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Q1. (a) Show that the eigenvalues of a self-adjoint operator are real. 2 (b) For the linear operator T: R³ → R³ defined by: T(x, y, z) =(x – y + 2z, 2x + y, -x – 2y + 2z) find range of T and kernel of T. 5 (c) Let the characteristic polynomial of a matrix A be λ³ – 5λ² + 2λ + 1. Is the matrix invertible? If yes, then express A⁻¹ as a linear combination of A², A and I. If no, then justify your answer. 3

Ans. (a) Let T be a self-adjoint operator on an inner product space V. This means, ⟨T(u), v⟩ = ⟨u, T(v)⟩ for all u, v ∈ V. Let λ be an eigenvalue of T and v ≠ 0 be the corresponding eigenvector. Then T(v) = λv. We need to show that λ is a real number, i.e., λ = λ̄ (the complex conjugate of λ). Consider the inner product ⟨T(v), v⟩. On one hand, we have: ⟨T(v), v⟩ = ⟨λv, v⟩ = λ⟨v, v⟩ = λ||v||² On the other hand, since T is self-adjoint: ⟨T(v), v⟩ = ⟨v, T(v)⟩ = ⟨v, λv⟩ = λ̄⟨v, v⟩ = λ̄||v||² Equating these two expressions, we get: λ||v||² = λ̄||v||² Since v is an eigenvector, v ≠ 0, hence ||v||² > 0. We can divide both sides of the equation by ||v||², which gives: λ = λ̄ This shows that the imaginary part of λ is zero, and therefore λ is a real number .

(b) The given linear operator is T(x, y, z) = (x – y + 2z, 2x + y, -x – 2y + 2z). Finding the Kernel: The kernel of T, ker(T), is the set of all vectors (x, y, z) such that T(x, y, z) = (0, 0, 0). x – y + 2z = 0 —(1) 2x + y = 0 —(2) -x – 2y + 2z = 0 —(3) From equation (2), we get y = -2x. Substituting y = -2x into equation (1): x – (-2x) + 2z = 0 ⇒ 3x + 2z = 0 ⇒ z = -3/2 x. Now, we check these values of y and z in equation (3): -x – 2(-2x) + 2(-3/2 x) = -x + 4x – 3x = 0. The equation is satisfied. Thus, vectors in the kernel are of the form (x, -2x, -3/2 x), for x ∈ R. ker(T) = {x(1, -2, -3/2) | x ∈ R}. A basis for ker(T) is {(2, -4, -3)}. Thus, dim(ker(T)) = 1.

Finding the Range: The range T(R³) = Im(T) is spanned by the vectors T(e₁), T(e₂), T(e₃), where {e₁, e₂, e₃} is the standard basis of R³. T(1,0,0) = (1, 2, -1) T(0,1,0) = (-1, 1, -2) T(0,0,1) = (2, 0, 2) By the Rank-Nullity Theorem, dim(Im(T)) + dim(ker(T)) = dim(R³). dim(Im(T)) + 1 = 3 ⇒ dim(Im(T)) = 2. So, we need two linearly independent vectors from the three spanning vectors above to form a basis for the range. The first two vectors, (1, 2, -1) and (-1, 1, -2), are clearly not scalar multiples of each other, hence they are linearly independent. Thus, the Range of T = span{(1, 2, -1), (-1, 1, -2)}.

(c) The characteristic polynomial of matrix A is p(λ) = λ³ – 5λ² + 2λ + 1. A matrix is invertible if and only if its determinant is non-zero. The determinant of a matrix is the constant term of its characteristic polynomial, multiplied by (-1)ⁿ where n is the size of the matrix. Alternatively, det(A) = p(0). p(0) = (0)³ – 5(0)² + 2(0) + 1 = 1. Since det(A) = 1 ≠ 0, the matrix A is invertible .

