finite dimensional $K$-vector space $V$ with linear endomorphism $T$ is a cyclic module over $K[x]$

Question

Problem: Suppose that $V$ is a finite dimensinal $K$-vector space with a linear endomorphism $T$. Then show that

i) Associated $K[x]$-module $V$ is such that $V\cong K[x]/(g)$ for some monic polynomial $g$ in $K[x]$ if and only if there exists $v\in V$ such that $V=\text{Span}\{T^i(v):i\ge 0\}$

ii) And that $g$ is both minimal characteristic polynomial of $T$

My attempt: I'm sorry to say this, but there's not much thing that I could do to call it an attempt. I'm basically stuck on first part, not to mention the second. I definitely feel like I need to use structure theorem, but I'm not very familiar with the theorem yet, and I cannot see how to apply it here.

Also, by "associated module $V$" I presume it means we take $f(x)\cdot v=f(T(v))$ for any $x$, is that right? (is this the scalar multiplication, that is?)

If anyone could explain them to me, it would be really nice.

Thanks in advancce,

Answer 1

Suppose $V \cong K[x]/(g)$ for a monic polynomial $g(x) \in K[x]$ (as $K[x]$-modules).

Let $\phi: K[x]/(g) \to V$ be such an isomorphism, and take $v = \phi(1 + (g))$.

Then $\phi(x^k + (g)) = \phi(x^k\cdot(1 + (g)) = x^k\cdot v = T^kv$, for any $k \in \Bbb N$.

Since $\{x^i + (g): i \geq 0\}$ span $K[x]/(g)$, we have that $\{T^iv: i \geq 0\}$ span $V$.

Conversely, suppose that for some $v \in V$ that $\{T^iv: i \geq 0\}$ spans $V$. Let $k$ be the smallest positive integer such that: $\{v,Tv,\dots,T^kv\}$ is $K$-linearly dependent ($k$ exists since $V$ is finite-dimensional over $K$).

It follows that $\{v,Tv,\dots,T^{k-1}v\}$ is a $K$-basis for $V$, Hence that:

$T^kv = a_0v + a_1Tv + \cdots + a_{k-1}T^{k-1}v$, for some $a_0,a_1,\dots,a_{k-1} \in K$.

Set $g(x) = -a_0 - a_1x - \cdots - a_{k-1}x^{k-1} + x^k$, and define:

$\psi: K[x] \to V$ by $\psi(f(x)) = f(T)v$. This is surjective because the $\{T^iv\}$ span $V$.

It should be clear that $(g) \subseteq \text{ker }\psi$.

For $f(x) \in K[x]$, write $f(x) = q(x)g(x) + r(x)$, where $\text{deg}(r) < \text{deg}(g)\ (= k)$, or $r = 0$.

If $f(x) \in \text{ker }\psi$, we have $0_V = \psi(f(x)) = r(T)v$. If $r \neq 0$, we have:

$\{v,Tv,...,T^{\text{deg}(r)}v\}$ is a $K$-linearly dependent set, contradicting our choice of $k$.

Hence $f(x) \in (g)$, so by the Fundamental Isomorphism Theorem (for modules), $V \cong K[x]/(g)$.

Now show $g(x)$ is the characteristic polynomial for $T$ because its degree is $\dim_K(V)$, and it's the minimal polynomial because the $\{T^iv\}$ exhibited above are a $K$-basis (unique, because $g$ is monic) (I've left some blanks for you to fill in).

Answer 2

I don't know how much you know about this, but we got the following.

First, the associated $K[X]$-module structure of $V$ is not free, more precisely it isn't torsion free and any $v\in V$ has a non-trivial annihilator. In fact the ideal

$Ann_{K[X]}(v):=\{p \in K[X] \ : \ p\cdot v = p(T)(v)=0\}$

is a principal ideal whose generator is called the minimal polynomial of the endomorphism $T$ with respect to $v$. We will denote it $\mu_{v}$.

Now, if we call $m=dim(V)$ then $deg(\mu_{v})\leq m$ for all $v\in V$.

We also know that the submodules are the invariant subspaces by the endomorphism $T$. We can use this fact to prove that the $k$-dimensional subspace $W(v,T)=K\langle v, T(v),\ldots ,T^{deg(\mu_{v})-1}(v) \ \rangle$ is a submodule.

Now if $\{\beta_{1},\ldots ,\beta_{m}\}$ is a basis of $V$ then $\mu_{T}=mcm(\mu_{\beta_{1}},\ldots ,\mu_{\beta_{m}})$ is the minimal polynomial of the endomorphism $T$, i.e., $\mu_{T}(T)=0$.

I think this should suffice to prove the desired statement, you just need to prove the steps you didn't know (or anything if you know all of this).