since xy   0 .

Thus, this set is not a subspace.

Option 3: W = 

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    and  , then

 .

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 =  
This set includes points where the x -coordinate is rational.

However, it is not closed under scalar multiplication.

For example, if (x, y, z)  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 =   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 consists of all matrices over .

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 .

Therefore, the dimension of V over   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, 

So, zero vector belongs to vector space

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

 = a²((y + z)2 + (2x − 3y)2) = a2.0 = 0

So a.(x, y, z) ∈ U

(iii) Let u =  and v  

u + v =  

Now, Check u + v belongs to vector space 

 +  = 0

Therefore, 

Hence it is a vector space. 

(1) is correct

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

Scalar multiplication does not hold 

 ∈ ℝ3 but 

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)  U as y.z = 1.1 = 1  

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  

So zero vector does not belongs to U

Hence, It is not a vector space 

(4) is false