Hello,

I have the following problem: Let $V$ be a vector space and $E$ an idempotent linear operator on $V$. Prove that $(I+E)$ is invertible. Find $(I+E)^{-1}$.

Since the problem is for a project not an assignment I cheated a bit and found a worked solution which declares $(I+E)^{-1}=(1-\frac{1}{2}E)$. Now if you know this is the inverse then everything is very easy. Just calculate $(I+E)^{-1}(I-\frac{1}{2}E)$ and you see that it is equal to $I$ and then find $(I-\frac{1}{2}E)(I+E)^{-1}=I$ and apply a theorem ($AB=I$ and $CA=I$ implies exists $A^{-1}$).

The problem is I can't figure out how to deduce that exact form of the inverse.

I tried $(I+E)(X+Y)$ constructions but could not find a second equation so I'm stuck with 1 equation in two unknowns.

I tried setting $X=I$, which seems reasonable in general but not so intuitive, but then in the process of solving for $Y$ I get an expression which requires division.

Like, $(I+E)(I+Y)=I^2+IY+EI+EY$. Since I want this to equal $I$, set $I^2+IY+EI+EY=I$ and solve for $Y$, $(I+E)Y=-E \iff Y=\frac{-E}{I+E}$. Perhaps this expression makes sense perhaps not I don't know. I know in general that under certain conditions you can define both left and right division of linear transformations but this seems too complicated for this.

I've also tried setting $IY+EI+EY=0$ since the truth of that would cause $I^2+IY+EI+EY=I$ as $I$ is idempotent. But then this appears only to tell me $(I+E)Y=-E \iff Y=\frac{-E}{I+E}$ again.

Regards