showing that $|u|$ is invertible whenever $u$ is in a C* algebra

Question

If I'm working in an inner product space then it is easy to show that

$$\left|\left|Av\right|\right|^2 = \langle Av,Av \rangle = \langle v,A^*Av\rangle = \langle v, \sqrt{A^*A}\cdot \sqrt{A^*A}v \rangle = \langle \sqrt{A^*A}v, \sqrt{A^*A}v \rangle = \left|\left| \sqrt{A^*A}v\right|\right|^2$$

and therefore $ker(A) = ker(|A|)$ where $|A|=\sqrt{A^*A}$. From here, we know an element is invertible iff its kernel is trivial.

However, I don't see why a C* algebra behaves anything like an inner-product space, a priori.

I'm new to C* algebras and I would appreciate hints.... but not a full solution.

Answer 1

If $A$ is a C$^*$-algebra and $u\in A$ is invertible, then $u^*u$ is invertible. Now since $u^*u$ is normal, you can apply a functional calculus argument to show that $\sqrt{u^*u}$ is invertible.

Answer 2

The precise answer to your question depends on the way you define C$^*$-algebras and how you progress in building the theory. For instance, as soon as you know that all C$^*$-algebras admit a faithful representation into $B(H)$, you can repeat the argument you already know.

In any case, to work with $|u|$, you need some way to obtain square roots, which is usually functional calculus.

No matter how, as soon as you know that you can take square roots of positive elements, you can prove that if $a$ and invertible, then $a^{1/2}$ is invertible. Indeed, if $b$ is the inverse of $a$, you have $ab=ba=I$. We can rewrite the two equalities as $$ a^{1/2}(a^{1/2}b)=(ba^{1/2})a^{1/2}=I. $$ So $a^{1/2}$ has a left and a right inverse, which will necessarily be equal and thus $a^{1/2}$ is invertible.