Linear Dependence, Basis & Dimension MCQ Quiz in தமிழ் - Objective Question with Answer for Linear Dependence, Basis & Dimension - இலவச PDF ஐப் பதிவிறக்கவும்

Explanation:

1. : The vector space of polynomials in the variable x with real coefficients,

of degree at most 11, including the zero polynomial.

2. and  

These are subsets of   , each containing 12 elements (polynomials).

3. Conditions:
For set E : .

For set F : 

Analysis of Sets E and F :

1. Set E :

Each polynomial   in E satisfies  .

This condition means that each polynomial in E has a root at x = 3 .

Since there are 12 polynomials in E ,

they cannot all be linearly independent in   ,

which has dimension 12.

A set of 12 polynomials in this space could potentially span the space

if they were linearly independent.

However, the root condition for all i imposes a dependency

because each polynomial in E must include (x - 3) as a factor.

This requirement limits the linear independence of E , making it linearly dependent.

2. Set F :

Each polynomial  in F satisfies .

This condition does not impose a common factor as it did with set E ;

instead, it specifies the value at x = 4 .

The condition does not constrain the degree or form of the polynomials in a way that would force a dependency.

Therefore, it is possible for F to be linearly independent,

as the polynomials can be chosen such that they span the space without a common factor or dependency among them.

Conclusion: 

Any such E is necessarily linearly dependent due to the constraint  ,

which imposes a root at x = 3 for each polynomial in E .

Any such F is not necessarily linearly dependent because the condition  does not force linear dependence.

Explanation:
1. Set E : The root condition  means each polynomial in E has (x - 3) as a factor,

which creates dependency among them.

2. Set F : The condition  only specifies the value of each polynomial at x = 4 but does not impose dependency.

Thus, F can be chosen to be linearly independent.

Hence, Any such E is necessarily linearly dependent but any such F is not necessarily linearly dependent.

Option(2) is the right answer.

Concept-

Basis of polynomial of degree less or equal to 2 is {1, x, x²}.

Explanation-

Take basis of polynomial of degree less or equal to 2 as  and calculate the required values. We choose basis of polynomial of degree less or equal to 2 because we need to calculate three unknown constants.

Take

Values of a and b in option (3) and (4) are not satisfy equation

So, option (3) and (4) are false.

Take

Values of a and c in option (2) are not satisfying equation

So, option (2) is false, and option (1) is true.

Explanation:

M4(ℝ) be the space of all (4 × 4) matrices over ℝ

So dimension of M4(ℝ) is 4 × 4 = 16

Number of conditions of W = 7

Hence dimension of W = 16 - 7 = 9

(3) is correct

Concept:

(i) The dimension of a subspace is the number of linearly independent vectors.

(ii) If Wornskian W(x) ≠ 0 at a point then W(x) ≠ 0 for all x

Explanation:

W be the subspace of V spanned by {sin(x), cos(x), tan(x)}

Wornskian =  = 

At x = π/4

Wornskian =  

                 =  (, )

               =  = 5 ≠ 0

So {sin(x), cos(x), tan(x)} is Linearly independent

Hence  the dimension of W over ℝ is 3

(3) is correct

Explanation:

T : P2 →  P3 by (Tf) (x) = 

Basis of  P2 is {1, x, x2} and  P3 is {1, x, x2, x3} 

T(1) =  + 0 = x + 0 = x = 0 + 1x + 0x² + 0x³

T(x) =  + 1 =  x² + 1 = 1 + 0x + x2 + 0x3

T(x²) =  + 2x =  x3 + 2x = 0 + 2x + 0x2 + x3

Therefore matrix representation of T is

(2) is correct

Concept:

(i) The null space of a matrix A, is the set of all solutions to the homogeneous equation Ax = 0

(ii) The column space of a matrix A is the span (set of all possible linear combinations) of its column vectors.

