Given $P$ idempotent, show that $I-P$ is idempotent.

Question

As title, given $V$ as a finite dimensional vector space over $F$. Suppose the linear transformation $P: V \rightarrow V$ is idempotent. I need to show that $I-P$ is idempotent.

I know that a matrix is idempotent if $P^2=P$, and that $P$ must be a square matrix.

So I need to show that $(I-P)^2 = (I-P)$? I don't have a clue on how I should proceed.

Thanks in advance!

Answer 1

$(I-P)^2=I-2P+P^2=I-2P+P=I-P$.

Answer 2

Direct brute force computation!

$(I-P)^2=(I-P)(I-P)$. Now expand it using distributive law, and keep in mind that $P^2=P$