To find A⁻¹, we use the Cayley-Hamilton theorem, which states that every matrix satisfies its own characteristic equation. p(A) = A³ – 5A² + 2A + I = 0 where I is the identity matrix. Multiplying this equation by A⁻¹ (which exists): A⁻¹(A³ – 5A² + 2A + I) = A⁻¹(0) A² – 5A + 2I + A⁻¹ = 0 Isolating A⁻¹: A⁻¹ = -A² + 5A – 2I This expresses A⁻¹ as a linear combination of A², A, and I.

Q2. (a) For the basis B={(0,0,-1), (0,1,0), (2,2,0)} of R³, find the dual basis of B. 4 (b) Find all the eigenvalues, their corresponding eigenvectors and dimensions of the eigenspaces for the matrix: A=[-9 4 4; -8 3 4; -16 8 7]. 6

Ans. (a) Let the given basis of R³ be B = {v₁, v₂, v₃}, where v₁ = (0, 0, -1), v₂ = (0, 1, 0), and v₃ = (2, 2, 0). The dual basis B* = {f₁, f₂, f₃} is a set of linear functionals from R³ to R that satisfies the property fᵢ(vⱼ) = δᵢⱼ (the Kronecker delta). Let any vector in R³ be (x, y, z). Each fᵢ can be written as fᵢ(x, y, z) = aᵢx + bᵢy + cᵢz.

Calculating f₁: f₁(v₁) = 1 ⇒ f₁(0,0,-1) = -c₁ = 1 ⇒ c₁ = -1 f₁(v₂) = 0 ⇒ f₁(0,1,0) = b₁ = 0 f₁(v₃) = 0 ⇒ f₁(2,2,0) = 2a₁ + 2b₁ = 0 ⇒ 2a₁ + 0 = 0 ⇒ a₁ = 0 So, f₁(x, y, z) = -z .

Calculating f₂: f₂(v₁) = 0 ⇒ f₂(0,0,-1) = -c₂ = 0 ⇒ c₂ = 0 f₂(v₂) = 1 ⇒ f₂(0,1,0) = b₂ = 1 f₂(v₃) = 0 ⇒ f₂(2,2,0) = 2a₂ + 2b₂ = 0 ⇒ 2a₂ + 2(1) = 0 ⇒ a₂ = -1 So, f₂(x, y, z) = -x + y .

Calculating f₃: f₃(v₁) = 0 ⇒ f₃(0,0,-1) = -c₃ = 0 ⇒ c₃ = 0 f₃(v₂) = 0 ⇒ f₃(0,1,0) = b₃ = 0 f₃(v₃) = 1 ⇒ f₃(2,2,0) = 2a₃ + 2b₃ = 1 ⇒ 2a₃ + 0 = 1 ⇒ a₃ = 1/2 So, f₃(x, y, z) = (1/2)x .

Therefore, the dual basis is B* = {f₁, f₂, f₃} where f₁(x,y,z) = -z, f₂(x,y,z) = -x+y, and f₃(x,y,z) = (1/2)x.

(b) The given matrix is A = [[-9, 4, 4], [-8, 3, 4], [-16, 8, 7]]. Finding Eigenvalues: We solve det(A – λI) = 0. However, we can first inspect for simple eigenvalues. Sum the coefficients of each row: Row 1: -9 + 4 + 4 = -1 Row 2: -8 + 3 + 4 = -1 Row 3: -16 + 8 + 7 = -1 Since all row sums are -1, λ₁ = -1 is an eigenvalue, and a corresponding eigenvector is (1, 1, 1).

Now, we know the sum of eigenvalues equals the trace of the matrix: Tr(A) = -9 + 3 + 7 = 1. λ₁ + λ₂ + λ₃ = 1 ⇒ -1 + λ₂ + λ₃ = 1 ⇒ λ₂ + λ₃ = 2.

The product of eigenvalues equals the determinant of the matrix: det(A) = -9(21 – 32) – 4(-56 – (-64)) + 4(-64 – (-48)) = -9(-11) – 4(8) + 4(-16) = 99 – 32 – 64 = 3. λ₁λ₂λ₃ = 3 ⇒ (-1)λ₂λ₃ = 3 ⇒ λ₂λ₃ = -3.

