Cyclic spaces ($\Bbbk[x]$-modules) and "faithful" evaluation of polynomial operators. Let $V$ be a $\Bbbk$-linear space and $\varphi\in \mathrm{End}_\Bbbk(V)$. A vector $v\in V$ is a $\varphi$-cyclic generator of $V$ if the following composite of $\Bbbk$-algebra morphisms is surjective.
$$\Bbbk[x]\overset{\operatorname{ev_\varphi}}{\longrightarrow}\mathrm{End}_\Bbbk(V)\overset{\operatorname{ev}_v}{\longrightarrow}V.$$
If $v$ is a $\varphi$-cyclic generator of $V$ then $\operatorname{Ker}(\operatorname{ev}_v\circ \operatorname{ev}_\varphi)=\operatorname{Ker}\operatorname{ev}_\varphi$.
Questions.


*

*What is some geometric intuition for the equality $\operatorname{Ker}(\operatorname{ev}_v\circ \operatorname{ev}_\varphi)=\operatorname{Ker}\operatorname{ev}_\varphi$? It says the evaluating polynomials in $\varphi$ at $v$ is "faithful", but I don't know what to visualize.

*What is an instructive example where we have equality for some $v\in V$ which is not a $\varphi$-cyclic generator? (Preferably such that $V$ does not posses any $\varphi$-cyclic generators at all.)

 A: *

*I don't know if this is really geometric, but let $M$ be the matrix $\text{ev}_\phi(x)$. Saying that $v$ isa cyclic generator says that you can get to any other vector in the vector space by taking finite linear combinations of the form
$$
a_0 v + a_1Mv + a_2M^2v + \ldots + a_nM^nv.
$$The equality of kernels, on the other hand, says that if some polynomial
$$
b_0 + b_1 Mv + b_2M^2v + \ldots + b_mM^m v = 0,
$$
then in fact $b_0 + \ldots + b_mM^m$ is already the zero matrix. 


Let's think of what this means. 
If we ignore the special case $v = 0$ (where the equality of kernels holds iff $M=0$), we may as well assume that we've picked a basis and $v$ is the first basis vector $e_1$. Then the annihilator of $v$ is the set of matrices whose first column is all $0$s. The claim is therefore that the only element of this space in the span of the powers of $M$ is the zero matrix itself.
Now for geometry. Let's picture $\text{End}(V)$ as a vector space in-and-of-itself, of dimension $n^2$ where $n = \text{dim} V$. We have the annihilator of $v$ on the one-hand, which is $n^2 - n$-dimensional. We also have the space spanned by powers of $M$, which is at most $n$-dimensional by the Cayley-Hamilton theorem. 
The equality of kernels says that these spaces, whose dimensions add up to (at most) the dimension of the ambient space, have trivial intersection in $\text{End}(V)$, i.e. only intersect at the 0 matrix. This is clearly a necessary condition for them to span $\text{End}(V)$, but we can't say it's sufficient because we don't know that the space spanned by powers of $M$ is actually $n$-dimensional. If we also knew that, we'd have a necessary and sufficient condition, i.e. with this additional hypothesis, $v$ is a cyclic vector iff
we have the equality of kernels.


*For an example of where these are different, $\text{ev}_\phi$ could be the zero map, i.e. the matrix $M_\phi$ could be zero. Now there are no $\phi$-cyclic generators (assuming the dimension of $V$ is greater than one) but the equality of kernels holds trivially.

