Self-adjoint elements in a C*-algebra

Question

I have a simple question which confused me.

Suppose $A$ is a C*-algebra. every $x\in A$ has a representation such as $x=a+ib$ where $a,b$ are self-adjoint elements of $A$. Also we claim that $x^*x$ is self-adjoint while $x^*x$ has the representation $x^*x= a^2+b^2+i(ab-ba)$ which is not self-adjoint in general. I know that it's self-adjoint iff $a,b$ commute. But I think always $a,b$ do not commute. So why do we always say $x^*x$ is self-adjoint?

Answer 1

If $a$ and $b$ are self-adjoint then so are $a^2 + b^2$ and $i ( ab - ba)$. Let's check the last one:

$$[i(ab - ba)]^* = (-i) ( (ab)^* - (ba)^*) = -i ( b^* a^* - a^* b^*) = -i( b a - a b) = i(ab - ba)$$

Answer 2

Well, $x^* x$ is clearly self-adjoint since $(x^* x)^* = x^* x^{**} = x^* x$. (Here we just use $x^{**}=x$ and $(xy)^* = y^* x^*$, which are rules in the definition of a $*$-algebra.)

Notice that your expression $x^* x = a^2 + b^2 + i(ab-ba)$ doesn't have the property that $c:=ab-ba$ is self-adjoint, in fact, $c^*=-c$. This implies that $ic$ is self-adjoint and therefore we see again that $x^* x$ is self-adjoint. (But this decomposition is overkill when one wants to check that $x^* x$ is self-adjoint.)