Questions: Linear Algebra and Matrices - Mathematical Methods of Physics, UGC - NET Physics | Physics for IIT JAM, UGC - NET, CSIR NET PDF Download

Question 1. Let M_n be the vector space of all n × n matrices with real entries under usual matrix addition and scalar multiplication. Find the dimensions of the fol lowing subspaces of M_n:

1. W₁ = {A ∈ M_n : A is symmetric}
2. W₂ = {A ∈ M_n : A is skew-symmetric}
3. W₃ = {A ∈ M_n : A is upper triangular}
4. W₄ = {A ∈ M_n : A is tridiagonal}
Sol:
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1. Subspace (W₁ = A ∈ M_n : A is symmetric)
A matrix A = [a_ij]∈ M_n  is symmetric if A = A^T , meaning  a_ij  = a_ji for all i, j. 
In a symmetric matrix: 

• The diagonal entries a_ii can take any value (n independent entries). 
• For i < j, the upper triangular entries a_ij determine the lower triangular entries a_ji , leading to independent entries.

Total number of independent entries:

2. Subspace (W₂ = A ∈ M_n :  A is skew-symmetric)
A matrix A= [a_ij]∈ M_n is skew-symmetric if A^T= −A, meaning a_ij  = −a_ji  for all i, j, and a_ii  = 0 (diagonal entries are zero). 

• In a skew-symmetric matrix: 
  The diagonal entries a_ii = 0 (n constraints). 
  For i < j, the upper triangular entries a_ij determine the lower triangular entries a_ji  = −a_ij, giving  independent entries.

Total number of independent entries:

3. Subspace (W₃ = A ∈ M_n :  A is uper triangular)
A matrix A = [a_ij] ∈ M_n is upper triangular if a_ij = 0 for all i > j. 
In an upper triangular matrix: 

• All entries a_ij for i ≤ j are independent (1 + 2+ ⋯ + n  entries).
• The remaining entries (a_ij for i > j) are zero and do not contribute to the dimension. 

Total number of independent entries:

4. Subspace (W₄ = A ∈ M_n :  A is tridiagonal)
A matrix A = [a_ij] ∈ M_n  is tridiagonal if a_ij = 0 for all ∣i − j∣ > 1. 
In a tridiagonal matrix: 
There are n diagonal entries (a_ii). 

• There are n − 1 upper diagonal entries (a_{i, i + 1}). 
• There are n − 1 lower diagonal entries (a_{i + 1,i}). 
• Total number of independent entries: 

Dimension of W₄ = n + (n − 1) + (n − 1) = 3n − 2.
Final Dimensions

Question 2. Let V be the subspace of M₂(R) consisting of matrices such that the entries of the first row add up to zero. Write down a basis for V.
Sol: 
Subspace V Definition
Let  The condition for A ∈ V is:
a + b = 0(the sum of the entries in the first row is zero). 
This implies b = −a. Thus, matrices in V can be written as:
, where .
Dimension of  V 
The general form of A shows that it depends on three parameters:  a, c, and  d. Therefore, dim ⁡ (V) = 3. 
Basis for V 
To find a basis, we express A as a linear combination of matrices where each parameter (a, c ,d) is varied independently. 
Write:
.

Thus, the following matrices form a basis for V:

Verification 

1. The three matrices are linearly independent because no linear combination of them produces the zero matrix unless all coefficients are zero. 
2. They span V, as any matrix in V can be written as a linear combination of these matrices.

Question 3. Write down a necessary and sufficient condition, in terms of a, b, c and d (which are assumed to be real numbers), for the matrix  not to have a real eigenvalue.
Sol:
To determine a necessary and sufficient condition for the matrix

not to have a real eigenvalue, we analyze the eigenvalues of A using the characteristic equation.

1. Eigenvalues of A: 
The eigenvalues of A are the roots of its characteristic polynomial, given by: 
det(A − λI) = 0, 
where I is the identity matrix. Explicitly: 

This simplifies to: 

The determinant is: 

(a − λ)(d − λ) − bc = 0. 

Expanding this, we get the quadratic equation: 

λ² − (a + d)λ + (ad − bc) = 0. 

2. Condition for Real Eigenvalues: 
The roots of this quadratic equation are real if and only if the discriminant of the quadratic equation is non-negative. The discriminant is: 
Δ = (a + d)² − 4(ad − bc). 
Simplifying: 
Δ = (a + d)² − 4ad + 4bc = (a − d)² + 4bc. 

• For the eigenvalues not to be real, the discriminant must be negative: 

Δ < 0

3. Necessary and Sufficient Condition: 
The necessary and sufficient condition for A not to have real eigenvalues is: 
(a − d)²  + 4bc < 0. 

4. Interpretation: 
(i) (a − d)² ≥ 0 (since it is a square term). 
(ii) For Δ < 0, the term 4bc must be negative and sufficiently large in magnitude to make (a − d)² + 4bc < 0. 
This implies: 

• bc < 0 (the product of b and c must be negative). 

