Multiplicative Inverse in Nicely Normed Star-Algebra

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 $a$.

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 $

$=$

$\ds \norm a^2 \cdot \dfrac 1 {\norm a^2}$

$\ds $

$=$

$\ds 1$

$\ds $

$=$

$\ds \dfrac 1 {\norm a^2} \cdot \norm a^2$

$\ds $

$=$

$\ds \dfrac 1 {\norm a^2} \cdot a^* \oplus a$

$\ds $

$=$

$\ds \dfrac {a^*} {\norm a^2}\oplus a$

$\blacksquare$

Note that this construction works whether $\oplus$ is associative or not.