Expression of a non-orthogonal projection in a $C^*$ algebra via an orthogonal one A paper I'm currently reading uses the following fact. If $A$ is a unital $C^*$-algebra, $P=P^2\in A$, then there are $T, F\in A$ s.t. $F$ is an orthogonal projection ($F=F^*=F^2$) and
$$P=F+FT(1-F).\tag{1}$$
The question:

Since no proof or further reference is given, I assume, that this fact
  is obvious from some point of view. Is it? What do I have to learn, so
  the similar facts would be obvious for me?

P.S. Here is a proof I came up with after some trial and error.
We would like to define $F=f(PP^*)$, where $f(x)=0$ if $x=0$ and $f(x)=1$ otherwise. To check, that $F$ is well-defined we need to check, that $f$ is continuous on the spectrum of $PP^*$. To do this it is enough to show, that the open interval $(0,1)$ doesn't intersect with its spectrum or, equivalently, that $(2PP^*-1)^2\geq 1$. Expanding we see, that this is equivalent to $PP^*PP^* \geq PP^*$.
Define $A=P-P^*$. Then $A^*A\geq 0$. On the other hand expanding we get $A^*A=PP^*+P^*P-P-P^*$, so $PP^*+P^*P\geq P+P^*$. Multiplying by $P$ on the left and by $P^*$ on the right we get $PP^* + PP^*PP^*\geq 2 PP^*$ or $PP^*PP^*\geq PP^*$. Thus $F=f(PP^*)$ is indeed well defined.
Since $f(x)$ is real valued and $f(x)=f(x)^2$ we have $F=F^*=F^2$. To show, that $FP=P$ notice, that $f(x)x=x$ for $x\geq 0$. Therefore $FPP^*=PP^*F=PP^*$ and
$(FP-P)(FP-P)^*=(F-1)PP^*(F-1)=0(F-1)=0$. Thus indeed $FP=P$.
Thus we have 
$$P=F+(P-F)=F+F(P-1).$$
Finally, notice, that
$(P-1)F=(P-1)PP^*F=0P^*F=0$, so $P-1=(P-1)(1-F)$ and
$$P=F+F(P-1)(1-F).$$
 A: Just a few remarks in addition to David Handleman's answer and your work. To make sense of ranges and nullspaces below, you can assume your $C^*$-algebra is sitting in $B(H)$. I just added that hoping it makes things more "obvious". But the whole thing works of course in an abstract $C^*$-algebra. Even the notion of range projection.


*

*If such an $F$ exists, it must be the projection (self-adjoint idempotent) onto the range of the idempotent $P$ (this is often called the range projection of $P$), since $PF=F \iff \mbox{im} F\subseteq \ker (I-P)=\mbox{im} P$ and $FP=P\iff \mbox{im} P\subseteq \ker (I-F)=\mbox{im} F$. Altogether, these two conditions are equivalent to $PA=FA$.

*Now for every idempotent $P$, the range projection of $P$ is given, for instance, by the formula $$F=P(P+P^*-I)^{-2} P^*$$ where the existence of the central factor follows from the fact that $(P+P^*-I)^2=I-(P-P^*)^2=I+H$ with $H=(P-P^*)^*(P-P^*)$ is a positive element. Simple algebraic manipulations show that this formula yields the range projection of $P$: that is $P^2=P=P^*$, $PF=F$, and $FP=P$. The formula also shows that $F$ lies in the $C^*$-algebra generated by $P$.

*To understand what $T$ could be, it is helpful to know about Peirce decomposition with respect to an orthogonal decomposition of the unit: here $I=F\oplus (I-F)$. In the latter, we have
$$ P=\pmatrix{A& B\\C&D} \quad A=FPF=F\quad\;B=FP(I-F)\quad C=(I-F)PF=0\quad D=(I-F)P(I-F)=0$$
So the most natural choice is $T=P$ as we have $P=F+FP(I-F)$.

*Hopefully, now things should be as obvious as in the following particular case. Take a rank one idempotent $P$ on $\mathbb{C}^2$. Then in $\mathbb{C}^2=\mbox{im} P\oplus (\mbox{im} P)^\perp$, we have
$$
P=\pmatrix{1& b\\ 0&0}\quad F=\pmatrix{1& 0\\ 0&0}\quad P=\pmatrix{1& 0\\ 0&0}+\pmatrix{0& b\\ 0&0}=F+FP(I-F)
$$
A: This appears in Kaplansky, $Rings\  of\ Operators$ (Benjamin, prehistoric). I don't have it in front of me, but it has a theorem (Kap called everything a theorem) asserting that if $(R,*)$ is a ring with involution such that for all $r \in R$, the element $1+ r^*r$ is invertible (C*-algebras obviously satisfy this), then whenever $e$ is an idempotent, there exists a projection $p$ such that $eR = pR$, and the construction yields the formula. The results nearby are also important.
