A question from Kreyszig: Let $X$ be complex Banach space with $T\in B(X,X)$ and a $p$ a polynomial. Show that the equation $p(T)x=y$ has a unique solution $x$ for every $y \in X$ if and only if $p(\lambda)\neq 0$ for all $\lambda \in \sigma(T)$.
$\Rightarrow :$ We have that $p(T)$ is invertible. Suppose $\exists\lambda\in \sigma(T)$ such that $p(\lambda)=0$. Then $p(z)=(z-\lambda)h(z)$ for some polynomial $h$. So we have $p(T)=(T-\lambda)h(T)$. Since $T-\lambda$ is not invertible, $p(T)$ is not invertible, contradiction.
But what about the converse? I think that I should use the spectral mapping theorem for polynomials, $\sigma(p(T))=p(\sigma(T))$ at somewhere. But how? Thanks!
We have $p(\lambda)=(\lambda-\lambda_1)\dots(\lambda-\lambda_n)$, where $\lambda_1,\dots,\lambda_n$ are the roots of $p$ (we assume the leading coefficient of $p$ is $1$). Then $p(T)=(T-\lambda_1)\dots(T-\lambda_n)$, and both implications are obvious.