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


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In particular: fill in $\text C^\ast x$ with template
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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

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