Thus, the matrix A will not have real eigenvalues if and only if:
bc < 0 and ∣4bc∣ > (a − d)².

Question 4. Given that the matrixhas 1 as an eigenvalue, compute its trace and its determinant.
Sol: 
We are given the matrix:

and the information that 1 is an eigenvalue of A. We need to compute the trace and determinant of A.

Step 1: Eigenvalue Condition 
The eigenvalues of A satisfy the characteristic equation: 
det(A − λI) = 0, 
where I is the 2 × 2 identity matrix. For A − λI, we have: 

The determinant of A − λI is: 

Using the determinant formula, we compute: 

det (A − λI) = (α − λ)(3 − λ) − (1)(2). 

Simplify: 

det (A − λI) = (α − λ)(3 − λ) − 2 = α(3 − λ) − λ(3 − λ) − 2. 

Expanding further: 

det (A − λI) = 3α − αλ − 3λ + λ² − 2. 

Rearranging: 

det(A − λI) = λ² −(α + 3)λ + (3α − 2). 

Since 1 is an eigenvalue, it satisfies the characteristic equation: 

1² −(α + 3)(1) + (3α − 2) = 0. 

Simplify: 

1 − α − 3 + 3α − 2 = 0. 

Solve for α: 

α = 2. 

Step 2: Compute the Trace 
The trace of a matrix is the sum of its diagonal elements: 
Trace (A) = α + 3. 
Substitute α = 2: 
Trace (A) = 2 + 3 = 5. 
Trace(A) = 2 + 3 = 5. 

Step 3: Compute the Determinant 
The determinant of A is: 
det (A) = α(3) − (1)(2). 
Substitute α = 2:
det (A) = 2(3) − 2 = 6 − 2 = 4. 

Final Answer: 

• Trace: 5 
• Determinant: 4

Question 5. Let A be a non diagonal 2 × 2 matrix with complex entries such that A = A⁻¹. Write down its characteristic and minimal polynomials.
Sol: 
We are tasked with finding the characteristic and minimal polynomials for a 2 × 2 non-diagonal matrix A with complex entries, satisfying the condition A = A⁻¹. 
Step 1: Properties of A = A⁻¹ 

1. The condition A = A⁻¹ implies: 
A² = I, 
where I is the 2 × 2 identity matrix. 
2. This means that A is involutory, i.e., squaring the matrix yields the identity matrix. 
3. The eigenvalues of A are the roots of its characteristic polynomial. Since A² = I, any eigenvalue λ of A satisfies: λ² = 1. 
Thus, the eigenvalues are λ = 1 and λ = −1. 

Step 2: Characteristic Polynomial 
The characteristic polynomial of a 2 × 2 matrix A is defined as: 
p(λ) = det(A − λI). 
From Step 1, the eigenvalues of A are 1 and −1. Thus, the characteristic polynomial is: 
p(λ) = (λ − 1)(λ + 1). 
Simplify: p(λ) = λ² − 1. 

Step 3: Minimal Polynomial 
The minimal polynomial of A is the monic polynomial of the smallest degree that annihilates A, i.e., m(A) = 0. 
From the property A² = I, it is clear that A satisfies
A²− I = 0. 
Thus, the minimal polynomial is: m(λ) = λ² −1. 
Step 4: Non-diagonal Condition 
For A to be non-diagonal, it must not commute with all matrices in . A simple example of such a matrix is: 

This matrix satisfies A² = I, is non-diagonal, and has eigenvalues 1and −1. 

Final Answer 

• Characteristic Polynomial: p(λ) = λ² −1. 
• Minimal Polynomial: m(λ) = λ² −1.

Question 6. For a fixed positive integer n ≥ 3, let A be the n × n matrix defined by A = I −, where J is the n × n matrix with all entries equal to 1. Which of the fol lowing statements is not true?

1. A^k = A for every positive integer k.
2. Trace(A) = n − 1
3. Rank(A) + Rank(I − A) = n
4. A is invertible.

Sol:
Let us analyze the properties of the given matrix , where J is the n × n matrix with all entries equal to 1.

Key Properties of J: 

1. J is an n × n matrix where all entries are 1. 
2. The rank of J is 1, because all rows (or columns) of J are identical. 
3. The eigenvalues of J are: 

• n, corresponding to the eigenvector [1, 1,…,1] ^T . 
• 0, with multiplicity n − 1, corresponding to eigenvectors orthogonal to [1, 1, …,1]^T. 

Eigenvalues of A: 
1. Since  the eigenvalues of A are obtained by shifting the eigenvalues of J:

• If λ is an eigenvalue of J, then   is an eigenvalue of A.

2. The eigenvalues of J are n (once) and 0 (multiplicity n − 1): 

• 
•  (multiplicity n − 1).