We now have two equations: λ₂ + λ₃ = 2 and λ₂λ₃ = -3. These are the roots of the quadratic equation t² – (sum)t + (product) = 0, which is t² – 2t – 3 = 0. Factoring, (t – 3)(t + 1) = 0 ⇒ t = 3, -1. So, the other two eigenvalues are λ₂ = 3 and λ₃ = -1 . Overall, the eigenvalues are -1 (with algebraic multiplicity 2) and 3 (with algebraic multiplicity 1) .

Eigenvectors and Eigenspaces: For λ = -1: We solve (A + I)v = 0. [[ -8, 4, 4 ], [ -8, 4, 4 ], [ -16, 8, 8 ]] [x, y, z]ᵀ = [0, 0, 0]ᵀ All rows reduce to the equation -2x + y + z = 0. This is the equation of a plane, which has dimension 2. Let x=1, y=2, then -2(1) + 2 + z = 0 ⇒ z = 0. One eigenvector is v₁ = (1, 2, 0) . Let x=1, z=2, then -2(1) + y + 2 = 0 ⇒ y = 0. Another linearly independent eigenvector is v₂ = (1, 0, 2) . The eigenspace E₋₁ = span{(1, 2, 0), (1, 0, 2)}. The dimension of E₋₁ is 2 .

For λ = 3: We solve (A – 3I)v = 0. [[ -12, 4, 4 ], [ -8, 0, 4 ], [ -16, 8, 4 ]] [x, y, z]ᵀ = [0, 0, 0]ᵀ From the second row: -8x + 4z = 0 ⇒ z = 2x. From the first row: -12x + 4y + 4z = 0 ⇒ -12x + 4y + 4(2x) = 0 ⇒ -4x + 4y = 0 ⇒ y = x. So, the eigenvectors are of the form (x, x, 2x). An eigenvector is v₃ = (1, 1, 2) . The eigenspace E₃ = span{(1, 1, 2)}. The dimension of E₃ is 1 .

Q3. (a) Check whether the vector (1,3,4) is in the linear span of vectors (4,0,-2),(0,1,1) and (1,1,2) . 2 (b) Reduce the quadratic form 3x² + 2xy + 3y² to a normal canonical form, giving the corresponding coordinate transformations. 6 (c) State the Cauchy-Schwarz inequality. Verify the inequality for the vectors (1,0,-1,1) and (1,1,0,1) in R⁴. 2

Ans. (a) To check if the vector w = (1,3,4) is in the linear span of v₁=(4,0,-2), v₂=(0,1,1), and v₃=(1,1,2), we need to see if there exist scalars a, b, c such that: a(4,0,-2) + b(0,1,1) + c(1,1,2) = (1,3,4) This leads to the system of linear equations: 4a + c = 1 b + c = 3 -2a + b + 2c = 4 We can solve this system using an augmented matrix: [ 4 0 1 | 1 ] [ 0 1 1 | 3 ] [-2 1 2 | 4 ] R3 → R3 + (1/2)R1: [ 4 0 1 | 1 ] [ 0 1 1 | 3 ] [ 0 1 2.5 | 4.5 ] R3 → R3 – R2: [ 4 0 1 | 1 ] [ 0 1 1 | 3 ] [ 0 0 1.5 | 1.5 ] The last row implies 1.5c = 1.5, which gives c=1. From the second row, b + c = 3 ⇒ b + 1 = 3 ⇒ b = 2. From the first row, 4a + c = 1 ⇒ 4a + 1 = 1 ⇒ 4a = 0 ⇒ a = 0. Since we found a unique solution for a, b, c (a=0, b=2, c=1), yes, the vector (1,3,4) is in the linear span of the given vectors.

