Multiplicative Inverse in Nicely Normed Star-Algebra
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Theorem
Let $A = \struct {A_F, \oplus}$ be a nicely normed $*$-algebra whose conjugation is denoted $*$.
Let $a \in A$.
Then the multiplicative inverse of $a$ is given by:
- $a^{-1} = \dfrac {a^*} {\norm a^2}$
where:
Proof
For the result to hold, we need to show that $a \oplus \dfrac {a^*} {\norm a^2} = 1 = \dfrac {a^*} {\norm a^2} \oplus a$.
\(\ds \) | \(\) | \(\ds a \oplus \dfrac {a^*} {\norm a^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a \oplus a^* \cdot \dfrac 1 {\norm a^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm a^2 \cdot \dfrac 1 {\norm a^2}\) | Definition of Nicely Normed Star-Algebra | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\norm a^2} \cdot \norm a^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\norm a^2} \cdot a^* \oplus a\) | Definition of Nicely Normed Star-Algebra | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {a^*} {\norm a^2}\oplus a\) |
$\blacksquare$
Note that this construction works whether $\oplus$ is associative or not.