Thus, the eigenvalues of A are 0 (once) and 1 (multiplicity n − 1). 

Analyzing the Statements: 

1. A^k = A for every positive integer k: 

Since A has eigenvalues 0 and 1, raising A to any power k does not change its eigenvalues. Specifically: 

This statement is true. 

2. Trace(A) = n − 1: 

• The trace of A is the sum of its eigenvalues. From the eigenvalue analysis, A has 0 (once) and 1 (n − 1 times). Thus: 
  Trace(A) = 0 + (n − 1)(1) = n − 1. 
• This statement is true. 

3. Rank(A) + Rank(I − A) = n: 

• The rank of A is equal to the number of nonzero eigenvalues, which is n − 1 (corresponding to eigenvalue 1). 
• , and its rank is 1 (since J has rank 1).
• Hence: Rank (A) + Rank (I − A) = (n − 1) + 1 = n. 
• This statement is true. 

4. A is invertible: 
A is not invertible because it has 0 0 as an eigenvalue, which makes its determinant 0 0. This statement is false. 

Final Answer: 
The statement that is not true is: 
4. A is invertible.

Question 7. Let A be a 5 × 4 matrix with real entries such that Ax = 0 if and only if x = 0, where x is a 4 × 1 vector. Then find the rank of A.
Sol: 
We are given that A is a 5 × 4 matrix with real entries, and the condition Ax = 0 if and only if x = 0, where x is a 4 × 1 vector. 
We need to find the rank of A. 

Key Observations: 
1. Condition Ax = 0 implies x = 0: 

This means that the null space (kernel) of A contains only the zero vector. 

Thus, the nullity of A is zero. 

2. Rank-Nullity Theorem: For any matrix A of size m × n, the rank-nullity theorem states: 

Rank(A) + Nullity(A) = n, 
where n is the number of columns of A. 
Here, n = 4, and since the nullity is zero: 
Rank (A) = 4. 

3. Rank of A: 

• The rank of A is 4, which means A has full column rank (all 4 columns are linearly independent). 
• Since A is a 5 × 4 matrix, having rank 4 4 is the maximum possible for the number of columns. 

Final Answer: 

The rank of A is: 4

Question 8. Consider the following row vectors

a₁ = (1, 1, 0, 1, 0, 0), a₂ = (1, 1, 0, 0, 1, 0),
a₃ = (1, 1, 0, 0, 0, 1) a₄ = (1, 0, 1, 1, 0, 0),
a₅ = (1, 0, 1, 0, 1, 0), a₆ = (1, 0, 1, 0, 0, 1)

What is the dimension of the real vector space spanned by these vectors?
Sol: 
To determine the dimension of the real vector space spanned by the given row vectors:

Step 1: Form the Matrix 
We arrange the vectors as rows in a matrix A:

Step 2: Row Reduce the Matrix To find the dimension of the vector space spanned by these vectors, we compute the rank of A by row reducing it to row echelon form. 
1. Subtract the first row from rows 2–6:

2. Simplify further by eliminating duplicate rows and linear combinations. After row reduction, the matrix reduces to:

Step 3: Count the Pivot Columns 
The row echelon form has 4 non-zero rows, corresponding to 4 pivot columns. This implies that the vectors span a 4-dimensional space. Final Answer: 
The dimension of the real vector space spanned by these vectors is: 4

Question 9. Leti, j = 1, 2, . . . , n for some distinct real numbers b₁ , b₂ , . . . , b_n . Then find det(A).
Solution. Performing the elementary operations, R_i → R_i − R₁ for each i = 2, 3, . . . , n we see that det(A) = 0. Note that det(A) may not be zero if n = 2.

Question 10. Number of matrices with real entries, of the formsuch that A² − A = 0 i
1. only one
2. only two
3. finitely many
4. infinitely many
Solution. Using the equation A² = A, it can be shown that any matrix A of the form A = where bc = a − a² , satisfies the given condition. So, there are infinitely many such matrices.

Question 11. Let V be a vector space of dimension n over the field Z_p where p is a prime. Then how many elements are there in V ?
Solution. Since dim(V ) = n, so let {e₁ , e₂ , . . . , e_n } be a basis of V . Then any element of V is a linear combination of the form α_i e_i where α_i ∈ Z_p . So,
the number of elements in V is pn .

Question 12. If T :be a linear transformation defined by T (1, 0, 0) = (1, 0, 1), T (0, 1, 0) = (0, 0, 1) and T (0, 0, 1) = (1, 0, 0). Then find the range space of T , null space of T , rank and nullity of T. 
Sol: 
To solve this problem, let  be a linear transformation defined by: 
T(1, 0, 0) = (1, 0, 1), T(0, 1, 0) = (0, 0, 1), T(0, 0, 1) = (1, 0, 0). 
We need to find: 
1. The range space of T, 
2. The null space of T, 
3. The rank of T, 
4. The nullity of T. 

