# Proof about Diagonalization of A

The question asks WHY is it true that $$A^{n} = PD^{n}P^{-1}$$

I can never do proper proving in algebra; what I almost know for sure is that a proof by induction is the way to go here. But how do you go about proving with induction, whenever I think of starting a proof, I just end up blocked at the initial hypothesis. How do you carry on?

Moreover, is it legitimate to say that for $$A^{1} = PD^{1}P^{-1}$$ It holds true simply because of the hypothesis/theorems with started out with?

Thanks

Let me answer your second question first: yes. When $n=1$, the statement is simply the diagonalizability of $A$. (Note that it does not hold for general matrix in a field which is not algebraically closed.)
This serves as a base case for induction. Following the usual steps, we set up an inductive hypothesis: Suppose for some $k \geq 1$ we have $$A^k = PD^kP^{-1}.$$
Then when $n = k+1$, we have $$A^{k+1} = A^k \,A = (PD^kP^{-1})(PDP^{-1}) = PD^k(P^{-1}P)DP^{-1} = PD^kDP^{-1} = PD^{k+1}P^{-1}.$$ This shows the statement is true when $n=k+1$. The proof by induction is completed.
initially when you suppose that $A^1=PD^1P^{-1}$ is because $A$ is diagonalizable. Then using induction method, you suppose that is true for the case $n$, then you have that prove that this is true for $n+1$:
$A^{n+1}=A^nA^1=(PD^nP^{-1})(PD^1P^{-1}),$