Positivity in $C^*$-algebra vs Riesz spaces If $A$ is a $C^*$-algebra, then a self-adjoint element $x\in A$ is a called positive if $sp(x)\subseteq [0,\infty)$. I know of the following result:

There exist positive elements $x_+,x_-\in A$ such that $x=x_+-x_-$ and $x_+x_-=0$.

Now I immediately think of a Riesz space when I look at this condition. So the following question has left me puzzled.

Let $A$ be a commutative $C^*$-algebra and let $x,h\in A$ be self-adjoint such that $h$ is positive and $h\geq x$. Show that $h\geq x_+$. Show that the  result fails if $A$ is not commutative. Hint: Take $h=\begin{bmatrix} 6&0\\0&1\end{bmatrix}$ and $x=\begin{bmatrix} 0&-2\\-2&0\end{bmatrix}$.

Now for the first part thought was that $h\geq 0, h\geq x$ implies $h\geq \sup\{0,x\}=x_+$.
For the second part, I calculated $x_+=0$ and $x_-=\begin{bmatrix} 0&2\\2&0\end{bmatrix}$. But we still have $h\geq x_+$. Then I realized that $sp(x_-)=\{-2,2\}$, so $x_-\not\geq 0$ (in the sense of $C^*$-algebra).
However the matrices form a Riesz space with respect to the pointwise order. This means that the positive cone does not coincide. Does this mean that the matrices (with the lattice cone) don't form a $C^*$-algebra? I feel I'm misunderstanding something subtle here.
 A: Your mixing up different partial orders on $M_2(\mathbb C)$, and of course the positive and negative part of and element depend on the chosen order. In general, the partial order on a $C^\ast$-algebra is not a lattice order (unless the $C^\ast$-algebra is commutative).
The positive cone of the $C^\ast$-algebra $M_2(\mathbb{C})$ is the set of all $2\times 2$ matrices $A$ that satisfy $\langle \xi,A\xi\rangle\geq 0$ for all $\xi\in \mathbb{C}^2$. As you noticed, this is not the same as the positive cone induced by the pointwise order.
As it happens, the entrywise order is also a $C^\ast$-algebra order on the vector space $M_2(\mathbb{C})$, namely the one where the product and adjoint are taken entrywise. This $C^\ast$-algebra is usually denoted by $\mathbb{C}^4$ - of course the underlying vector spaces are isomorphic. Of course, the spectrum of an element depends on the multiplication in the algebra. With respect to the pointwise algebra structure, the spectrum of your $x_-$ is actually $\{0,2\}$.