Step 1: Representing T as a Matrix 
The action of T on the standard basis vectors e₁  =(1, 0, 0), e₂  = (0, 1, 0), e₃ = (0, 0, 1) gives us the columns of the matrix representation of T.
T(e₁) = (1, 0, 1), T(e₂) = (0, 0, 1), T(e₃) = (1, 0, 0). 
Thus, the matrix representation of T is:

Step 2: Range Space of T 
The range space (or column space) of T is spanned by the columns of the matrix:

These vectors are linearly dependent, as:

Thus, the range space is spanned by:

The range space is 2-dimensional. 

Step 3: Null Space of T 
The null space (or kernel) of T is the set of vectors x ∈ R³ such that T(x) = 0. Solving:
T(x) = [T]⋅x = 0, 
where:

This gives the system of equations:

1. x₁ + x₃ = 0

2. 0 = 0, (no condition),
3. x₁ + x₂ = 0.

From these equations: x₃ = −x₁  , x₂  = −x₁ . 

Let x₁ = t. Then: 

Thus, the null space is:

Step 4: Rank and Nullity

From the Rank-Nullity Theorem: 

Rank (T) + Nullity (T) = n, 

where n = 3. 

• The rank is the dimension of the range space: Rank (T) = 2. 
• The nullity is the dimension of the null space: Nullity(T) = 1. 

Final Answers: 

1. Range space of T: 

2. Null space of T:

3. Rank of T: 
Rank (T) = 2. 
4. Nullity of T: 
Nullity (T) = 1.

Question 13. Let A ∈ M₂(R) such that tr(A) = 2 and det(A) = 3. Write down the characteristic polynomial of A⁻¹ .
Sol: 
We are given a matrix  such that: 
tr(A) = 2, det(A) = 3. 
We are tasked with finding the characteristic polynomial of A⁻¹. 

Step 1: Characteristic Polynomial of A 
The characteristic polynomial of A is given by: 
pA (λ) = det(A − λI). 
For a 2 × 2 matrix A, the characteristic polynomial is:
pA (λ) = λ² − tr(A)λ + det(A). 
Substitute the given values tr (A) = 2 and det (A) = 3: 
p_A (λ) = λ² − 2λ + 3. 

Step 2: Relationship between A and A⁻¹ 
If λ is an eigenvalue of A, then 1/λ is an eigenvalue of A⁻¹. The characteristic polynomial of A⁻¹ can be derived from that of A. 
Let the eigenvalues of A be λ₁  and λ₂. Then: 

The trace of A satisfies: λ₁ + λ₂ = tr(A) = 2, 

The determinant of A satisfies: λ₁ λ₂ = det(A) = 3. 
The characteristic polynomial of A⁻¹ has eigenvalues so:

Step 3: Expressing p_A−1 (λ) 

1. The trace of A ⁻¹ is the sum of its eigenvalues:

Substitute λ₁ + λ₂ = 2 and λ₁ λ₂ = 3: 
2. The determinant of A⁻¹ is the product of its eigenvalues:

Substitute λ₁λ₂ = 3:

Thus, the characteristic polynomial of A⁻¹ is: 

Substitute tr (A⁻¹) = 2/3 and det (A⁻¹) = 1/3: 

Final Answer: 

The characteristic polynomial of A⁻¹ is:

Question 14. Let a 3 × 3 matrix A have eigenvalues 1, 2, −1, then find the trace of the matrix B = A − A⁻¹ + A² .
Sol: 
We are tasked with finding the trace of the matrix B = A − A⁻¹ + A², where the 3 × 3 matrix A has eigenvalues 1, 2, −1. 

Step 1: Properties of Trace and Eigenvalues 

1. Trace of A: The trace of a matrix is the sum of its eigenvalues. Since the eigenvalues of A are 1, 2, −1: 
tr(A) = 1 + 2 − 1 = 2. 

2. Eigenvalues of A⁻¹: If λ is an eigenvalue of A, then 1/λ  is an eigenvalue of  A⁻¹. Thus, the eigenvalues of A⁻¹ are: 

3. Eigenvalues of A² : If λ is an eigenvalue of A, then λ² is an eigenvalue of A². 

Thus, the eigenvalues of A² are: 

1² = 1, 2² = 4, (−1)² = 1. 

Step 2: Trace of  B 
The trace of B is the sum of the traces of A, A⁻¹ , and A² : 
tr(B) = tr(A) − tr(A⁻¹) + tr(A²). 
1. Trace of A: tr(A)=2. 
2. Trace of A⁻¹: The trace of A⁻¹ is the sum of its eigenvalues: 

3. Trace of A²: The trace of  A² is the sum of its eigenvalues: 

tr(A²) = 1 + 4 + 1 = 6. 

