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.
Best Answer
Guide:
\begin{align}(I-cP)(I+dP)&=I-cP+dP-cdP^2 \\ &=I+(-c+d-cd)P\end{align}
We just have to solve for $d$ in
$$-c+d-cd=0$$
I will leave this task to you.