# For each $x$ in a Hilbert $A$-module $X$, there exists a unique $y\in X$ such that $x=y\langle y,y \rangle$.

How do I prove this lemma?

Lemma: Suppose that $X$ is a Hilbert $A$-module. For each $x\in X$ there exists a unique $y\in X$ such that $x=y \langle y,y \rangle$.

I don't know if this is needed but we already know Cohen's factorization theorem that states If $A$ is a Banach algebra with a bounded left or right approximate identity, then for all $a\in A$ there exist $b,c \in A$ such that $a=bc$.

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If $\langle x,x\rangle$ is invertible you can set $y=x\langle x,x\rangle^{1/3}$ –  Norbert Jun 26 '12 at 21:26
@Norbert: You mean $-1/3$, right? –  Jonas Meyer Jun 26 '12 at 21:49
@JonasMeyer Yes, you are right. Do you have any ideas for the general case? –  Norbert Jun 26 '12 at 21:50
@Norbert: I cheated, i.e., I knew where to look it up. It is Proposition 2.31 in Raeburn and Williams's Morita equivalence and continuous trace $C^*$-algebras. They cite the article at the following link as the source of the proof they present (link to large pdf, where the result is found on the sixth page, a.k.a. page 145): archive.numdam.org/ARCHIVE/BSMF/BSMF_1996__124_1/… –  Jonas Meyer Jun 26 '12 at 21:54
@JonasMeyer Thanks! –  Norbert Jun 26 '12 at 22:06