# Show that the orthogonal decomposition for a hermitian element of a $C^*$-algebra is unique

I am reading about $C^*$-algebras from Chapter VIII in Conway's A Course in Functional Analysis. I've come across the following proposition which describes the "orthogonal decomposition" for hermitian elements of a $C^*$-algbera:

If $\mathcal{A}$ is a $C^*$-algebra with identity $1$ and $a \in \mathcal{A}$ is hermitian, then there exist unique positive elements $u,v \in \mathcal{A}$ such that $a = u - v$ and $0 = uv = vu$.

Here, an element $b \in \mathcal{A}$ is positive if $b$ is hermitian and the spectrum $\sigma(b) \subseteq [0, \infty)$.

Conway constructs $u$ and $v$ using the continuous functional calculus, and I understand the construction well enough.

However, my difficulty is understanding the uniqueness part of the proof.

Here's is my attempt to understand Conway's uniqueness proof. Suppose we also have positive $u_1, v_1$ that satisfy $a = u_1 - v_1$, $0 = u_1v_1 = v_1u_1$. Then one can use functional calculus to show that $a, u, v, u_1, v_1$ are pairwise commuting elements of $\mathcal{A}$. If we then let $\mathcal{B}$ be the $C^*$-algebra generated by $1, a, u,v, u_1,$ and $v_1,$ then $\mathcal{B}$ is abelian and so we know that the Gelfand transform $\gamma: \mathcal{B} \to C(\Sigma)$, $\mathcal{B} \ni b \mapsto \hat{b}$, is an isometric $\ast$-isomorphism. Here $\Sigma$ is the maximal ideal space of $\mathcal{B}$ (i.e., the set of all non-zero homomorphisms $h: \mathcal{B} \to \mathbb{C}$, equipped with the weak$^*$ subspace topology inherited from $\mathcal{B}^*$).

Conway then asks the reader to use uniqueness in $C(\Sigma)$ to finish the proof, but I am not sure how to do this. One thing I am aware of is that since $u, v, u_1, v_1$ are all positive elements, their Gelfand transforms are nonnegative functions on $\Sigma$ (since $\hat{b}$ takes values in $\sigma(b)$).

Hints or solutions are greatly appreciated.

In the $C^*$-algebra $C(\Sigma)$ every self-adjoint (=real valued) function $f$ has the unique decomposition $f=f_+-f_-,$ where $f_+,f_-$ are positive and $f_+f_-=0.$
Now use the fact that the Gel'fand transform $\hat b$ of $b\in\mathcal B$ is a positive function $\Longleftrightarrow$ $b$ is a positive element of $\mathcal B$ $\Longleftrightarrow$ $b$ is a positive element of $\mathcal A.$