For b), the equation 12) is

1. (40 points) When discretizing the differential equation = -u" (2) u(0) f(x), re [0,1 u(1) = 0, = (1) using second order centered differences, we obtained the tridiagonal matrix -1 0 0 0 0 N -1 2 -1 0 0 0 -1 2 -1 0 0 A= 1 h2 ... ... ... ... (2) 0 0 0 -1 2 -1 0 0 0 0 0 -1 2 -1 0 0 0 0 0 -1 2 (3) In components, remember that this means Ui+1 - 2u; + wi-1 h2 f(xi) i = 1,2,...,n, where h = 1/(n + 1). (a). (10 points) For each k = 1, 2, ..., n, show that the vector u(k) given by (k) ki u i = 1, 2, ..., n+1 = sin (n+1) (4) is an eigenvector of the matrix A, and determine the corresponding eigenvalue Ak. (b). (5 points) The Jacobi iteration matrix is defined as T, =D (L+U), (5) where A = L +D+U. Show that the vectors (12) are also eigenvectors of Ty and find its eigenvalues. Hint: Prove that since D = dI (constant multiple of the identity), Tjx=yx Ax = d(u +1)x. (c). (5 points) Determine the spectral radius of TJ, (6) p(TJ) = max ukl. k=1,...,n (7) (d). (5 points) The Jacobi iteration can be written as x(k+1) = Tjx(k) + c. (8) From the previous steps, we know that Tj is symmetric and diagonalizable. Use this fact to show that if x* is the (unique) fixed point of (8), then || x) x*||2