Please help me with parts (d) and (e), thanks.

Eigenvectors and Diagonalization (a) Let A be an n x n matrix with n linearly independent eigenvectors 71, 72, ..., Un, and corre- sponding eigenvalues A1, A2, ..., An. Define V to be a matrix with 71, 32, ..., Un as its columns, V = 51 32 ... Un . Show that AV = VA, where A = diag (A1, 12,..., An), a diagonal matrix with the eigenvalues of A as its diagonal entries. (b) Argue that V is invertible, and therefore A = VAV-1. (9) (HINT: What condition on a matrix's columns means that it would be invertible? It is fine to cite the appropriate result from 16A.) (c) Write A in terms of the matrices A, V, and V-1. (d) A matrix A is deemed diagonalizable if there exists a square matrix U so that A can be written in the form A = UDU- for the choice of an appropriate diagonal matrix D. Show that the columns of U must be eigenvectors of the matrix A, and that the entries of D must be eigenvalues of A. (HINT: Recall the definition of an eigenvector (i.e., AU = Av). Then, recall what U U is. Lastly, consider how matrix multiplication works column-wise.) The previous part shows that the only way to diagonalize A is using its eigenvalues/ eigenvectors. Now we will explore a payoff for diagonalizing A - an operation that diagonalization makes much simpler.PAGE - 1 solution - ( a. ) Here is given - In AUK = dKUK V K-- 1, 2, 3,... .. let A = [aijJ Isojan and Vic = EV1,, Ukz, . - - VICK, isn-0 aijoki = AKUKP Now am Unz AV 92n Viz 921 any Uin Onn - - Sims . . ani 12j ani Uni Un V21 O 4 2 VA Unz\f(e) For a matrix A and a positive integer k, we define the exponent to be Ak A . . . . . . A . A (10) k times Let's assume that matrix A is diagonalizable with eigenvalues A1, 12, ..., An, and corresponding eigenvectors 71, 32, . .., Un (i.e. the n eigenvectors are all linearly independent). Show that A* has eigenvalues 14, 12, ..., A, and eigenvectors 71, 72, ..., Un. Conclude that Ak is diagonalizable.\f