Thus: 

Step 3: Simplify the Result 

Final Answer: 

tr (B) = 15/2

Question 15. The eigenvectors of a 3 × 3 matrix A corresponding to the eigenvalues 1, 1, 2 are (1, 0, −1)T , (0, 1, −1)T and (1, 1, 0)T . Find the matrix A.
Sol: 
We are tasked with finding the 3 × 3 matrix A, given its eigenvalues and corresponding eigenvectors. 
The eigenvalues are 1, 1 ,2, and the corresponding eigenvectors are: 
1. For eigenvalue 1:  , 
2. For eigenvalue 2:

Step 1: Diagonalization 
The matrix A can be diagonalized as: A = PDP ⁻¹, 
where:

• P is the matrix whose columns are the eigenvectors of A, 
• D is the diagonal matrix with eigenvalues on its diagonal. 

Step 2: Construct P and D

1. Construct the matrix P using the eigenvectors as columns: 

2. Construct the diagonal matrix D using the eigenvalues: 

Step 3: Compute P ⁻¹ 
To compute P ⁻¹ , use the formula for the inverse of a 3×3 matrix: 

Compute det (P): 

Expanding along the first row: 

det(P) = 1⋅1 + 1⋅1 = 2.
Compute P⁻¹: 

The adjugate of P, adj (P), is computed by finding the cofactors of P. For brevity: 

Thus:

Step 4: Compute A = PDP ⁻¹ 
Substitute P, D, and P⁻¹ into  A = PDP⁻¹:

After performing the matrix multiplication:

Final Answer: 

The matrix A is:

Question 16. What is the null space of the matrix A =

Sol: 

We are tasked with finding the null space of the matrix: 
A = [ 1 0 2 5 1 4 ]. 

Step 1: Definition of Null Space 
The null space of a matrix A consists of all column vectors x such that:
Ax = 0.
Here, x is a vector in  (since A has 6 columns), so . The equation becomes:

This gives a single linear equation: 

or: x₁ + 2x₃ + 5x₄ + x₅ + 4x₆ = 0. . ..(1)

Step 2: Parametrize the Solution 
We express x₁ in terms of the free variables  x₃, x₄, x₅, x₆  
x₁ = −2x₃  −5x₄  − x₅  − 4x₆. 
Let x₂ = t₂, x₃  = t₃, x₄  = t₄, x₅ = t₅, x₆ = t₆ , where t₂, t₃, t₄, t₅ ,t₆ are free parameters. 
Then:

Split this into a linear combination of the free variables:

Step 3: Basis for the Null Space 
The null space of A is spanned by the vectors:

Question 17. Let V be the vector of all n × n matrices over the fieldof real numbers. A subset W of V is not a subspace of V if W consists of all matrices for which

1. A² = A
2. A² ≠ A
3. A = A^T
4. A = −A^T

Sol: 
We are tasked with determining which of the given subsets W of the vector space V (all n × n matrices over the field of real numbers) is not a subspace of V. 
Definition of a Subspace: 
For W to be a subspace of V, it must satisfy: 
1. W is non-empty (contains the zero matrix). 
2. W is closed under matrix addition: If A, B ∈ W, then A + B ∈ W. 
3. W is closed under scalar multiplication: If A ∈ W and c ∈ R, then c A ∈ W. 

Analyzing Each Case: 
1. W = A ∈ V : A² = A (Idempotent Matrices): � 2 = � 

• A² = A implies that A satisfies the quadratic equation A(A − I) = 0, where  I is the identity matrix. 
• The zero matrix is in W, and it is closed under addition and scalar multiplication: 
  If A, B ∈ W, then (A + B)² = (A + B) does not necessarily hold because cross-terms like AB or BA might not satisfy the idempotent condition. 
• Conclusion: W is not closed under addition, so it is not a subspace. 

2. W = A ∈ V : A² ≠ A: 

• This subset consists of matrices that do not satisfy A² = A. 
• The subset is not closed under addition or scalar multiplication because combining two matrices A and B in W could result in a matrix A + B for which (A + B)² = A + B, violating the condition. 
• Conclusion: W is not closed under addition or scalar multiplication, so it is not a subspace. 

3. W = A ∈ V : A ≠ A^T (Symmetric Matrices): 

A=A T means A is symmetric. 

The zero matrix is symmetric, and the set is closed under addition and scalar multiplication: 

• If A, B ∈ W, then (A + B)^T = A^T + B^T = A + B, so A + B ∈ W. 
• If c ∈ R and A ∈ W, then (cA)^T = cA^T = cA^T,  so c A ∈ W. 

Conclusion: W is a subspace. 
4. W = A ∈ V : A = - A^T ( (Skew-Symmetric Matrices): 

• A = −A^T means A is skew-symmetric. 
• The zero matrix is skew-symmetric, and the set is closed under addition and scalar multiplication: 
  If A, B ∈ W, then(A + B)^T = A^T + B^T = −A −B = −(A + B), so A + B ∈ W. 
  If c ∈ R and A ∈ W, then (cA)^T = cA^T =−cA, so cA ∈ W. 
• Conclusion: W is a subspace.

