Question

Let A be a diagonalizable n × n matrix and let P be an invertible n × n matrix such that B = P−1AP is the diagonal form of A. Prove that Ak = PBkP−1, where k is a positive integer. Use the result above to find the indicated power of A. A = −4 0 4 −3 −1 4 −6 0 6 , A5

Answer 1

1 solution We have A be diggenalizable nan matrix such that B=P'AP where P cu an invertible nan matix Pre multiply both sides by P, we get PB = P(P-AP) PB= (PP-1) AP = PB = AP Post multiply both sidel by pl we get PBPT= (AP) P-1 PBP-1= A (PP-1) 1 PBP-1 A A= PBP - clearly the result is true for kal

Assume that result w true far kam 1.e. AM PBM pl we will prove the result for komt Par Kamal Amt AMA (Pomp-11.A [by (2) (P8"p-).CPBP-') [By ()] = PB (pp) BP1 = PB (I) BP- P gmtip- 10 브 Anti Pguntip- Therefore, by Principle of Mathematical Induction result in true all positive integer k m Pom

ie la = PBkpl Now, we have 1-4 As -3 -1 4 4 -6 6 The eigen value of a are given by IA-dl=0 4 4 1-4-0 -3 -6 o =0 6-d 20 20 >> (42)[(4-1)(6-d) +243 =) (-1-d] [-24 +40-6d to +24]- 5) - (1+0) [12_213-0 (1+r) (d-2)=0

A =) d= 0,-1,2 24 Let x= X2 be the eigen vector cosoresponding to do- 93 Then (A +1 )x=0 -3 3 - 6 4 4 7 X₂ Applying & Be A and Rgr R3-24, weget O o o ú' ::00 0 0 - 3x +483=0 and - X2=0 34= 483 and ag ao Ta =) x=0 and X z=0

24 NE 1: w the basis of eigen sbare Quociated coich ds-1 Agein let X= be dhe egen veetcon cosetes pending to dzo *3 Then AX=0 24 -4 -3 4 o 6 o opplying & try and Ry - 11 l we get 1 0 - 1 -3 - 4 IND 0 24

Applying Rg Rg + + 8R and Rg Ry -R, we get 1 1 0 13 4 - 1320 and g- X3 20 =) x = x3 and 2 =23 x = x 2 X3 ng 1) - 8 16. in the basis of eigenspace assoucted with d=0

4 Again, let X= be the elgonveces comesponding Ma 33 00 d=2 Then (A-21 )X =0 -6 o 24 = ) - 3 -3 L-6 INI o Applying Rg + Pg - LR and Rg > R3 - Q, we get -6 24 o wo 0 2 10 e o - 624 +44₂ = 0 and -34 +2X₃20 0 64 = U23 and 34 = 223 Take azot 24 = pt and X &t and az at

8 쁜 213 2/3 t *3 1 {Bly in the basis of eigenspace corresponding to d=2 Then we have three .L.P. eigen vectors so that A i diagonalizable 0 - 2 Also il P= then > - 0 1 3 0 0 P"Apa diag (0,-1, 2) = O o -B 0 A= PBPI at p Bk pl * A3 = PBS pl

6 [com 0 0 P 0 o Ipt 0 0 Q35 O 0 0 = CHE P 0 1 - 0 py LO o 32 NBLO P= =) pia 3 0 -२ 0 3 0 o O a) AS o -1 o 3 0-2 2 3 o 32 0 0 -1 -) 0 3 > AS 64 64 96 -2 0 - o O - 67 0 66 85 -67 O 66 -99 0 98