Is the hypothesis that $V$ be finite dimensional needed in this exercise? I am confused if the hypothesis that $V$ be finite dimensional is required in this exercise as I never use that hypothesis in my proof. I have typed the exercise and my own attempted proof below.

Exercise: Suppose $V$ is finite-dimensional, $T \in L(V)$, and $v \in V$ with $v \ne 0$.
Let $p$ be a nonzero polynomial of smallest degree such that $p(T)v = 0$.
Prove that every zero of $p$ is an eigenvalue of $T$.
Proof: Let $\lambda$ be a zero of $p$. Then there exists a polynomial $q$ such that $$p(z)=(z-\lambda)q(z)$$
Evaluating $p(T)$ for $v$ we get $$p(T)v=((T-\lambda I)q(T))v=0$$
Given that $p$ is the polynomial of the smallest degree that satisfies $p(T)v=0$, we have that $q(T)v\ne 0$ as $\deg q <\deg p$. Then the above equation implies that $$(T-\lambda I)(q(T)v)=0$$
Because $q(T)v\ne 0$, the above equation implies that $T-\lambda I$ is not injective. Which is equivalent to $\lambda$ being an eigenvalue of $T$.

I am not sure where I use the hypothesis that $V$ is finite dimensional. The condition that $$\text{$\lambda$ is an eigenvalue of $T\iff T-\lambda I$ is not injective}$$ is true on any vector space and not just for finite dimensional vector spaces. Am I correct in believing that the hypothesis that $V$ be finite dimensional is not needed?
 A: Your proof is fine. The hypothesis that $V$ is finite dimensional ensures the existence of the polynomial $p$. But the argument doesn't need such existence. If such a polynomial exists, then its roots are eigenvalues.
Indeed, if $\dim V=n$, the the vectors $v,Tv,T^2v,\dots,T^nv$ aren't linearly independent, so there are scalars $a_0,a_1,\dots,a_n$ not all zero such that
$$
a_0v+a_1Tv+\dots+a_nT^nv=0
$$
So if $q(z)=a_0+a_1z+\dots+a_nz^n$, we have $q(T)v=0$. A nonzero polynomial exists, so there's a nonzero one of minimal degree.
It's not difficult to find a linear map $T\colon V\to V$ that has no eigenvalue (with $V$ infinite dimensional, of course), so for no vector $v\ne0$ such a polynomial exists.
A: I think $\mathscr{dim}~V<\infty$ guarantees the existence of the polynomial $p$.
Consider $V=\mathbb{R}^{N}$,$ \forall ~x=(x_n)\in \mathbb{R}^N$, we define $A(x)=(k_nx_n)$.
Assume $0\ne f$ is a polynomial, then $f(A)(x)=(f(k_n)x_n)$.
Let $v=(1,1,\cdots,1,\cdots)$,$f(A)v=(f(k_1),f(k_2),\cdots,f(k_n),\cdots)=0$.
So $f$ has infinite roots.
A contradictions
