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
$$P=F+F(P-1)(1-F).$$



Expression of a non-orthogonal projection in a $C^*$ algebra via an orthogonal one
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](http://en.wikipedia.org/wiki/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
$$
P=\pmatrix{1& b\\ 0&0}\quad F=\pmatrix{1& 0\\ 0&0}\quad P=\pmatrix{1& 0\\ 0&0}+\pmatrix{0& b\\ 0&0}=F+FP(I-F)
$$


