$T:H\to H$ bounded operator where $H$ is Hilbert space. How to show that $p(T)x=y$ has solution $x$ for all $y$ iff $ p(\lambda)\neq 0$ for all $\lambda$ on the specter of $T$ ($\sigma(T)$). My attempt
$\rightarrow$ By contradiction, suppose that exist $\lambda\in \sigma(T)$ such that $p(\lambda)=0$, so i think (not sure) $\lambda$ is a eigenvalue of $T$, hence
$T(v)=\lambda v$ for some $v\in H$, then when $y=0$ by hypothesis $p(T)x=0$ has only one solution, but that is not true because in the left side we have $p(\lambda v)x=0$, and this si true for infinitely $x\in H$, i am not sure of this!
The other implication i do not have idea, can somebody help me please?thank you
I'm assuming you're working with a polynomial $p$. Suppose $p(\lambda)=0$. Then $$ p(T)=p(T)-p(\lambda)I=(T-\lambda I)q(T,\lambda). $$ where $q$ is a polynomial in $T$ and $\lambda$. From this it follows that $$ \mathcal{R}(p(T))\subseteq\bigcap_{\{\lambda : p(\lambda)=0\}}\mathcal{R}(T-\lambda I). $$ Therefore, if $\mathcal{R}(p(T))=H$ then $\mathcal{R}(T-\lambda I)=H$ for all $\lambda$ in the zero set of $p$. Conversely, if $\mathcal{R}(T-\lambda I)=H$ for all $\lambda$ in the zero set of $p$, then $p(T)$ is the product of surjective factors, which makes $p(T)$ surjective.