The result stated in the title is thoroughly standard - or that's the impression I got. I seem to remember seeing it stated somewhere in a book I was reading in the library, and then reverse-engineering a proof from the hints given.

For a preprint I'm working on, it would be preferable to give a precise citation from a "standard text", rather than spend time giving the proof "for the reader's convenience". Any suggestions?

If anyone's interested, an outline of a proof is as follows: consider an idempotent P in B(H), with H a Hilbert space, and note that we can always decompose H as an orthogonal sum with respect to which P has the block matrix form

$$ P= \left(\begin{matrix} I & R \\\\ 0 & 0 \end{matrix}\right) $$

Then it's not hard to see that conjugating $P$ by the invertible operator

$$ S= \left( \begin{matrix} I & R \\\\ 0 & I \end{matrix} \right) $$

will give

$$ E = \left(\begin{matrix} I & 0 \\\\\ 0 & 0 \end{matrix} \right) $$

Since $S= I+P-E$, it suffices to show that $E$ is in the C*-algebra generated by I and P (for then S will also lie in that algebra, and then we're done). This follows by messing around with various combinations of P, its adjoint, and their products.