I'm reading chapter 4 of Gerard Murphy's C*-algebras book and am confused by a statement in his proof of theorem 4.1.10. In his proof, he says, "$A'$ is the linear span of its unitaries" (where $A'$ denotes the commutant of $A$). I can't find any reference to why this is true anywhere earlier in the chapter. Am I missing something, or is this referring to some other result?

For context, 4.1.10 states:

Let $A$ be a von Neumann algebra on a Hilbert space $H$ and $v$ an element of $A$ with polar decomposition $v = u|v|$. Then $u \in A$.

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    $\begingroup$ Does Murphy show in a previous chapter (perhaps an exercise) that in a unital C*-algebra, every element is a linear combination of 4 unitary elements? (If $x=x^*$ and $\|x\|\leq 1$, then $x\pm i\sqrt{1-x^2}$ are unitary and $x$ is their average$.) $A'$ is a unital C*-algebra. $\endgroup$ Oct 16, 2012 at 22:06
  • $\begingroup$ Thanks Jonas. Murphy showed decomposition into hermitian elements and into positive elements and that has been used heavily in the last chapter. However, I don't recall coming across that decomposition into unitaries (and I have done all the HW problems). Thanks for providing it, I figured I was missing something simple. $\endgroup$ Oct 17, 2012 at 15:33

1 Answer 1


See Remark 2.2.2 on p.45 of Murphy's book. There he shows every element in a unital C*-algebra may be written as a linear combination of four unitary elements.


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