Dgdrxrt

Let $V$ be a $Bbbk$-linear space and $varphiin mathrmEnd_Bbbk(V)$. A vector $vin V$ is a $varphi$-cyclic generator of $V$ if the following composite of $Bbbk$-algebra morphisms is surjective.

$$Bbbk[x]oversetoperatornameev_varphilongrightarrowmathrmEnd_Bbbk(V)oversetoperatornameev_vlongrightarrowV.$$

If $v$ is a $varphi$-cyclic generator of $V$ then $operatornameKer(operatornameev_vcirc operatornameev_varphi)=operatornameKeroperatornameev_varphi$.

Questions.

1. What is some geometric intuition for the equality $operatornameKer(operatornameev_vcirc operatornameev_varphi)=operatornameKeroperatornameev_varphi$? It says the evaluating polynomials in $varphi$ at $v$ is "faithful", but I don't know what to visualize.


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

1. I don't know if this is really geometric, but let $M$ be the matrix $textev_phi(x)$. 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 $textEnd(V)$ as a vector space in-and-of-itself, of dimension $n^2$ where $n = textdim V$. 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 $textEnd(V)$, i.e. only intersect at the 0 matrix. This is clearly a necessary condition for them to span $textEnd(V)$, 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.

2. For an example of where these are different, $textev_phi$ 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.