Positive elements of a $C^*$-algbera form a poset

Question

My knowledge of $C^*$-algebras is very little.

We call an element positive if $a=b^*b$ for some $b$ and make a relation on all positive elements by saying $$ b \geqslant a \iff b-a \text{ is positive}. $$ I can't figure out why this gives us a poset.

Answer 1

Only anti-symmetry seems to be non-trivial here. Let us apply the spectral theorem.

Suppose that $b\leqslant a$ and $a\leqslant b$. Since $a-b$ is positive, $C^*(a-b)$ is commutative and of course $b-a\in C^*(a-b)$. But now you can think of $a-b$ and $b-a$ as continuous functions on $\sigma(a-b)$ which are both positive (non-negative, strictly speaking). Since $a-b = -(b-a)$, necessarily $a-b=0$.

Answer 2

Below is one avenue.

Since $a=b^*b$, we have $a^*=a$. Then $C^*(a)$, the C$^*$-subalgebra generated by $a$, is commutative. In a commutative Banach algebra, we have $$ \sigma(a)=\{\tau(a):\ \tau \text{ is a character }\}. $$ So, when $a=b^*b$, $\tau(a)=\tau(b^*)\tau(b)=\overline{\tau(b)}\tau(b)=|\tau(b)|^2\geq0$ for any character $\tau$. Thus, $\sigma(a)\subset[0,\infty)$.

If $a\geq0$ and $a\leq0$, we deduce that $\sigma(a)=\{0\}$. Being selfadjoint, the spectral theorem implies that $a=0$.