Question about the converse of a well known result from Linear Algebra

Solution 1:

Let $V$ be a complex vector space, let $T$ be an operator on $V$ and let $p\in \mathbb{C}[x]$. The following steps lead to a solution:

Exercise 1: Prove that the result holds if $p$ is a constant polynomial.

(1) (We can assume, without loss of generality, that $p$ is non-constant by Exercise 1.) Let $\lambda$ be an eigenvalue of $p(T)$. Note that $p(T)-\lambda I$ is not injective and we can factor the polynomial $p(z)-\lambda = c(z-c_1)\cdots (z-c_k)$ for some positive integer $k$ and scalars $c,c_1,\cdots,c_k\in\mathbb{C}$.

Exercise 2: Prove that $T-c_iI$ is not injective for some $1\leq i\leq k$. Deduce that $c_i$ is an eigenvalue of $T$.

Exercise 3: Prove that $p(c_i)=\lambda$.

You should now be able to solve your question. Let me give another couple of Exercises:

Exercise 4: Prove that there exists an operator $T:\mathbb{R}^2\to \mathbb{R}^2$ with characteristic polynomial $p(x)=x^2-1$. Is this operator unique? If $T$ is any such operator, prove that there is an eigenvalue of $p(T)$ that is not of the form $p(\lambda)$ for an eigevalue $\lambda$ of $T$. Therefore, the answer to your question is negative in the context of real vector spaces.

Exercise 5: Let $V$ be an odd-dimensional real vector space, let $T$ be an operator on $V$ and let $p\in\mathbb{R}[x]$. If $\lambda$ is an eigenvalue of $p(T)$, is it true that there is an eigenvalue $a$ of $T$ such that $p(a)=\lambda$? Prove or give a counterexample.

I hope this helps!