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


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 $$ P=\pmatrix{1& b\\ 0&0}\quad F=\pmatrix{1& 0\\ 0&0}\quad P=\pmatrix{1& 0\\ 0&0}+\pmatrix{0& b\\ 0&0}=F+FP(I-F) $$


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.