$I-cP$ Invertible Matrix Question: Assume P is a nonzero $n$ x $n$ matrix, $n \ge 2$, such that $P^2=P$. Let $c\in\mathbb R, c\ne1$. Show that the matrix $I-cP$ is invertible and find its inverse.
I'm having trouble going about this question. By manipulating $Av=\lambda v$, I get that P has eigenvalues $0$ and $1$. I know that if the determinant of $I-cP$ is nonzero then it is invertible, and that if it has a trival kernel it is invertible. But I don't know how to proceed. Any help would be appreciated.

If $c = 0$ then $I-cP$ is clearly invertible. So assume $c \neq 0$.
If $I-cP$  is not invertible, then there is a vector $x \neq 0$ with $(I-cP)x =0$, that is, $x = cPx$. This implies $Px \neq 0$. Multiplying both sides by $P$ we have $Px = cPx$, i.e., $(1-c)Px = 0$, since $Px \neq 0$ we must have $1-c=0$, i.e., $c=1$. A contradiction.
