Element of *-Algebra Uniquely Decomposes into Hermitian Elements
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Theorem
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$.
Let $a \in A$.
Then there exists unique Hermitian elements $b, c \in A$ such that:
- $a = b + i c$
Proof
Proof of Existence
This needs considerable tedious hard slog to complete it. In particular: fill in $\text C^\ast x$ with template To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Let:
- $b = \dfrac 1 2 \paren {a + a^\ast}$
and:
- $c = \dfrac 1 {2 i} \paren {a - a^\ast}$
Then we have using $(\text C^\ast 2)$ and $(\text C^\ast 1)$:
- $b^\ast = \dfrac 1 2 \paren {a^\ast + a^{\ast \ast} } = \dfrac 1 2 \paren {a + a^\ast} = b$
and:
\(\ds c^\ast\) | \(=\) | \(\ds -\frac 1 {2 i} \paren {a^\ast - a^{\ast \ast} }\) | $(\text C^\ast 4)$, $(\text C^\ast 2)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 {2 i} \paren {a^\ast - a}\) | $(\text C^\ast 1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 i} \paren {a - a^\ast}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds c\) |
We therefore have:
\(\ds b + i c\) | \(=\) | \(\ds \frac 1 2 \paren {a + a^\ast} + \frac i {2 i} \paren {a - a^\ast}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {a + a^\ast + a - a^\ast}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a\) |
$\Box$
Proof of Uniqueness
Suppose that $b_1, c_1, b_2, c_2 \in A$ are Hermitian elements such that:
- $a = b_1 + i c_1 = b_2 + i c_2$
Then, we have:
- $b_1 - b_2 = i \paren {c_2 - c_1}$
Using $(\text C^\ast 4)$ and $(\text C^\ast 2)$, we have:
- $b_1 - b_2 = -i \paren {c_2 - c_1} = -\paren {b_1 - b_2}$
So we have $b_1 = b_2$.
Hence we have $i \paren {c_2 - c_1} = 0$.
Hence $c_1 = c_2$.
Hence we have the desired uniqueness.
$\blacksquare$
Sources
- 1990: Gerard J. Murphy: C*-Algebras and Operator Theory ... (previous) ... (next): $2.1$: $C^\ast$-Algebras