Finite field extension $L\setminus K$: Minimal polynomial of transformation matrix equals minimal polynomial of $\alpha \in L$. I want to prove the following statement:

Consider the finite field extension $L\setminus K$ and $\alpha \in L$. Prove that the minimal polynomial of $\varphi_\alpha: L \to L, x \to \alpha x$ is equal to the minimal polynomial $g_\alpha$ of $\alpha$ over $K$.

My attempt:
Since $L \setminus K$ is finite we have $[L:K] = n < \infty$ and $L \cong K^n$. The transformation matrix of $\varphi_\alpha$ should be given by $\alpha\cdot E_n$. Therefore the characteristic polynomial is given by $\chi_{\varphi_\alpha}(\lambda) = (\alpha - \lambda)^n$. We know that $f_\alpha$ divides $\chi_{\varphi_\alpha}(\lambda)$. So $f_\alpha$ is of the form $(\alpha - \lambda)^m$ for some $m \leq n$.
My first (and only) idea is to show that $h_\alpha := f_\alpha - g_\alpha$ is equal to $0$. Right now I only do know that $h_\alpha(\alpha) = 0$.
Firstly, is this a good start? Secondly, can anyone help finish the argument or provide a good strategy how to prove the statement?
 A: Let $g_{\alpha}=\sum_{i=0}^n b_ix^i$ be the minimal polynomial of $\alpha$ over $K$. Also, let $f_{\alpha}$ be the minimal polynomial of $\varphi_{\alpha}$. Note that the map $\varphi_{\alpha}^i$ is defined by $\varphi_{\alpha}^i=\alpha^ix$. Hence, for each $x\in L$ we have:
$g_{\alpha}(\varphi_{\alpha})(x)=\sum_{i=0}^n b_i\varphi_{\alpha}^i(x)=\sum_{i=0}^n b_i\alpha^ix=(\sum_{i=0}^n b_i\alpha^i)x=0x=0$
Hence $g_{\alpha}(\varphi_{\alpha})=0$, and so $f_{\alpha}|g_{\alpha}$.
Conversely, let's write $f_{\alpha}=\sum_{i=0}^m c_ix^i$. Then $f_{\alpha}(\varphi_{\alpha})$ is the zero transformation. In particular:
$f_{\alpha}(\alpha)=\sum_{i=0}^m c_i\alpha^i=\sum_{i=0}^m c_i\varphi_{\alpha}^i(1)=f_{\alpha}(\varphi_{\alpha})(1)=0$
And so $g_{\alpha}|f_{\alpha}$ as well.
A: This is immediate from the definitions once the existence of both minimal polynomials is known (which follows from the finite dimensionality you cited). By definition $g_\alpha$ is the unique unitary polynomial in $K[X]$ that generates the ideal $I_\alpha$ of polynomials annihilating $\alpha$, i.e., $I_\alpha=\{\,P\in K[X]\mid P[\alpha]=0\,\}$, while $g_{\varphi_\alpha}$ is similarly the unique unitary polynomial in $K[X]$ that generates the ideal $I_{\varphi_\alpha}$ of polynomials annihilating $\varphi_\alpha$. To show they are the same it suffices to show each one is in the ideal generated by the other, so that they mutually divide each other, which ensures equality given that both are unitary.
So we must show that $g_\alpha[\varphi_\alpha]=0$ and $g_{\varphi_\alpha}[\alpha]=0$; to do this one uses the fairly obvious fact that for any $l\in L$ and $P\in K[X]$, and with $\varphi_l$ the operation $x\mapsto lx$ in$~L$ of multiplying by$~l$, one has $P[\varphi_l]:x\mapsto P[l]x$ (those who doubt can prove it for all $P=X^k$ by induction on $k$, and use linearity).
The $K$-vector space endomorphism $g_\alpha[\varphi_\alpha]$ of $L$ is that of multiplication in $L$ by $g_\alpha[\alpha]=0\in L$, which is obviously the zero endomorphism. The operation on $L$ of multiplying by $g_{\varphi_\alpha}[\alpha]$ is $g_{\varphi_\alpha}[\varphi_\alpha]$ which is the zero endomorphism, and $g_{\varphi_\alpha}[\alpha]$ itself is the result of applying this endomorphism to $1\in L$, which is clearly$~0$.
