Let $E$ be an idempotent operator: If $E^2 = E$ is an operator on a finite dimensional vector space $V$. If $E^*E=EE^*$,  find $$(E + E^* -I)^{-1}$$
I know this operator is invertible, but I cannot find an explicit formula for its inverse.
 A: (the original answer is below)
The condition $E^*E=EE^*$ is not needed; neither is finite dimension. 
Let us do the finite-dimensional case first. Suppose that there exists $x$ with $(E+E^*)x=x$. Writing $x=Ex+(I-E)x $ we get 
$$
E^*x=(I-E)x,
$$
so $EE^*x=0$. It follows that $E^*x=0$. The roles of $E$ and $E^*$ are symmetric, so we also get that $Ex=0$. Then $x=(E+E^*)x=0$. Thus $E+E^*-I$ has trivial kernel, and so it is invertible. 
In the arbitrary case, we can still use the above idea if $E$ is bounded. Note that $E+E^*$ is selfadjoint, so all the elements in its spectrum are approximate eigenvalues. Suppose that 
$$\tag1
(E+E^*)x_n-x_n\xrightarrow[n\to\infty]{}0.
$$
Since $(E+E^*)x_n-x_n=E^*x_n-(I-E)x_n$ and $E$ is bounded, we get $EE^*x_n\to0$. Then 
$$
\|E^*x_n\|^2=\langle E^*x_n,E^*x_n\rangle=\langle EE^*x_n,x_n\rangle\to0, 
$$
so $E^*x_n\to0$. As the roles of $E$ and $E^*$ are symmetric, we also get $Ex_n\to0$. But now $(1)$ becomes 
$$\tag2
x_n\xrightarrow[n\to\infty]{}0.
$$
To have $1$ in the spectrum of $E+E^*$, we would need $(1)$ to hold for a sequence with $\|x_n\|=1$, and by $(2)$ this is impossible. So $E+E^*-I$ is invertible. 

(old answer)
This is a weird formulation. If $E^2=E$, then its eigenvalues are necessarily $0$ and $1$. As it is normal, it is unitarily equivalent to a diagonal operator $D$, where the diagonal consists of $0$ and $1$ according to their multiplicites. The point is that $E$ is then selfadjoint. So you are looking at $(2E-I)^{-1}$. 
If you square,
$$
(2E-I)^2=4E+I-4E=I. 
$$
So $$(2E-I)^{-1}=2E-I.$$ 
