Linear Dependence, Basis & Dimension MCQ Quiz in తెలుగు - Objective Question with Answer for Linear Dependence, Basis & Dimension - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

rank-nullity theorem

\(\dim(V) = \dim(\text{ker}(T)) + \dim(\text{Im}(T))\)

Explanation:

Option 1: By rank-nullity theorem,

Since \(\ker(T) = \text{Im}(T)\), let  . Then

\(\dim(V) = k + k = 2k\)

Hence, the dimension of \(V\) must be even. So, Option 1 is true.

Option 2: The trace of an operator is the sum of its eigenvalues (counting multiplicities).

When \(\ker(T) = \text{Im}(T) \), \(T\) behaves somewhat like a nilpotent matrix (though this is not explicitly said).

If \(T\) has a minimal polynomial of the form \(x^2\), it indicates that the trace (sum of eigenvalues) could be zero.

Thus, based on this reasoning, Option 2 is true.

Option 3: The minimal polynomial of a linear operator describes the smallest polynomial such

that \(T\) satisfies it. If  this suggests that \(\ker(T) = \text{Im}(T) \),  \(T\) 

behaves similarly to a nilpotent operator, whose minimal polynomial is typically \(x^2\) or some

higher power of \(​​x\) and it cannot have distinct roots.

Thus, Option 3 is true.

Option 4: The characteristic polynomial of an operator generally has the same degree as the

dimension of the vector space, while the minimal polynomial is a divisor of the characteristic

polynomial and could have lower degree.

For example, if the characteristic polynomial is \((x - 0)^n \), the minimal polynomial could still

be \(x^2\) in some cases, where \( n > 2\) .

Thus, Option 4 is false,

Consider a linear operator \(T\) on a 4-dimensional vector space such that its minimal polynomial is \(x^2\), but its characteristic

polynomial is \( x^4\) . This shows that the minimal polynomial is not equal to the characteristic polynomial.

Hence, correct options are 1), 2) and 3).

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 \(f(x) \) 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 \(f(x)=1 \Rightarrow \int_{-h}^{h} 1 dx=ah+bh+ah+0+0\)

\(\Rightarrow2h=2ah+2bh+2ah \Rightarrow2=2a+b ....(i)\)

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

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

Take \(f(x)=x^2\Rightarrow \int_{-h}^{h} x^2dx=ah^3+ah^3-2ch^3-2ch^3\)

\(\Rightarrow\frac{2h^3}{3}=2ah^3-4ch^3 \)

\(\Rightarrow\frac{1}{3}=a-2c.....(ii)\)

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

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

\(\mathrm{W}=\left\{\left(a_{i j}\right) \in M_{4}(\mathbb{R}) \mid\sum_{i+j=k} a_{i j}=0, k = 2, 3, 4, 5, 6, 7, 8 \right\}\)

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 = \(\begin{vmatrix}f_1&f_2&f_2\\f_1'&f_2'&f_3'\\f_1''&f_2''&f_2''\end{vmatrix}\) = \(\begin{vmatrix}\sin x&\cos x&\tan x\\\cos x&-\sin x&\sec^2x\\-\sin x&-\cos x&2\sec^2x\tan x\end{vmatrix}\)

At x = π/4

Wornskian = \(\begin{vmatrix}\frac1{\sqrt2}&\frac1{\sqrt2}&1\\\frac1{\sqrt2}&-\frac1{\sqrt2}&2\\-\frac1{\sqrt2}&-\frac1{\sqrt2}&4\end{vmatrix}\) 

                 = \(\begin{vmatrix}\frac1{\sqrt2}&\frac1{\sqrt2}&1\\0&\frac2{\sqrt2}&-1\\0&0&5\end{vmatrix}\) (\(R_2\rightarrow R_2-R_1\), \(R_3\rightarrow R_3+R_1\))

               = \(​​\frac1{\sqrt2}(​​\frac{10}{\sqrt2}-0)\) = 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) = \(\int_0^x f(t) d t+f^{\prime}(x)\)

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

T(1) = \(\int_0^x 1dt\) + 0 = x + 0 = x = 0 + 1x + 0x² + 0x³

T(x) = \(\int_0^x t dt\) + 1 =  \(\frac12\)x² + 1 = 1 + 0x + \(\frac12\)x2 + 0x3

T(x²) = \(\int_0^x t^2 dt\) + 2x =  \(\frac13\)x3 + 2x = 0 + 2x + 0x2 + \(\frac13\)x3

Therefore matrix representation of T is

\(\left(\begin{array}{ccc}0 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{3}\end{array}\right)\)

(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 \(A=\begin{bmatrix}1&0&0&0&0&0\\0&1&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\\0&0&0&0&0&0\end{bmatrix}\)

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.

 Explanation:

Orthogonal Matrix Condition:

Each nonzero matrix A \in W is orthogonal, meaning \(A^T A = I \)  , where \( A^T \) is the transpose of A .

Orthogonal matrices preserve lengths and angles 

⇒ each matrix in W represents a transformation with special properties like rotations and reflections.

Symmetric Matrix Condition:

Each matrix in W is also symmetric, so \(A = A^T \)  for any \(A \in W \) .

Symmetric orthogonal matrices in \(M_8(\mathbb{R}) \)  must have eigenvalues that are either +1 or -1

because symmetry ensures that eigenvalues are real, and orthogonality restricts them to lie on the unit circle.

Eigenvalue Structure of Symmetric Orthogonal Matrices:

Since each symmetric orthogonal matrix in W has eigenvalues of \(\pm 1 \)  ,

each matrix essentially performs a reflection (flipping signs of some components) or the identity transformation.

In \( M_8(\mathbb{R}) \) , the dimension of the space of symmetric matrices is \(\frac{8 \times 9}{2} = 36 \)  .

However, due to the orthogonality constraint, only certain reflections (diagonal matrices with \pm 1 entries) are allowed.

Commutativity in W :

If every matrix in W is symmetric and orthogonal, the matrices in W commute.

This restricts the variety of matrices allowed in W

since commuting symmetric matrices are generally diagonalizable in the same basis.

Thus, any subspace W with commuting symmetric orthogonal matrices can only contain scalar multiples of a very limited set of matrices.

Dimension Analysis of W :

For each matrix in W , we can independently select +1 or -1 along each diagonal entry

(since the matrices must be symmetric and orthogonal).

This constraint limits W to a maximum of two independent dimensions:

one representing the identity matrix (with all +1 entries) and

another representing a matrix with some of its diagonal entries as -1 ,

which effectively spans a 2-dimensional space.

Given the constraints of orthogonality, symmetry, and commutativity, the maximal possible dimension for W is 2.

Thus, the answer is  2 .

Hence Option(3) is the correct answer.