(b) The quadratic form is 3x² + 2xy + 3y². It can be written in matrix form as Q = XᵀAX, where X = [x, y]ᵀ and A = [[3, 1], [1, 3]]. The first step is to find the eigenvalues and eigenvectors of A. det(A – λI) = (3-λ)² – 1 = 0 ⇒ 3-λ = ±1 ⇒ λ = 3±1. The eigenvalues are λ₁=4 and λ₂=2 . Now we find the normalized eigenvectors: For λ₁=4: (A-4I)v = 0 ⇒ [[-1, 1], [1, -1]]v = 0 ⇒ -x+y=0. An eigenvector is (1,1). Normalized: p₁ = (1/√2, 1/√2)ᵀ . For λ₂=2: (A-2I)v = 0 ⇒ [[1, 1], [1, 1]]v = 0 ⇒ x+y=0. An eigenvector is (1,-1). Normalized: p₂ = (1/√2, -1/√2)ᵀ . The coordinate transformation is given by X = PX’, where P = [p₁ p₂] is an orthogonal matrix and X’ = [x’, y’]ᵀ. P = [[1/√2, 1/√2], [1/√2, -1/√2]] This transformation is: x = (1/√2)x’ + (1/√2)y’ y = (1/√2)x’ – (1/√2)y’ Under this transformation, the quadratic form becomes diagonal: Q = λ₁(x’)² + λ₂(y’)² = 4(x’)² + 2(y’)² This is the canonical form. The normal canonical form has coefficients that are 1, -1, or 0. To achieve this, we make another transformation: x” = √4 x’ = 2x’ y” = √2 y’ With this transformation, the normal canonical form is: Q = (x”)² + (y”)² . The full transformation relating x, y to x”, y” is: x’ = x”/2, y’ = y”/√2 x = (1/√2)(x”/2) + (1/√2)(y”/√2) = x”/(2√2) + y”/2 y = (1/√2)(x”/2) – (1/√2)(y”/√2) = x”/(2√2) – y”/2

(c) Cauchy-Schwarz Inequality: For any two vectors u and v in an inner product space V, |⟨u, v⟩| ≤ ||u|| ||v||. Where ⟨u, v⟩ is the inner product and ||u|| = √⟨u, u⟩ is the norm of a vector.

Verification: Let u = (1, 0, -1, 1) and v = (1, 1, 0, 1) in R⁴ with the standard Euclidean inner product. Calculate the inner product: ⟨u, v⟩ = (1)(1) + (0)(1) + (-1)(0) + (1)(1) = 1 + 0 + 0 + 1 = 2. So, |⟨u, v⟩| = 2 . Now calculate the norms: ||u|| = √(1² + 0² + (-1)² + 1²) = √(1 + 0 + 1 + 1) = √3 . ||v|| = √(1² + 1² + 0² + 1²) = √(1 + 1 + 0 + 1) = √3 . The product of the norms is: ||u|| ||v|| = √3 × √3 = 3 . Check the inequality: 2 ≤ 3. Since 2 is less than 3, the Cauchy-Schwarz inequality is verified for the given vectors.

Q4. (a) For A = [-1 2 -2; 1 2 1; -1 -1 0] find A⁻¹ by using row-reduction method. 4 (b) Define the following and give an example: 4 (i) Norm of a vector in an inner product space (ii) Symmetric matrix (c) Define the signature of a quadratic form. Find the signature of the quadratic form x₁² + x₂² – x₃² – x₄². 2

Ans. (a) To find A⁻¹, we form the augmented matrix [A | I] and row-reduce it to the form [I | A⁻¹]. [A | I] = [ -1 2 -2 | 1 0 0 ] [ 1 2 1 | 0 1 0 ] [ -1 -1 0 | 0 0 1 ]

R1 → -R1: [ 1 -2 2 | -1 0 0 ] [ 1 2 1 | 0 1 0 ] [ -1 -1 0 | 0 0 1 ]

