Question

5. A 3 × 3 matrix is given by A=1020 -i 0 1 (a) Verify that A is hermitian. (b) Calculate Tr (A) and det (A), where det (A) represents the determinant of A (c) Find the eigenvalues of A. Check that their product and sum are consistent with Prob. (5b) (d) Write down the diagonalized version of A (e) Find the three orthonormal eigenvectors of A. (f) Construct the unitary matrix U that diagonalizes A, and show explicitly that the similarity transformation using U reduces A to the appropriate diagonal form. Did this similarity transformation yield the same diagonalized matrix you wrote down in Prob. (5d). If not, why not?

Answer 1

Given a 3 times 3 matrix A = (1 0 -i 0 2 0 i 0 1) a) Condition for Hermiston is A^t = A or A^t = (A*)^T = A Now A* = (1 0 i 0 2 0 -i 0 1) (A* means conjugate from of the matrix A.) \r\n (A*)^T = (1 0 -i 0 2 0 i 0 1) [(A*)^T means transpose of the matrix A*) Thus A^t = (A*)^T = A which is a Hermitian b) A is a Hermitian matrix tr(A) is the sum of diagonal elements i.e tr(A) = 1 + 2 + 1 = 4 det (A) = det (1 0 -i 0 2 0 i 0 1) = Vertical-bar 1 0 -i 0 2 0 i 0 1 Vertical-bar = 1[2 - 0] - 0[0 - 0] + i[0 + 2i] (i times i = -i) = 2 + i times 2i = 2 - 2 = 0 therefore det (A) = 0