Find an expression for $(\lambda\textbf{1}-a)^{-1}$. Let $A$ be a unital Banach algebra, and suppose $a\in A$ has the property that $a^2=1$, but that $a\neq \textbf{1}$ and $a\neq -\textbf{1}$.
By the spectral mapping theorem, $\sigma(a^2) = \{\lambda^2 : \lambda \in \sigma(a) \}.$
Therefore $\sigma(a^2)= \sigma(\textbf{1})= \{ 1 \}$, i.e. $\sigma(a)\subseteq \{-1,1\}.$
It can easily be shown (by an argument of contradiction) that equality holds, i.e. $\sigma(a)=\{-1,1\}$.
Now suppose that $\lambda \not\in \sigma(a)$, then $(\lambda\textbf{1}-a)^{-1}$ exists.

I am trying to find an expression for $(\lambda\textbf{1}-a)^{-1}$ in terms of $a$ and $\lambda$ which holds for all $\lambda \not\in \sigma(a)$.

Can anyone please assist? I am currently self-studying some work on spectral theory, and this is one of the problems I am trying to solve as part of the process.
 A: Solve for real values $\alpha,\beta,$ $(\lambda I-a)(\alpha I+\beta a)=I.$ Since $a^2=I,$ the equation becomes: $$\lambda \alpha-\beta=1,\\\lambda\beta-\alpha=0.$$
Solving this gives: $$\beta=\frac{1}{\lambda^2-1}\\\alpha=\frac{\lambda}{\lambda^2-1}.$$

In general, if $a\in A$ has a polynomial $p(x),$ with $p(a)=0,$ and if $\lambda\in\mathbb C$ is not a root of $p(x),$ then we can divide $p(x)$ by $x-\lambda$ to get:
$$p(x)=(x-\lambda)q(x)+r,$$ where $q(x)$ is a polynomial of one lower degree and $r=p(\lambda)\in\mathbb C\setminus\{0\}.$ Then, substituting $x=a,$ you get $$(a-\lambda I)q(a)+p(\lambda)I=0,$$ or $$(\lambda I-a)^{-1}=\frac1{p(\lambda)}q(a).$$
In your case, $p(x)=x^2-1,$ and $$p(x)=p((x-\lambda)+\lambda)=(x-\lambda)^2+2\lambda(x-\lambda)+(\lambda^2-1)$$ so $q(x)=x-\lambda+2\lambda=x+\lambda,$ and you get: $$\frac{1}{\lambda^2-1}\left(a+\lambda I\right),$$ again.

Even more generally, if $p_1,p_2$ are relatively prime polynomials (so, they share no roots) we can apply the Euclidean algorithm and solve for polynomials $q_1,q_2$ such that:
$$p_1(x)q_1(x)+p_2(x)q_2(x)=1.$$
Then if $p_1(a)=0,$ then $p_2(a)^{-1}=q_2(a).$