R2 → R2 – R1; R3 → R3 + R1: [ 1 -2 2 | -1 0 0 ] [ 0 4 -1 | 1 1 0 ] [ 0 -3 2 | -1 0 1 ]

R2 → R2 + R3 (to simplify calculation by getting a 1 in the pivot): [ 1 -2 2 | -1 0 0 ] [ 0 1 1 | 0 1 1 ] [ 0 -3 2 | -1 0 1 ]

R1 → R1 + 2R2; R3 → R3 + 3R2: [ 1 0 4 | -1 2 2 ] [ 0 1 1 | 0 1 1 ] [ 0 0 5 | -1 3 4 ]

R3 → (1/5)R3: [ 1 0 4 | -1 2 2 ] [ 0 1 1 | 0 1 1 ] [ 0 0 1 | -1/5 3/5 4/5 ]

R1 → R1 – 4R3; R2 → R2 – R3: [ 1 0 0 | -1 – 4(-1/5) = -1/5 | 2 – 4(3/5) = -2/5 | 2 – 4(4/5) = -6/5 ] [ 0 1 0 | 0 – (-1/5) = 1/5 | 1 – 3/5 = 2/5 | 1 – 4/5 = 1/5 ] [ 0 0 1 | -1/5 | 3/5 | 4/5 ]

Thus, the inverse matrix is: A⁻¹ = [[-1/5, -2/5, -6/5], [1/5, 2/5, 1/5], [-1/5, 3/5, 4/5]] or (1/5) * [[-1, -2, -6], [1, 2, 1], [-1, 3, 4]]

(b) (i) Norm of a vector: In an inner product space V, the norm of a vector v ∈ V, denoted ||v||, is defined as the non-negative square root of the inner product of the vector with itself. ||v|| = √⟨v, v⟩ It generalizes the notion of “length” or “magnitude” of a vector. Example: In R³ with the standard Euclidean inner product, the norm of the vector v = (1, 2, -2) is: ||v|| = √(1² + 2² + (-2)²) = √(1 + 4 + 4) = √9 = 3.

(ii) Symmetric matrix: A square matrix A is called symmetric if it is equal to its transpose, i.e., A = Aᵀ. This means that the entry in the i-th row and j-th column is equal to the entry in the j-th row and i-th column, for all i and j (aᵢⱼ = aⱼᵢ). Example: The following matrix is symmetric: A = [[5, -2, 1], [-2, 0, 7], [1, 7, 3]] because Aᵀ = A.

(c) Signature of a quadratic form: The signature of a quadratic form is a pair of integers (p, q), where ‘p’ is the number of positive square terms in the normal canonical form and ‘q’ is the number of negative square terms. It is defined by Sylvester’s Law of Inertia, which states that p and q are invariants of the quadratic form, regardless of the choice of diagonalizing basis. Sometimes the signature is also defined as the difference p-q.

Signature of the given quadratic form: The quadratic form is x₁² + x₂² – x₃² – x₄². This is already in its normal canonical form. Number of positive square terms (p) = 2 (from x₁² and x₂²). Number of negative square terms (q) = 2 (from -x₃² and -x₄²). Therefore, the signature is (p, q) = (2, 2) . If the signature is defined as p-q, it would be 2-2 = 0.

Q5. (a) Check whether the set: S= {(a₁, a₂,…,aₙ) ∈Rⁿ | aᵢ ≥ 0} is a subspace of Rⁿ or not. 3 (b) Check whether the following system of equations can be solved, using Cramer’s rule: x+y+z=3, 2x-y+z=1, x+2y+2z=4. If yes, apply Cramer’s rule to solve the system of equations. If no, use Gaussian elimination to solve the system of equations. 5 (c) If v₁,v₂ and v₃ are linearly independent vectors in a vector space over C, show that v₁ +v₂, v₂ +v₃ and v₁ +v₃ are also linearly independent over C. 2

Ans. (a) For a set to be a subspace of a vector space, it must satisfy three conditions: 1. It must contain the zero vector. 2. It must be closed under vector addition. 3. It must be closed under scalar multiplication.