Final Answer: 
The subset W is not a subspace of V if it consists of matrices satisfying:
A² = A or A² ≠ A. 

Question 18. Let A be a 4 × 4 invertible matrix. Which of the following is NOT true?
1. The rows of A form a basis of
2. Null space of A contains only the zero vector
3. A has four distinct eigenvalues
4. Image of the linear transformation x → Ax on

Sol: 
Let A be a 4 × 4 invertible matrix. We need to determine which of the given statements is NOT true. 

Statement 1: The rows of A form a basis of  .

• Since A is invertible, its rank is 4, meaning its rows are linearly independent. 
• Linearly independent rows of a 4 × 4 matrix span .
• This statement is true. 

Statement 2: The null space of A contains only the zero vector. 

• For an invertible matrix A, the null space contains only the zero vector because Ax = 0 implies x = 0. 
• This statement is true. 

Statement 3: A has four distinct eigenvalues. 

• An invertible matrix is not guaranteed to have four distinct eigenvalues. The matrix can have repeated eigenvalues, or some eigenvalues may have algebraic multiplicity greater than 1. 
• For example, the identity matrix I has a single eigenvalue 1 repeated four times. 
• This statement is NOT necessarily true. 

Statement 4: The image of the linear transformation x ↦ Ax is . 

• Since A is invertible, the transformation x ↦ Ax is surjective, meaning it maps all of  onto itself.
• The image of the transformation is  .
• This statement is true. 

Final Answer: 
The statement that is NOT true is: A has four distinct eigenvalues. 

Question 19. Let Cbe the linear space of all continuous functions from R to R. Let Wc be the set of differentiable functions u(x) that satisfy the differential equation u′ = xu + c. For which value(s) of the real constant c is this set a linear subspace of C ()?

Sol: 
To determine the values of c for which the set W_c is a linear subspace of , let's analyze the given conditions and the properties of subspaces. 
Step 1: Definition of W_c 
The set W_c is defined as: 
W_c = {u(x) ∈ :u ′ (x)=xu(x)+c}. 
Here, u(x) is a differentiable function satisfying the given first-order differential equation. 

Step 2: Conditions for W_c to be a Subspace 

For W_c to be a subspace of , it must satisfy the following conditions:

• Zero function belongs to W_c : The zero function u(x) = 0 must satisfy the differential equation. 
• Closed under addition: If u₁(x), u₂ (x) ∈ W_c , then u₁ (x) + u₂ (x) ∈ W_c . 
• Closed under scalar multiplication: If u(x) ∈ W_c and , then αu(x)∈W_c. 

Step 3: Check the Zero Function 

Substitute u(x) = 0 into the differential equation: 
u′ (x) = xu(x) + c ⟹ 0 = 0 + c. 
For the zero function to satisfy the equation, we require: c = 0. 

Step 4: Closure under Addition and Scalar Multiplication 

Let u₁ (x), u₂ (x) ∈ W_c. This means: 

Consider their sum v(x)= u₁ (x) + u₂ (x). Then: 

For v(x) to satisfy the differential equation v′ (x) = xv(x) + c, we must have: 

2c = c ⟹ c = 0. 

Similarly, for scalar multiplication w(x) = αu(x), where , we find: 
w′ (x) = αu′ (x) = α(xu(x) + c) = x(αu(x)) + αc. 
For w(x) to satisfy w ′ (x) = xw(x) + c, we require: 
αc = c ⟹ c(1 − α) = 0. 
This holds for all α only if c = 0. 

Step 5: Conclusion 
The set W_c is a linear subspace of  if and only if c = 0. 
Final Answer: c = 0. 

Question 20. Let U and V both be two-dimensional subspaces ofand U ≠ V , and let W = U ∩ V . Find all possible values for the dimension of W.
Sol: 
We are tasked with finding all possible values for the dimension of W = U∩V, where U and V are two-dimensional subspaces of a vector space  , and U ≠ V.

Step 1: Key Properties 

1. dim (U) = dim (V) = 2 because U and V are two-dimensional subspaces. 
2. The dimension of the intersection W = U ∩ V satisfies: 
dim (W) ≤ min (dim (U), dim(V)) = 2. 
3. The sum of the dimensions of U and V is related to their intersection by the dimension formula: 
dim (U + V) = dim (U) + dim (V) − dim (U ∩ V). 
Since U + V is a subspace of  , we know: 
dim (U + V) ≤ n. 

Step 2: Possible Dimensions of W 

The intersection W = U ∩ V can have dimension 0, 1, or 2: 
1. Case dim(W) = 0: This happens when U and V have no non-zero vectors in common, meaning they are disjoint except for the zero vector. 
2. Case dim(W) = 1: This happens when U and V intersect along a one-dimensional subspace (a line through the origin in).
3. Case dim (W) = 2: This happens when U = V, which is ruled out by the condition U ≠ V . 

