Hermitian Element of C*-Algebra Decomposes into Positive Elements

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra. 
Let $b \in A$ be Hermitian.
Then there exists positive elements $b^+$ and $b^-$ such that:
:$b^+ b^- = {\mathbf 0}_A$
:$b^+ - b^- = b$ 
:$\norm {b^+} \le \norm b$ and $\norm {b^-} \le \norm b$. 
=== Corollary ===
== Proof ==
First let $A$ be unital.
Let $B$ be be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, b}$. 
Let $\Phi_B$ be the spectrum of $B$.
By C*-Algebra Generated by Commutative Self-Adjoint Set is Commutative, $B$ is commutative.
Let $\struct {\map {\CC_0} {\Phi_B}, \overline \cdot, \norm {\, \cdot \,}_\infty}$ be the $\text C^\ast$-algebra of continuous complex-valued functions vanishing at infinity.
From Gelfand-Naimark Theorem: Commutative Case, there exists an isometric $\ast$-algebra isomorphism $G : B \to \map {\CC_0} {\Phi_B}$.
Let $f = \map G b$. 
Since $G$ is an $\ast$-algebra isomorphism, $f$ is Hermitian. 
Hence $\overline f = f$. 
From Complex Number equals Conjugate iff Wholly Real, we have $\map f \phi \in \R$ for each $\phi \in \Phi_B$.
From Positive Part of Continuous Function is Continuous and Positive Part of Real-Valued Function Vanishing at Infinity Vanishes at Infinity, we have $f^+ \in \map {\CC_0} {\Phi_B}$.
Similarly from From Negative Part of Continuous Function is Continuous and Negative Part of Real-Valued Function Vanishing at Infinity Vanishes at Infinity, we have $f^- \in \map {\CC_0} {\Phi_B}$.
From Inverse of *-Algebra Isomorphism is *-Algebra Isomorphism and Inverse of Linear Isometry is Linear Isometry, $G^{-1} : \map {\CC_0} {\Phi_B} \to B$ is an isometric $\ast$-algebra isomorphism.
Hence we can take $b^+ = \map {G^{-1} } {f^+}$ and $b^- = \map {G^{-1} } {f^-}$.
Then:
Since $f^- \ge 0$, we also obtain:
:$\map {\sigma_A} {b^-} \subseteq \hointr 0 \infty$
Since $B$ is commutative, $b^+$ and $b^-$ are both normal and hence both Hermitian from Normal Element of C*-Algebra is Hermitian iff Spectrum is Real.
Since further $\map {\sigma_A} {b^+} \subseteq \hointr 0 \infty$ and $\map {\sigma_A} {b^-} \subseteq \hointr 0 \infty$, $b^+$ and $b^-$ are both positive.
Finally, since $G^{-1}$ is an algebra homomorphism, we have:
and:
Finally, since $G^{-1}$ is an isometry, we have:
and:
Now suppose that $A$ is not unital.
Let $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$ be the unitization of $A$. 
From the definition of the unitization, we have $\norm {\tuple {x, 0} }_\ast = \tuple x$ for each $x \in A$. 
Applying the unital case to $\tuple {b, 0} \in A_+$, there exists $\tuple {b^+, \lambda}, \tuple {b^-, \mu} \in A_+$ positive such that:
:$\tuple {b^+, \lambda} \tuple {b^-, \mu} = \tuple {b, 0}$
:$\tuple {b, 0} = \tuple {b^+, \lambda} - \tuple {b^-, \mu}$
:$\norm {\tuple {b^+, \lambda} }_\ast \le \norm {\tuple {b, 0} }_\ast = \norm b$ and $\norm {\tuple {b^-, \mu} }_\ast \le \norm b$
From the second equation we have:
:$\tuple {b, 0} = \tuple {b^+ - b^-, \lambda - \mu}$
Hence $\lambda - \mu = 0$, giving $\lambda = \mu$. 
We have:
:$\tuple {b, 0} = \tuple {b^+, \lambda} \tuple {b^-, \mu} = \tuple {b^+ b^- + \lambda b^- + \mu b^+, \lambda \mu}$
Since $\lambda = \mu$, we have $\lambda^2 = 0$, and hence $\lambda = \mu = 0$. 
So we have:
:$\tuple {b, 0} = \tuple {b^+ b^-, 0}$
and:
:$\tuple {b, 0} = \tuple {b^+ - b^-, 0}$ 
We have $\norm {\tuple {b^+, \lambda} }_\ast = \norm {\tuple {b^+, 0} }_\ast = \norm {b^+}$, so we have $\norm {b^+} \le \norm b$ also.
Similarly we get $\norm {b^-} \le \norm b$. 
Hence:
:$b = b^+ - b^-$
:$b^+ b^- = {\mathbf 0}_A$
:$\norm {b^+} \le \norm b$ and $\norm {b^-} \le \norm b$.
From Element of C*-Algebra is Positive iff Positive in Unitization, $b^+$ and $b^-$ are positive.
Hence the demand.
Category:Positive Elements of C*-Algebras
Category:Hermitian Element of C*-Algebra Decomposes into Positive Elements