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
Let $b \in A$.
Then there exists positive elements $b_1, b_2, b_3, b_4 \in A$ such that:
- $b = b_1 - b_2 + i \paren {b_3 - b_4}$
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:
| \(\ds \map {\sigma_A} {b^+}\) | \(=\) | \(\ds \map {\sigma_B} {b^+}\) | Spectrum of Element of Unital C*-Subalgebra of Unital C*-Algebra | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map {\sigma_{\map \CC {\Phi_B} } } {f^+}\) | Spectrum of Image of Element of Unital Algebra under Unital Algebra Homomorphism | |||||||||||
| \(\ds \) | \(=\) | \(\ds \set 0 \cup \set {\map {f^+} \phi : \phi \in \Phi_B}\) | Spectrum of Element of Space of Continuous Functions Vanishing at Infinity | |||||||||||
| \(\ds \) | \(\subseteq\) | \(\ds \hointr 0 \infty\) | since $f^+ \ge 0$ |
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:
| \(\ds b^+ - b^-\) | \(=\) | \(\ds \map {G^{-1} } {f^+ - f^-}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map {G^{-1} } f\) | Difference of Positive and Negative Parts | |||||||||||
| \(\ds \) | \(=\) | \(\ds b\) |
and:
| \(\ds b^+ b^-\) | \(=\) | \(\ds \map {G^{-1} } {f^+ f^-}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds {\mathbf 0}_A\) | Product of Positive and Negative Parts is Zero |
Finally, since $G^{-1}$ is an isometry, we have:
| \(\ds \norm {b^+}\) | \(=\) | \(\ds \norm {\map {G^{-1} } {f^+} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {f^+}_\infty\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sup_{\phi \in \Phi_B} \cmod {\map {f^+} \phi}\) | Definition of Supremum Norm | |||||||||||
| \(\ds \) | \(\le\) | \(\ds \sup_{\phi \in \Phi_B} \cmod {\map f \phi}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \norm f_\infty\) | Definition of Supremum Norm | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {\map {G^{-1} } f}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \norm b\) | $G^{-1}$ is an isometry |
and:
| \(\ds \norm {b^-}\) | \(=\) | \(\ds \norm {\map {G^{-1} } {f^-} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {f^-}_\infty\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sup_{\phi \in \Phi_B} \cmod {\map {f^-} \phi}\) | Definition of Supremum Norm | |||||||||||
| \(\ds \) | \(\le\) | \(\ds \sup_{\phi \in \Phi_B} \cmod {\map f \phi}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \norm f_\infty\) | Definition of Supremum Norm | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {\map {G^{-1} } f}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \norm b\) | $G^{-1}$ is an isometry |
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.
$\blacksquare$