Bounded self adjoint operator can be written as difference of positive operators

Question

This is supposedly quite a simple fact, yet I haven't been able to show it. If we suppose that $T$ is a bounded self-adjoint operator, then $$\langle Tv,v \rangle = \langle v,Tv \rangle$$ for all $v \in \mathcal{H}$. I don't see how we are allowed to then decompose it into the form $A - B$ for $A$ and $B$ both positive operators.

Answer 1

Since $T$ is self-adjoint, $C^*(T)$ (the $C^*$-algebra generated by $1$ and $T$) is commutative, and therefore the Gelfand transform $\phi$ is an isometric isomorphism of $C^*(T)$ onto $C(X)$ for some compact Hausdorff space $X$. Furthermore, $\phi(T)(X)=\sigma(T)\subset\mathbb{R}$ so $\phi(T)^+,\phi(T)^-\in C(X)$ (the positive and negative parts of $\phi(T)$, respectively), and thus there are $A,B\in C^*(T)$ such that $\phi(A)=\phi(T)^+$ and $\phi(B)=\phi(T)^-$. From this, it is clear that $A$ and $B$ are positive and that $T=A-B$.