Let’s check the set S = {(a₁, a₂,…,aₙ) ∈ Rⁿ | aᵢ ≥ 0 for all i}. 1. Zero vector: The zero vector (0, 0,…, 0) is in S, since all its entries are 0, which satisfies the condition aᵢ ≥ 0. 2. Closure under addition: Let u = (u₁, …, uₙ) and v = (v₁, …, vₙ) be in S. This means uᵢ ≥ 0 and vᵢ ≥ 0. Their sum is u+v = (u₁+v₁, …, uₙ+vₙ). Since uᵢ and vᵢ are both non-negative, their sum uᵢ+vᵢ will also be non-negative. Thus u+v ∈ S. 3. Closure under scalar multiplication: Let u = (1, 1,…, 1). Clearly u ∈ S since all entries are 1 ≥ 0. Now, choose a scalar c ∈ R. Let c = -1. c u = -1 (1, 1,…, 1) = (-1, -1,…, -1). All components of the resulting vector are -1, which does not satisfy the condition ≥ 0. Therefore, c*u ∉ S.

Since the set is not closed under scalar multiplication, S is not a subspace of Rⁿ .

(b) The given system of equations is: x + y + z = 3 2x – y + z = 1 x + 2y + 2z = 4 Cramer’s rule can be used if and only if the determinant of the coefficient matrix is non-zero. The coefficient matrix A is: A = [[1, 1, 1], [2, -1, 1], [1, 2, 2]] Calculate det(A): det(A) = 1((-1)(2) – (1)(2)) – 1((2)(2) – (1)(1)) + 1((2)(2) – (-1)(1)) = 1(-2 – 2) – 1(4 – 1) + 1(4 + 1) = -4 – 3 + 5 = -2 Since det(A) = -2 ≠ 0, yes, Cramer’s rule can be used .

Now we solve the system using Cramer’s rule: x = det(Aₓ)/det(A), y = det(Aᵧ)/det(A), z = det(A₂)/det(A) Aₓ (replace 1st column of A with [3,1,4]ᵀ): det(Aₓ) = det([[3,1,1],[1,-1,1],[4,2,2]]) = 3(-2-2) – 1(2-4) + 1(2-(-4)) = 3(-4) – 1(-2) + 1(6) = -12 + 2 + 6 = -4. x = (-4)/(-2) = 2.

Aᵧ (replace 2nd column of A with [3,1,4]ᵀ): det(Aᵧ) = det([[1,3,1],[2,1,1],[1,4,2]]) = 1(2-4) – 3(4-1) + 1(8-1) = -2 – 3(3) + 7 = -2 – 9 + 7 = -4. y = (-4)/(-2) = 2.

A₂ (replace 3rd column of A with [3,1,4]ᵀ): det(A₂) = det([[1,1,3],[2,-1,1],[1,2,4]]) = 1(-4-2) – 1(8-1) + 3(4-(-1)) = -6 – 7 + 3(5) = -13 + 15 = 2. z = 2/(-2) = -1.

Thus, the solution is x = 2, y = 2, z = -1 .

(c) To show that the vectors w₁ = v₁ + v₂, w₂ = v₂ + v₃, and w₃ = v₁ + v₃ are linearly independent, we consider their linear combination to the zero vector and show that all coefficients must be zero. Let a, b, c ∈ C be scalars such that: a(w₁) + b(w₂) + c(w₃) = 0 a(v₁ + v₂) + b(v₂ + v₃) + c(v₁ + v₃) = 0 Group the terms by v₁, v₂, and v₃: (a + c)v₁ + (a + b)v₂ + (b + c)v₃ = 0 Since v₁, v₂, and v₃ are given to be linearly independent, the only way their linear combination can be zero is if each coefficient is zero: a + c = 0 —(1) a + b = 0 —(2) b + c = 0 —(3) From equation (2), b = -a. From equation (1), c = -a. Substituting these into equation (3): (-a) + (-a) = 0 ⇒ -2a = 0 ⇒ a = 0. Since a = 0, then b = -0 = 0 and c = -0 = 0. Because the only solution is a = b = c = 0, the vectors v₁ + v₂, v₂ + v₃, and v₁ + v₃ are linearly independent .

