If we know that $P^2=P$, show that $(I+P)^{-1}=I-\frac{1}{2}P$

Question

the problem is as follows. given that P is a matrix with the property $P^2=P$ show that $(I+P)^{-1}=I-\frac{1}{2}P$. I tried squaring both quantities and using the fact that I can interchange the exponents on the RHS but it did not yield anything helpful.

Answer 1

Since $(I+P)(I-\tfrac12P)=I+\tfrac12P-\tfrac12P^2=I$, $(I+P)^{-1}=I-\tfrac12P$.

Answer 2

$(I+P)(I-\frac{1}{2}P) = I - \frac{1}{2}P + P - \frac{1}{2}P^2 = I - \frac{1}{2}P + P - \frac{1}{2}P = I$.

Thus $I-\frac{1}{2}P$ is an inverse of $I+P$. But the inverse is uniquely determined and therefore $(I+P)^{-1} = I-\frac{1}{2}P$.

Answer 3

If you have a candidate matrix that you'd like to show is the inverse of $(I+P)$, simply multiply it by $(I+P)$ and verify that you get the identity matrix $I$.

Try it yourself, then click to reveal the calculation:

! $$ \begin{align*} (I - \tfrac{1}{2}P) (I + P) &= I + P - \tfrac{1}{2}P - \tfrac{1}{2}\color{red}{P^2} \\ &= I + P - \tfrac{1}{2}P - \tfrac{1}{2}\color{red}{P} = I \end{align*} $$