Jawaharlal Nehru Technological University Hyderabad B.Tech I Year MATHEMATICS-I Previous Question Paper (October/November - 2020 - R15)

Let A be a skew-Hermitian matrix, which satisfies the condition A^θ = -A, where A^θ is the conjugate transpose of A. Let λ be an eigenvalue of A and x be the corresponding non-zero eigenvector, such that Ax = λx.

Premultiplying by x^θ, we have x^θAx = x^θλx = λ(x^θx). Taking the conjugate transpose of both sides of Ax = λx gives x^θA^θ = λ^*x^θ. Substituting A^θ = -A, we get -x^θAx = λ^*x^θ, or x^θAx = -λ^*(x^θx).

Equating the two expressions for x^θAx, we have λ(x^θx) = -λ^*(x^θx). Since x is a non-zero vector, x^θx = Σ|x_i|² > 0. Thus, λ = -λ^*, or λ + λ^* = 0. If λ = a + ib, then (a + ib) + (a - ib) = 2a = 0, implying a = 0. Therefore, λ is purely imaginary or zero.

Let A = [[i, 0, 0], [0, 0, i], [0, i, 0]]. A matrix is unitary if A^θA = I. Calculating A^θ = [[-i, 0, 0], [0, 0, -i], [0, -i, 0]], we find A^θA = [[(-i)(i), 0, 0], [0, (-i)(i), 0], [0, 0, (-i)(i)]] = [[1, 0, 0], [0, 1, 0], [0, 0, 1]] = I. Thus, A is unitary.

To find eigenvalues, solve det(A - λI) = 0: (i - λ)((-λ)(-λ) - i²) = 0, which simplifies to (i - λ)(λ² + 1) = 0. The eigenvalues are λ = i, λ = i, and λ = -i.

For λ = i, (A - iI)x = 0 gives [[0, 0, 0], [0, -i, i], [0, i, -i]]x = 0. This yields x₂ = x₃, so the eigenvector is [1, 0, 0]^T and [0, 1, 1]^T. For λ = -i, (A + iI)x = 0 gives [[2i, 0, 0], [0, i, i], [0, i, i]]x = 0. This yields x₁ = 0 and x₂ = -x₃, so the eigenvector is [0, 1, -1]^T.