Question: I need the problem 4 Please Problem 1 (20 %): For a dynamic equation x(k+1) = A x(k) with x(0) = xo = -1 and

I need the problem 4 Please Problem 1 (20 %): For

I need the problem 4 Please

Problem 1 (20 %): For a dynamic equation x(k+1) = A x(k) with x(0) = xo = -1 and A = A= 1.5 0.5 1 1 -1 complete the following tasks: 0.5 0.5 (a) For the given x(0) and A, calculate x(1) and x(3); 0 0 Problem 3 (20%): For a dynamic equation x(k+1)=Ax(k) with x(0) = xo, consider the Jordan canonical form J = diag{Im (N1),...,Jm, (aq)} of A, and complete the following tasks: (a) for J2(21) = 1 calculate (J2 (21)) and (J2 (21)); and 1 (b) for J3 (21) = 1 calculate (J3(21)). 0 0 2 (c) derive the closed-form expression of (J2 (21))* in terms of and k; A kak {k{k 1)24-2 (d) verify (J3 (21) * = 0 khk k-1 for k=1,2,3,4; and 0 M (e) formulate a lemma to summarize some properties of Jk in the general case. 0 Remark: In the case when some hi are complex, the matrix S is complex (and not dependent on k), and with the expressions of (Im;(2))", we can still get the expression Ak M,(R) in terms of k. Problem 4 (20%): For a dynamic equation x(k+1) = Ax(k) with x(0)=xo, consider the Jordan canonical form J = diag{Jm, (21),...,Jm. (g) } of A and complete the following tasks: (a) for J= diag{J. (1), (22),J. (23)}, find the condition on ; to ensure limx-jk = 03x3; (b) for a matrix A EM3(R) with J in Part (a), find the condition on A to make limk-Ak=03x3; (c) for A= A in Part (a) of Problem 1, check whether lim-Ak=03x3 or not; (d) for J = diag{12(1), (1)}, find the condition on, to ensure limx-toyk = 03x3; and (e) formulate a theorem to specify the necessary and sufficient conditions on a matrix A E M.(R) to ensure limkAk Onxn, and explain why? Problem 5 (20 %): To further study the solution x(k) of a dynamic equation x(k+1) = Ax(k) with x(k) = (x1(k),...,xn(k)]' R" and x(0) = xo, using the result of Part (a) of Problem 2, complete the following tasks: (a) for A = A1 in Problem 1, find a nonzero xo = [X01.X02.X0317 ER such that x(k) remains as a

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