induction for idempotent matrix : $P^n = P$

Question

Given that $P^2 = P$ how do i prove by induction that $P^n = P$?

I have tried the following: we know that $P^k = P$ holds for $k = \{1,2\}$. If we now take $k=3$: $$ \begin{align} P^3 &= P^2P \\ &=PP \tag*{($P$ is idempotent)} \\ \\&= P^2 \\&=P \end{align} $$ therefore $P^k = P$ holds for all natural numbers.

however, this seems... incomplete for me... Am I missing something?

Answer 1

Base case: $P^2 = P$. Given.

Induction Hypothesis: Assume $P^k = P$ for some $k > 1$.

Induction Step: We have to prove $P^{k+1} = P$

Now, $P^{k+1} = P^k P = P P = P^2 = P$, where we used induction hypothesis while going from 2nd to 3rd equality. Thus, it is proved for all $k \ge 1$.

Answer 2

Suppose $P^{n-1}=P$. Then $$\begin{align*}P^n&=P(P^{n-1})\\ &=PP\\ &=P^2\\ &=P. \end{align*} $$ We're given $P^2=P$, so by induction on $n$, we're done.

Thinking of induction as reaching back to the previous cases instead of reaching forward to the next case can be insightful.