if A ,P,D are square matrix and A=P*D*P^-1 prove that a)A^2=P*D^2*P^-1
b)prove with induction for every m=0 we have A^m=P*D^m*P^-1

Question

if A ,P,D are  square matrix and A=P*D*P^-1  prove that a)A^2=P*D^2*P^-1
b)prove with induction for every m>=0 we have A^m=P*D^m*P^-1

jamesj · Answer

Have you tried calculating?  Part a is very easy.  Just do the calculation.

anonymous · Answer

yes IS IT  this way A^2 =P*P*D^2*P^-1*P^-1 SINCE P*P^-1=I  A^2=P*D^2*P^-1?

jamesj · Answer

yes

anonymous · Answer

can you give me a little help for part b) i know how to prove it for m=1 for m=kbut i don't know hoe to continue

anonymous · Answer

how*

jamesj · Answer

Let P(j) be the hypothesis that $$ A^j = PD^jP^{-1} $$

Now P(0) is true, and you also know that P(1) and part (a) shows P(2) is true.

jamesj · Answer

Now you need to show that if P(k) is true, that this implies P(k+1).

jamesj · Answer

Suppose then P(k).  That is,
$$ A^k = PD^kP^{-1} $$
Now $$A^{k+1} = A^k A $$
$$ = (PD^kP^{-1})A, \ \ \ \ \hbox{by hypothesis} $$
$$ = ... $$
you take it from here.

anonymous · Answer

in the end it is A^k+1=P*P^-1*D^k+1

anonymous · Answer

can you give me an advice for this exercise too if C is a square matrix  that proves this C^2+C=I then PROVE THAT  C^4=2I-3C

jamesj · Answer

$$ (PD^kP^{-1})A $$
$$= (PD^kP^{-1})(PDP) $$
$$ = PD^k(P^{-1}P)DP^{-1} $$
$$ = PD^kIDP^{-1} $$
$$ = PD^{k+1}P^{-1} $$

anonymous · Answer

i did it this way