Multiplicative Inverse in Nicely Normed 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:

$a^*$ is the conjugate of $a$
$\norm a$ is the norm of $a$.


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.


Sources