Thus, the possible dimensions of W are: d
dim(W) ∈ {0, 1}. 

Step 3: Verification 
1. For dim(W) = 0: 

• If dim(W) = 0, then dim ⁡(U + V)dim (U) + dim (V) = 4, provided n ≥ 4. 

2. For dim ⁡(W) = 1: 

• If dim(W) = 1, then dim ⁡(U + V)dim (U) + dim (V) - dim (W) = 3, provided n ≥ 3. 

Both cases are consistent with the dimension formula and the structure of subspaces in .

Final Answer: 
The possible dimensions of W = U ∩ V are: dim(W) ∈ {0, 1}.

Question 21. Let P_n be the vector space of all polynomials of degree at most n ≥ 6 with real coefficients. Consider the following subspaces of P_n :
W₁ = {p(x) ∈ P_n : p(1) = 0, p(2) = 0, p(3) = 0, p(4) = 0}
W₂ = {p(x) ∈ P_n : p(3) = 0, p(4) = 0, p(5) = 0, p(6) = 0}
Find the dimensions of W₁, W₂ and W₁ ∩ W₂

Sol: 
We are tasked with finding the dimensions of the subspaces W₁ , W₂  , and W₁  ∩ W₂  of the vector space P_n  , where P_n is the vector space of all polynomials of degree at most n ≥ 6 with real coefficients. 
Step 1: Vector Space of Polynomials P_n    
The vector space P_n has dimension n + 1, as a general polynomial of degree at most n is written as: 

which involves n + 1 coefficients (a₀, a₁  ,…, a_n). 

Step 2: Subspace W₁  
The subspace W₁ is defined as: 

This means that the polynomial p(x) has roots at x = 1, 2, 3, 4. Hence, p(x) must be divisible by: 
(x−1)(x−2)(x−3)(x−4). 
Since p(x) is a polynomial in P_n , it can be written as: 
p(x) = (x−1)(x−2)(x−3)(x−4)q(x), 
where q(x) is a polynomial of degree at most n−4. 
Thus, the dimension of W₁ is the number of free coefficients in q(x), which is: 
dim(W₁) = (n − 4) + 1 = n − 3. 

Step 3: Subspace W₂  
The subspace W₂ is defined as: 

This means that the polynomial p(x) has roots at x = 3, 4, 5, 6. Hence, p(x) must be divisible by: 
(x−3)(x−4)(x−5)(x−6). 
Similarly, p(x) can be written as: 
p(x) = (x−3)(x−4)(x−5)(x−6)q(x), 
where q(x) is a polynomial of degree at most n − 4. 
Thus, the dimension of W₂ is: 
dim(W₂) = (n − 4) + 1 = n − 3. 

Step 4: Intersection W₁ ∩W₂  
The subspace W₁ ∩ W₂ contains polynomials p(x) that satisfy both W₁ and W₂. 
Such polynomials have roots at: x = 1, 2, 3, 4, 5, 6. 
Thus, p(x) must be divisible by:
(x−1)(x−2)(x−3)(x−4)(x−5)(x−6). 
Write p(x) as: 
p(x)=(x−1)(x−2)(x−3)(x−4)(x−5)(x−6)q(x), 
where q(x) is a polynomial of degree at most n−6. 
The dimension of W₁ ∩W₂ is
dim(W₁ ∩W₂) = (n−6)+1=n−5. 

Final Answers: 

1. dim(W₁)= n − 3, 
2. dim(W₂) = n − 3, 
3. dim(W₁ ∩ W₂) = n − 5.

Question 22. Let P₃ denote the real vector space of all polynomials with real coefficients having degree less than or equal to 3, equipped with the standard ordered basis {1, x, x₂ , x₃ }. Write down the matrix, w.r.t this basis,of the fol lowing linear transformation:

L(p) = p′′ − 2p′ + p, p ∈ P₃

Also, find the unique polynomial p such that L(p) = x³.

Solution. Here,

Thus matrix of L =

Now let p = ax³ + bx² + cx + d ∈ P₃ be such the L(p) = x³ . Then, using the definition of L and equating the coefficients on both sides, we get p = x³ + 6x² + 18x + 24.

Question 23. A non-zero matrix A ∈ M_n(R) is said to be nilpotent if A^k = 0 for some positive integer k ≥ 2. If A is nilpotent, which of the fol lowing statements are true? 1. k ≤ n for the smal lest such k. 2. The matrix I + A is invertible. 3. Al l the eigenvalues of A are zero.

Sol: 
We are tasked with analyzing the properties of a nilpotent matrix . A matrix A is said to be nilpotent if A^k = 0 for some positive integer k ≥ 2. Let’s examine the given statements: 

Statement 1: k ≤ n for the smallest such k.  

Explanation:

• The minimal polynomial of A, which divides the characteristic polynomial, determines the smallest integer k such that A^k = 0 
• A k =0. 
• The degree of the minimal polynomial cannot exceed n, the size of the matrix A. Hence, the smallest k such that A^k = 0 satisfies  k ≤ n. 