Explanation:

null space N(A) of A is

{(x, y, z, w) ∈ ℝ4 : x + y + z = 0, x + y + w = 0}

Number of independent constraints  = 2

So dim(null space (A)) = 2

Given A is 4 × 4 matrix

so dim(column space(A)) = 4 - 2 = 2

(2) is correct

Concept:

Let A and B are two subspace of V. Then

(i) dim(A ∩ B) ≤ dim(A)

(ii) dim(A ∩ B) ≤ dim(B)

(iii) dim(A + B) = dim(A) + dim(B) - dim(A ∩ B)

Explanation:

dim(V) = 100, dim(A) = 60. dim(B) = 63

dim(A ∩ B) ≤ dim(A) ⇒ dim(A ∩ B) ≤ 60

dim(A ∩ B) ≤ dim(B) ⇒ dim(A ∩ B) ≤ 60

Combinition both 

dim(A ∩ B) ≤ 60

Option (2) is correct

Also

dim(A + B) = dim(A) + dim(B) - dim(A ∩ B)

⇒ dim(A ∩ B) = dim(A) + dim(B) - dim(A + B)

⇒ dim(A ∩ B) = 60 + 64 -100

⇒ dim(A ∩ B) = 23

Option (3) is correct

Concept:

(i) Basis: A basis for a vector space is a sequence of vectors that form a set that is linearly independent and that spans the space. 

(ii) Dimension: The dimension of a vector space V is the cardinality (i.e., the number of vectors) of a basis of V over its base field

Explanation:

Here, it is given that vector space V over the field of real numbers spanned by the set

S = {(0,1,0,0), (1,1,0,0), (1,0,1,0), (0,0,1,0), (1,1,1,0), (1,0,0,0)}

⇒ (1,1,1,)) = (0,1,0,0) + (1,0,0,0) + (0,0,1,0)

So, it can be omitted.

(1,1,0,0) = 1(1,0,0,0) + 1(0,1,0,0) + 0(0,0,1,0)

It can also be omitted.

Similarly, (1,0,1,0) will be omitted.

Since, {(1,0,0,0), (0,1,0,0), (0,0,1,0)} are linearly independent and span whole set V.

It is a of given set

Hence, Dimension = 3

Option (3) is correct

Concept:

The rank of matrix A is the order of the largest subsquare matrix that is invertible.

Explanation:

A be an n × n matrix such that the first 3 rows of A are linearly independent and the first 5 columns of A are linearly independent.

So rank A ≥ 5 so A has at least 5 linearly independent rows

Option (1) and option (3) are correct. 

If we take A = I_n then A be an n × n matrix such that the first 3 rows of A are linearly independent and the first 5 columns of A are linearly independent. But rank A = n.

So option (2) is incorrect.

Let 

A be an 6 × 6 matrix such that the first 3 rows of A are linearly independent and the first 5 columns of A are linearly independent.

Rank A = 5 but Rank A² ≤ 4

Option (4) is incorrect.

∴ Option (1) and option (3) are correct.

Concept:

• Vector Space V: The set of all real 5 × 5 matrices, denoted ℝ^5×5, is a 25-dimensional real vector space.
• Commutator of Matrices: For matrices A, B ∈ V, the commutator is defined as [A, B] = AB − BA.
• Set S and Subspace W: Let S = {AB − BA | A, B ∈ V} and W = span(S). Then W is the subspace spanned by all commutators in V.
• Linear Transformation T: T: V → ℝ is defined by T(A) = trace(A), which maps a matrix to the sum of its diagonal elements.
• Kernel of T: ker(T) = {A ∈ V | trace(A) = 0} is the set of all matrices in V having trace zero.
• Key Result in Linear Algebra: The space of commutators AB − BA for square matrices of size n × n spans the space of trace-zero matrices, i.e., every trace-zero matrix is a linear combination of commutators.

Calculation:

Let A ∈ ker(T) ⇒ trace(A) = 0

Every matrix with trace zero can be expressed as a linear combination of commutators: A = ∑ c_i(A_iB_i − B_iA_i)

⇒ A ∈ span(S) = W

⇒ ker(T) ⊆ W

Also, ∀ A, B ∈ V:

⇒ trace(AB − BA) = trace(AB) − trace(BA) = 0

⇒ AB − BA ∈ ker(T)

⇒ W ⊆ ker(T)

So, W ⊆ ker(T) and ker(T) ⊆ W

⇒ W = ker(T)

∴ Correct answer is Option 1: W = ker(T)