Eigenvalue of multiplication in a number field. Let $\mathbb{K}$ be an algebraic number field. $b \in \mathbb{K}$ defines a linear transformation $\phi(b)$ on $\mathbb{K}/\mathbb{Q}$ by multiplication - $\phi(b)(x):=bx$ for all $x\in\mathbb{K}$. One way to calculate the eigenvalues of this matrix is by taking a suitable basis for $\mathbb{K}$ and calculating the determinant. This gives us that the eigenvalues are the conjugates of $b$ over $\mathbb{Q}$.
Is there some non computational way of proving this fact, maybe even showing what the eigenvectors should be? 
This answer does refer to something like what I am asking for, unfortunately I do not really understand it.
 A: Here's an argument you might like:
Since $\phi = \phi(b)$ is a linear map on a $\Bbb Q$ vector space, its minimal polynomial $m(x)$ is a polynomial with coefficients in $\Bbb Q$ (i.e. $m \in \Bbb Q[x]$).  Note that the $\Bbb K$-eigenvalues of $\phi$ are precisely the roots of $m(x)$ in $\Bbb K$ (not necessarily up to multiplicity).
Now, because $\Bbb K$ is algebraic, there exists a (monic) polynomial $p \in \Bbb Q[x]$ such that $p(x) = 0$ iff $x$ is conjugate to $b$ (that is, we take $p$ to be the minimal polynomial of $b$).  Because $p(b) = 0$, we may conclude that $p(\phi) = 0$.  Thus, $m(x) \mid p(x)$.  Because $p$ is irreducible over $\Bbb Q$, we may conclude that $m(x) = p(x)$
Note, however, that the degree of $m$ is equal to the dimension of $\Bbb K$ as a $\Bbb Q$-vector space.  Since $m$ divides the characteristic polynomial of $\phi$ and because $\phi$ has the same degree, we conclude that $m$ is the characteristic polynomial of $\phi$.
Thus, the eigenvalues of $\phi$ are precisely the conjugates of $b$ over $\Bbb Q$.