Q6. (a) Show that the set of all 3×3 upper triangular matrices with entries from R is a vector space over R under usual addition and scalar multiplication of matrices. Find a basis for the vector space. 8 (b) Show that |[1, 2, 2], [a, 1, 0], [0, b, b]| = 0 for infinitely many values of a and b. 2

Ans. (a) Let V be the set of all 3×3 upper triangular matrices with entries from R. We will show that V is a subspace of M₃(R) (the vector space of all 3×3 real matrices), which will prove that V is a vector space over R itself. A general form of an upper triangular matrix is: A = [[a, b, c], [0, d, e], [0, 0, f]] where a,b,c,d,e,f ∈ R.

Verify the subspace conditions: 1. Non-emptiness and Zero Vector: V is non-empty because the 3×3 zero matrix, O = [[0,0,0],[0,0,0],[0,0,0]], is an upper triangular matrix and is in V. O acts as the zero vector for the vector space.

2. Closure under Addition: Let A₁, A₂ ∈ V. A₁ = [[a₁, b₁, c₁], [0, d₁, e₁], [0, 0, f₁]] A₂ = [[a₂, b₂, c₂], [0, d₂, e₂], [0, 0, f₂]] A₁ + A₂ = [[a₁+a₂, b₁+b₂, c₁+c₂], [0, d₁+d₂, e₁+e₂], [0, 0, f₁+f₂]] The resulting matrix also has zeros below the main diagonal, so it is an upper triangular matrix. Thus, A₁ + A₂ ∈ V.

3. Closure under Scalar Multiplication: Let A ∈ V and k ∈ R be a scalar. k A = k [[a, b, c], [0, d, e], [0, 0, f]] = [[ka, kb, kc], [0, kd, ke], [0, 0, kf]] The resulting matrix also has zeros below the main diagonal. Thus, kA ∈ V.

Since all three subspace conditions are met, V is a vector space over R .

Basis for the vector space: Any matrix in V is defined by the six independent entries a, b, c, d, e, f. We can write any such matrix as a linear combination of six basis matrices: A = a[[1,0,0],[0,0,0],[0,0,0]] + b[[0,1,0],[0,0,0],[0,0,0]] + c[[0,0,1],[0,0,0],[0,0,0]] + d[[0,0,0],[0,1,0],[0,0,0]] + e[[0,0,0],[0,0,1],[0,0,0]] + f[[0,0,0],[0,0,0],[0,0,1]] This shows that the following six matrices span V: E₁₁ = [[1,0,0],[0,0,0],[0,0,0]], E₁₂ = [[0,1,0],[0,0,0],[0,0,0]], E₁₃ = [[0,0,1],[0,0,0],[0,0,0]] E₂₂ = [[0,0,0],[0,1,0],[0,0,0]], E₂₃ = [[0,0,0],[0,0,1],[0,0,0]], E₃₃ = [[0,0,0],[0,0,0],[0,0,1]] This set is also linearly independent because none of them can be written as a linear combination of the others. Therefore, a basis for V is {E₁₁, E₁₂, E₁₃, E₂₂, E₂₃, E₃₃} . The dimension of this vector space is 6.

