## Answers

Given: A = [3 1 1 1 3 1 1 1 3] Eigen values lambda_1 = lambda_2 = 2, lambda_3 = 5 (a) since the matrix A is symmetric [A^T = A] therefore it is orthogonally diagonalizable. (b) To find Eigen space for lambda_1 = lambda_2 = 2. Solve the system (A - lambda I)X = 0 (A - 2I)X = 0 implies (1 1 1 1 1 1 1 1 1)(x_1 x_2 x_3) = (0 0 0) implies x_1 + x_2 + x_3 = 0 [single equation & 3 unknowns] therefore Let x_2 = t, x_3 = 5 therefore x_1 = -t - s therefore X = (-t - s t s) = (-1 1 0)t + (-1 0 1)s therefore x_1 = (-1 1 0), x_2 = (-1 0 1) \r\n (c) The Eigen space for lambda_3 = 5 is solution to (A - lambda_3 I)X = 0 implies (-2 1 1 1 -2 1 1 1 -2)(x_1 x_2 x_3) = (0 0

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