Question

Let A be a diagonalizable n x n matrix and let P be an invertible n x n matrix such that B = p-1AP is the diagonal form of A. Prove that A* = Pokp-1, where k is a positive integer. Use the result above to find the indicated power of A. 10 18 A = -6 -11 18].46 A = 11

Answer 1

Let A be a diagonalize nan matsia and let p be an investible nan mation such that B = P'AP B is the diagonal forms of A. multiply with matrise p in equation both side from left, PB= PPTAP (ppl= I) PB = AP e both side, from right, multiply with matrix pt in equation PB pt = App! = PBPT = squaring both side, - (PBP1) 2 - AS (PBP1) ( PB PY) = AY - Poup = A² multiply 1 with matrix A. PB²ptA= - PB"p" PB Pl= AB (A = PBP", uringo) PB²p' = A (prip= I) ARA P 3 feeing 0.0 W we say that similarly, NK, where k is an integer. Ak - PB kpl Here, B = diagonal mation; which contains eigen values of A P- estrogenat matisc, which has columns as yeigen vectos. (linearly independent)

10 16 for eigen values (d), det (A-d I) TO 10-1 1 A-dl= -6 11-d (10-1) (-11-d) + 10b=0 d+d-110 + 108=0 +1800 dtd-250 i de-a do = 1 Eigen vector corresponding to do = -2 (A-(-2) I)x=0 B00 - 2x+3y=0 let not, y=- 200 = 3 Eigen space = {(1, -24) | FER) Eigen vector corresponding to d = 1 (A-I) X=0 "]016 - 6 x + 2y =o (AER) let x=24 - y = - = - 2+ = + Eigen space = { (2t', -t) I t'e R}

let t=3 and t=1 we get 5 3 P th & IP= -3+4=1 e pt = I adj p = adj p = & BE PI 3 A = PB pt - AO = PBR (26 o 3 3 192 lo -18 -192+4 -384 +6 -188 ~378 128-2 256-3 253 Hence, A -188 -378 126 253