Question

If A= diag[a1, a2, a3], then for any integer n > 1 show that A^n= diag[a1^n, a2^n, a3^n].

Answer 1

PLEASE MARK AS BRAINLIST

Answer 2

Step-by-step explanation:

If a = diag[a1, a2, a3,] n=1

Claim $a^{n}$ = [a1^n, a2^n, a3^n]

Let m = 2

a^2 = [a1 0 0, 0 a2 0, 0 0 a3] x [a1 0 0, 0 a2 0, 0 0 a3] = [a1^2 0 0, 0 a2^2 0, 0 0 a3^2]

Let if it is true m=n-1 then

$A^{n-1}$= [a1^n-1 0 0, 0 a2^n-1 0, 0 0 a3^n-1]

Now m=n then,

$A^{n}$ = $A^{n-1}$ x A = [a1^n-1 0 0, 0 a2^n-1 0, 0 0 a3^n-1] x [a1 0 0, 0 a2 0, 0 0 a3] = [a1^n 0 0, 0 a2^n 0, 0 0 a3^n]

$A^{n}$ = diag[a1^n, a2^n, a3^n]