# Square root in Banach algebra

Suppose we are given a unital Banach algebra $A$ and an element $a\in A$ such that the spectrum is a subset of the positive reals $\mathbb{R}_{>0}$. Then by a theorem (see for example W. Rudin 10.30) we know that there exists a "square root" of $a$, i.e. an element $b\in A$ such that $b^2 = a$.

Question: does there exist such a $b$ with a spectrum $\sigma(b)$ also contained in the positive reals $\mathbb{R}_{>0}$?

My thoughts:

• From the spectral mapping theorem we know that since the square root is holomorphic in a neighborhood of the spectrum, $\sigma(\sqrt a)=\sqrt{\sigma(a)}$. However, is it trivial that $\sqrt a= \pm b$ as in the complex case?
• Also, if the spectrum of $b$ had to be connected then it would be clear, but I don't think this is true in general.
• What you said about the spectral mapping answers your question. The element $b:=\sqrt{a}$ constructed by the holomorphic functional calculus in this case has positive spectrum. (this does not mean the only possible square roots are then $\pm b$, but that's a different question). – Julien Nov 25 '13 at 22:35

## 1 Answer

Question: does there exist such a $b$ with a spectrum $\sigma(b)$ also contained in the positive reals $\mathbb{R}_{>0}$?

Yes, that's an important point that you have a positive square root of a positive element.

You take the principal branch of the square root, defined in $\mathbb{C}\setminus (-\infty,0]$, with positive real part to construct it.

• Ah, the existence of $\sqrt(a)$ is just immediate from the application of the definition of $A$-valued functions with the contour integral, rather than from the theorem that a $b$ exists such that $b^2=a$, is that correct? And spectral mapping then gives the result immediately. – ScroogeMcDuck Nov 25 '13 at 22:52