Conclusion: 

This statement is true. 

Statement 2: The matrix I + A is invertible. 

Explanation: 

• If A is nilpotent, then A^k = 0 for some k ≥ 2. 
• The matrix  I + A can be inverted using the formula for a Neumann series: 

• Since A^k = 0, the series terminates at A^k−1 , ensuring that I + A is invertible. 

Conclusion: 

• This statement is true. 

Statement 3: All the eigenvalues of A are zero. 

Explanation: 

• If λ is an eigenvalue of A, then there exists a nonzero vector v such that Av = λv. 
• Applying A^k = 0 to this eigenvector: 
• A^kv = λ^kv = 0. 
• Since v ≠ 0 , we conclude that λ^k = 0, implying λ = 0. 

Conclusion: 

• This statement is true. 

Final Answer: 
All three statements are true: 
1 , 2 , and 3 are true.

Question 24. Justify whether the fol lowing are true or false: 
1. There exist n × n matrices A and B such that AB − BA = I .
2. Let A and B be two arbitrary n × n matrices. If B is invertible, then tr(A) = tr⁻¹ (B − 1AB) where tr(M ) denotes the trace of an n × n matrix M.

Sol: 
Statement 1: There exist n × n matrices A and B such that AB − BA = I. 
Analysis: 
1. The matrix AB − BA (called the commutator of A and B) is always traceless, meaning tr(AB − BA) = 0. 

• This is because the trace operator satisfies tr tr(AB) = tr(BA) for any n × n matrices A and B. 
• Thus: tr(AB−BA)=tr(AB)−tr(BA)=0. 

2. The identity matrix I has tr(I)=n, which is nonzero for any n ≥ 1. 
3. Since AB − BA is traceless, it is impossible for AB − BA = I, as their traces would contradict. 

Conclusion: 
The statement is false. 

Statement 2: Let A and B be two arbitrary n × n matrices. If B is invertible, then tr(A) = tr(B ⁻¹ AB), where tr(M) denotes the trace of an n × n matrix M. 

Analysis: 
1. For any n × n matrices A and B, the trace operator satisfies the cyclic property: 

where X,Y,Z are n × n matrices. 

2. Apply this property to  tr(B ⁻¹ AB): 
tr(B ⁻¹ AB)=tr((AB)B ⁻¹)=tr(A(BB ⁻¹))=tr(A⋅I)= tr(A). 
Hence, the equation tr(A) = tr(B ⁻¹ AB) is always true when B is invertible. 

Conclusion: 
The statement is true. 

Final Answer:

1. False: There do not exist n×n matrices A and B such that AB−BA=I. 
2. True: If B is invertible, then tr(A)=tr(B ⁻¹ AB).

Question 25. Let V be the real vector space of all polynomials in one variable with real coefficients and of degree less than, or equal to 5. Let W be the subspace defined by W = {p ∈ V |p(1) = p′(2) = 0}. What is the dimension of W ?
Sol: 
We are tasked with finding the dimension of the subspace W of the real vector space V of polynomials of degree at most 5, where: 
V = {p(x) ∣ deg(p(x)) ≤ 5}, 
and: 
W={p(x)∈V∣p(1)=0 and p ′ (2)=0}. 

Step 1: Dimension of V 

The vector space V consists of all polynomials of degree at most 5. A basis for V is 

dim(V) = 6. 

Step 2: Conditions Defining W 
1. Condition 1:  p(1) = 0: 

• This implies that (x−1) is a factor of p(x). 
• Thus, p(x) can be written as:
  p(x)  =(x−1)q(x), 
  where q(x) is a polynomial of degree at most 4 (since deg ⁡(p(x))≤5). 
  This condition reduces the dimension of  V by 1, so the subspace of polynomials satisfying p(1) = 0 has: 
  dim=6−1=5. 

2. Condition 2: p ′ (2) = 0: 

• Differentiate p(x): 
  p(x)=(x−1)q(x)⟹p ′ (x)=(x−1)q ′ (x)+q(x). 
• Substitute x = 2 into p′ (x) = 0: 
  p ′ (2)=(2−1)q ′ (2)+q(2)=q ′ (2)+q(2)=0. 
• This imposes a linear constraint on the coefficients of q(x). 

Since q(x) is a polynomial of degree at most 4, it has 5 coefficients. Imposing q′(2)+q(2)=0 reduces the dimension by 1. 

Step 3: Dimension of W 

Starting from dim(V) = 6, the two conditions reduce the dimension as follows: 

1. p(1) = 0 reduces the dimension by 1 ( dim ⁡ = 5 dim=5). 
2. p ′ (2)=0 reduces the dimension by 1 ( dim ⁡ = 4 dim=4). 
Thus, the dimension of W is: dim(W) = 4.