# To prove $P^2 = P$ meaning $P^n = P$ and $(I-P)^n = I-P$

So the question is that when a square matrix $$P$$ fulfills the equality that $$P^2 = P$$

Prove that $$P^n = P$$ and $$(I-P)^n = I -P$$

So for first $$P^n = P$$

I tried to prove by

$$P^n = P^{n-2}P^2$$

So given $$P^2 = P$$

$$P^n = P^{n-2}P^2 = P^{n-2}P = P^{n-1}$$

By same way it we can deduce

$$P^{n-1} = P^{n-2} = P^{n-3} = ... = P^2 = P$$

or my second attempt is that as $$P^2 = P$$

$$P^{-1}PP = P^{-1}P$$

$$P = I$$

Hence that

$$P^n = I^n = I = P$$

Which do you think is more suitable as proving???

And for second question now that we know $$P = I$$

For $$(I-P)^n$$,

I thought that $$I - P = 0$$, a zero matrix

and hence that $$0^n = 0$$

Is my proving, incomplete or not logical?? Please give feedback

• In general, $P$ is a singular matrix, which means that its inverse $P^{-1}$ does not exist. Use a) to prove b) Commented Apr 23, 2022 at 8:52
• For the second eqaulity, note that $(I-P)^2=I^2-2P+P^2=I-P$. Commented Apr 23, 2022 at 8:54
• Do you really think $P=I$ is correct? Note that the zer0 matrix also satisfies the hypothesis. Commented Apr 23, 2022 at 8:56
• hints : For the first one, use induction on $n$. For the second one, use the binomial theorem on $(I-P)^n$. Commented Apr 23, 2022 at 8:58
• Yes, that's a good idea. You're right, if $P$ is regular, then $P=I$. Commented Apr 23, 2022 at 9:00

Given that $$P^2 = P$$.

To prove that $$P^n = P, \ \ \ (n \in \mathbf{N})$$ we can use the principle of induction.

Clearly, $$P^n = P$$ for $$n = 1$$.

Assume that $$P^m = P$$ for some positive integer $$m$$.

Then $$P^{m + 1} = P^m P = P P = P^2 = P$$

This completes the induction.

Hence, $$P^n = P, \ \ \mbox{for all} \ \ n \in \mathbf{N}$$

Next, when $$n = 1$$, $$(I - P)^n = (I - P)^1 = I - P$$

Assume that $$(I - P)^m = I - P$$ for some positive integer $$m$$.

Then we find that $$(I - P)^{m + 1} = (I - P)^m (I - P) = (I - P) (I - P) = I - 2 P + P^2$$

Since $$P^2 = P$$, we get $$(I - P)^{m + 1} = I - 2 P + P = I - P$$

This completes the induction.

Hence, we conclude that $$(I - P)^n = (I - P) \ \ \mbox{for all} \ \ n \in \mathbf{N}$$

• There's no need to repeat the same inductive proof for the second case. Instead. simply verify that $\,(I-P)^2 = I-P,\,$ so the claim follows from the first proof, by replacing $\,P\,$ by $\,I-P.\ \$ Commented May 12, 2022 at 16:18