(b) We need to show that the determinant of the given matrix is zero for infinitely many values of a and b. Let’s calculate the determinant of the matrix: A = [[1, 2, 2], [a, 1, 0], [0, b, b]] Expanding along the first row: det(A) = 1 |[1, 0], [b, b]| – 2 |[a, 0], [0, b]| + 2 * |[a, 1], [0, b]| det(A) = 1 ((1)(b) – (0)(b)) – 2 ((a)(b) – (0)(0)) + 2 * ((a)(b) – (1)(0)) det(A) = 1 (b) – 2 (ab) + 2 * (ab) det(A) = b – 2ab + 2ab det(A) = b The determinant will be zero if and only if b = 0. det(A) = 0 ⇔ b = 0. This equation does not depend on the value of ‘a’. This means that as long as b = 0, ‘a’ can be any real number, and the determinant will always be zero. Since there are infinitely many choices for ‘a’ (a ∈ R), there are infinitely many pairs (a, b) of the form (a, 0) for which the determinant is zero. For example, (1,0), (2,0), (π,0), etc., all result in a determinant of zero. Thus, the statement is true.

Q7. Which of the following statements are true and which are false? Justify your answers with a short proof or a counter-example, whichever is appropriate: 2×5=10 (i) If a matrix is diagonalisable all its eigenvalues are distinct. (ii) If V is a vector space of dimension 7 and U and W are two subspaces of V such that dim(U)=2, dim(W)=6 and dim(U∩W)=1, then V=U⊕W. (iii) Diagonal elements of a Hermitian matrix are always real. (iv) If T: R³→R³ is a surjective linear operator, then ker(T) ≠ {0}. (v) If T₁ and T₂ are invertible linear transformations from R² to R², then T₁+T₂ is also an invertible linear transformation.

Ans. (i) False. Counter-example: Consider the 2×2 identity matrix I₂ = [[1, 0], [0, 1]]. This matrix is already in diagonal form, hence it is diagonalizable. However, its eigenvalues are λ₁ = 1 and λ₂ = 1, which are not distinct. For a matrix to be diagonalizable, it is necessary that for every eigenvalue, the geometric multiplicity equals its algebraic multiplicity, not that all eigenvalues are distinct.

(ii) False. Justification: For a direct sum V = U⊕W to hold, two conditions must be met: 1. V = U + W 2. U ∩ W = {0} Let’s check the first condition: dim(U + W) = dim(U) + dim(W) – dim(U∩W) = 2 + 6 – 1 = 7. Since dim(V) = 7, it is true that V = U+W. However, the second condition, U ∩ W = {0}, is not met. We are given that dim(U∩W) = 1, which means the intersection is not just the zero vector. By definition of a direct sum, the intersection must be trivial. Therefore, V is not the direct sum of U and W.

(iii) True. Proof: A matrix A is Hermitian if A = Aᴴ, where Aᴴ = (Aᵀ)̄ (the conjugate transpose). This means aᵢⱼ = āⱼᵢ. For a diagonal element, i = j. So, the condition becomes aᵢᵢ = āᵢᵢ. For a complex number z = x + iy, its conjugate is z̄ = x – iy. If z = z̄, then x + iy = x – iy, which implies 2iy = 0, and therefore y = 0. This means the imaginary part of the number is zero. Thus, aᵢᵢ must be a real number.

(iv) False. Justification: If T: R³ → R³ is a surjective linear operator, it means its range is the entire codomain, Range(T) = R³. Therefore, dim(Range(T)) = 3. By the Rank-Nullity Theorem: dim(ker(T)) + dim(Range(T)) = dim(Domain) dim(ker(T)) + 3 = 3 This implies that dim(ker(T)) = 0. The only subspace with dimension 0 is the trivial subspace {0}. Thus, ker(T) = {0}, which contradicts the statement ker(T) ≠ {0}.

(v) False. Counter-example: Let T₁: R² → R² be the identity transformation, T₁(v) = v. This is invertible (its inverse is itself). Let T₂: R² → R² be the transformation T₂(v) = -v. This is also invertible (its inverse is also itself). Now consider their sum, T = T₁ + T₂. T(v) = (T₁ + T₂)(v) = T₁(v) + T₂(v) = v + (-v) = 0, for all v ∈ R². T is the zero transformation, which maps every vector to the zero vector. The zero transformation is not invertible because it is not one-to-one (injective); its kernel is the entire space R², not just {0}.