Question

l. (20 pulita total) Given is A-| 1 3 . Tins matrix has eigruvalunAi-Ag-2 and la-s. (a) (2 points) Explain wby the matrix A can be orthogonally dingoaleed (b) (4 points) Show that i is a l-for the eigenspace eurrespcoding to 0 Xi- "xz- -2 (c) (3 points) Show thatxis a basis for the eigeneqpace responding to A (d) (5 points) plc by an orthoyonal basis Hint: This can be dose by applying the Gra- by Schmidt process to (x. Xl to get (vi.V).Or, alternatively, by setting vx and replncing -x+bxa and determining e and b so that voLv (e) (3 points) Nonwalee the vectors vi. va, and va s to replace the orthogonal basis v VaVs by an orthosormal basisu u () (3 points) Use the veetors obtained in part (e) to make an orthogonal matrix P which allows one to orthogonally dingomalize A. Verify that D PTAP, where D is a diagonal matrix

Answer 1

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