I am reading about field norms and I have run into an issue which I first thought to involve elementary linear algebra, but now I'm not so sure. A little bit of background is needed to get to the heart of my question.
Let $L/K$ be a separable finite field extension of degree $n$. Let $\alpha \in L$ and define the map
$$ \begin{align} m_{\alpha}: \ L &\to L \\ l &\mapsto \alpha l \end{align} $$
Then clearly $m_{\alpha}$ is a linear transformation of the vector space $L$ with field of scalars $K$.
Throughout the course of a certain proof that I am trying to construct, I have written out a basis for the vector space $L$ over $K$ and have constructed the matrix corresponding to the linear transformation $m_{\alpha}$. I'll denote this matrix by $[m_{\alpha}]$. I am all but certain that $\alpha$ is a root of the characteristic polynomial of $[m_{\alpha}]$. My initial proof of this just used basic well known ideas from linear algebra: Since $\alpha$ is an eigenvalue of $m_{\alpha}$ it is therefore an eigenvalue of $[m_{\alpha}]$ and a root of the characteristic polynomial of $[m_{\alpha}]$.
However, I then realized that though $\alpha$ is an eigenvalue of $m_{\alpha}$ it is not necessarily in $K$, so there is a mistake in my above reasoning. Since my basis is for the vector space $L$ over $K$ there is only a correspondence between eigenvalues of $m_{\alpha}$ in $K$ with eigenvalues of $[m_{\alpha}]$ in $K$. However, in this case the correspondence is useless to me because there are no eigenvalues in $K$. Thus, the correspondence gives me no information about $\alpha$. So, now I am stuck trying to reason why $\alpha$ is a root of the characteristic polynomial of $[m_{\alpha}]$. So:
Question: Why is $\alpha$ a root of the characteristic polynomial of $[m_{\alpha}]$ even though $\alpha$ need not be in $K$?
Finally, here is an example to make things concrete, in case I was not clear enough. Take $L = \mathbb{Q}(\sqrt{2})$ and $K = \mathbb{Q}$ and let $\alpha = 1 + \sqrt{2}$. In this case, it is clear by definition of $m_{\alpha}$ that $m_{\alpha}$ has no eigenvalues in $\mathbb{Q}$ (since there is only one eigenvalue, namely $\alpha$, and this is not in $\mathbb{Q}$). This fact is also reflected when we pick a basis $\beta = \{1, \sqrt{2}\}$ for $L$ over $K$. Using this basis
$$ [m_{\alpha}] = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} $$
The characteristic polynomial is $p(x) = (1 - x)^2 - 2$ which has roots $1 \pm \sqrt{2}$. These are certainly not in $\mathbb{Q}$, as we would expect from our analysis of the transformation. Yet, we can plainly see that $\alpha$ is a root, so it seems that there is a relationship between the eigenvalue $\alpha$ that lies in $L$ and the roots of the characteristic polynomial. I am trying to understand why this is.
One way to say it: $m_{\alpha}$, pretty much by definition, satisfies any polynomial that $\alpha$ does, in the sense that if $f(\alpha) = 0$ then $f(m_{\alpha}) = 0$. If we take $f$ to be the minimal polynomial of $\alpha$ then it follows that $f(m_{\alpha}) = 0$ and $f$ is irreducible, from which it follows that $f$ divides the characteristic polynomial of $m_{\alpha}$.
Another way to say it is to work, not with $L$, but with the extension of scalars $L \otimes_K L$. This is a vector space over $L$ with $\dim_L L \otimes_K L = \dim_K L$, and $m_{\alpha}$ acts on it with a genuine eigenvalue of $\alpha$ (you may want to introduce notation to distinguish the two copies of $L$ at play here). In general the roots of a characteristic polynomial correspond to eigenvalues which may only exist after extending scalars in this way.