Proving $(I -cP)^{-1} = I+ \left(\frac{c}{1-c}\right)P$ , $P$ idempotent matrix.

Question

Given that a matrix $P$ is idempotent how to prove the following relation: $$(I -cP)^{-1} = I+ \left(\frac{c}{1-c}\right)P$$ $c$ is any real constant.

Answer 1

Let's multiply these two matrices together: $$(I-cP)\left(I+\frac{c}{1-c}P\right)$$ If we get $I$ as a result, then we know that they are inverses. Use the distributive property: $$I^2-cP+\frac{c}{1-c}P-\frac{c^2}{1-c}P^2$$ Use idempotence of $I$ and $P$: $$I-cP+\frac{c}{1-c}P-\frac{c^2}{1-c}P$$ Factor out a $P$ from the last three terms: $$I+\left(-c+\frac{c}{1-c}-\frac{c^2}{1-c}\right)P$$ Use algebra to simplify: $$I+0P=I$$ Thus, the two matrices we started with in the original product are inverses.

Answer 2

$$\begin{aligned} (I - cP)^{-1} &= I + cP + c^2P^2 + c^3P^3 + \ldots &\text{(Maclaurin series)}\\ &= I + (c + c^2 + c^3 + \ldots) P &\text{(idempotence)}\\ &= I + \frac{c}{1-c} P &\text{(geometric series)} \end{aligned}$$