Let $V$ be a vector space over $\Bbb C$, $p$ is a polynomial over $\Bbb C$ and $a\in \Bbb C$ .
Prove that $a$ is an eigen value of $p(T)$ where $T$ is a linear operator on $V$ $\iff$ $a=p(\lambda$ ) for some eigen value $\lambda$ of $T$.
Atempt:$a$ is an eigen value of $p(T)$ where $p(z)=(z-\lambda_1)(z-\lambda_2)\ldots (z-\lambda_n)$ say.
$\implies p(T)v=av\implies (z-\lambda_1)(z-\lambda_2)\ldots (z-\lambda_n)(v)=av$
But how to show that $a=p(\lambda)$.
Please help.
Suppose that $\lambda\in\mathbb{C}$ and write $p(z)-\lambda=\beta(z-\alpha_1)(z-\alpha_2)\cdots(z-\alpha_n)$ for some elements $\beta,\alpha_1,\alpha_2,\ldots,\alpha_n\in\mathbb{C}$. From here, $$p(T)-\lambda I=\beta(T-\alpha_1I)(T-\alpha_2I)\cdots(T-\alpha_nI).$$ Notice that $p(T)-\lambda$ is non-invertible if and only if some factor $T-\alpha_j I$ is non-invertible (this isn't too hard to see since all the factors above commute). That means that $\lambda$ is an eigenvalue of $p(T)$ if and only if $\alpha_j$ is an eigenvalue of $T$ for some $j$. Since these $\alpha_j$'s are exactly the roots of $p(z)-\lambda$, this says that $\lambda$ is an eigenvalue of $p(T)$ if and only if $p(\alpha)-\lambda=0$ for some eigenvalue $\alpha$ of $T$. That is, if and only if $\lambda=p(\alpha)$ where $\alpha$ is an eigenvalue of $T$.