Let $F$ be a field and let $K$ be a unitary $F$-algebra. For each $v ∈ K$, let $S(v)$ be the set of all $a ∈ F$ satisfying the condition that $v−a1_K$ (lets denote $a1_K$ just as $a$) does not have an inverse with respect to the operation $•$. If $v ∈ K$ has a multiplicative inverse $v^{−1}$ with respect to this operation, show that either $S(v) = ∅ = S(v^{−1})$ or $S(v) \neq ∅$ and $S(v^{−1}) = \{a^{−1} | a ∈ S(v)\}$.
I have problems showing that if $S(v) \neq ∅\Rightarrow S(v^{−1}) = \{a^{−1} | a ∈ S(v)\}$. So how can I show that if $(v^{-1}-a^{-1})$ has a multiplicative inverse then $(v-a)$ has one too?
This is an exercise from the book: The Linear Algebra a Beginning Graduate Student Should Know by Golan.
Hint: In the commutative algebras, $$v^{-1}-a^{-1}1 = \frac{a1-v}{av}$$ Use Since $a1$ and $v$ commute, use this form to compute an inverse to $v-a$ explicitly in terms of $v^{-1}-a^{-1}$.
This is true for any $w,v$ that commute with $w,v$ invertible and $v^{-1}-w^{-1}$ is invertible and the ossociative law for $w$ and $v$.
(I am assuming your definition of $F$-algebra has associativity of multiplication. You will need to know that som limited associativity occurs with $aI$ and $v$.)