How large can the entries of the matrix representation of an element of a number field be?

Question

Let $K$ be a finite extension of $\mathbb{Q}$. For each $\alpha \in K$, multiplication by $\alpha$ is a linear map from $K$ to $K$. Fix an integral basis $\omega_1,\ldots,\omega_n$ for $K$ over $\mathbb{Q}$. Let $M_{\alpha}$ be matrix for the "multiplication by $\alpha$" map with respect to the given basis.

If $\alpha = \sum_{i=1}^{n} a_i \omega_i$, set $$\| \alpha \|_{1} = \max_{1 \leq i \leq n} \| a_i \|$$ and set $$ \| \alpha \|_{2} = \max_{1 \leq i,j \leq n} \| (M_{\alpha})_{ij} \| $$ where $(M_{\alpha})_{ij}$ is the entry in the $i$-th row and $j$-th column of $M_{\alpha}$.

Both $\| \cdot \|_1$ and $\| \cdot \|_2$ are norms on $K$, which is finite-dimensional, thus the norms are equivalent in the sense that there are constants $c,C > 0$ such that $$ c \|\alpha \|_1 \leq \| \alpha \|_2 \leq C \|\alpha \|_1 $$ for all $\alpha \in K$.

Update Since $K$ is over $\mathbb{Q}$ (as opposed to $\mathbb{R}$ or $\mathbb{C}$), it is not necessarily true that all norms on $K$ must be equivalent. So that gives a preliminary question: Are the two norms equivalent?

Update 2: I posted an answer that resolves the question in the first update. In summary, the norms are equivalent because they are equivalent when extended to the vector space given by the formal span $\text{span}_{\mathbb{R}}(\omega_1,\ldots,\omega_n)$. The questions below still stand.

Question How do the constants depend on $K$? Do they depend only on the degree $n$ of $K$, if so how? For a given degree $n$, can we always find a $K$ so that $c$ and $C$ are as small and large, respectively, as we desire?

Side Question Is there any good reference for properties of matrix representations of algebraic numbers? Every book I've seen hardly goes beyond saying that the determinant of $M_{\alpha}$ is the norm of $\alpha$.

Answer 1

EDIT: It appears I have misunderstood the question. I'll leave this up, in case what's in it can be used somehow (and I'll delete it, if folks think I ought to).

Take $K={\bf Q}(\sqrt2)$, $\alpha=\sqrt2$, $\omega_1=a+\sqrt2$, $\omega_2=a-1+\sqrt2$. Then (details can be found in the comments) $\|\alpha\|_1=a$, $\|\alpha\|_2\ge a^2-2$. So there is no constant $C>0$ such that $\|\alpha\|_2\le C\|\alpha\|_1$ holds for all $\alpha \in K$.

Answer 2

I have figured out that the norms are equivalent. Here is my argument.

Viewing $K$ as a vector space over $\mathbb{Q}$, it is a subset of $V = \text{span}_{\mathbb{R}}(\omega_1,\ldots,\omega_n)$.

The norm $\| \cdot \|_1$ extends to $V$ with the same formula.

Note $\omega_i \omega_j = \sum_{k=1}^{n} c_{i,j,k} \omega_k$ with $c_{i,j,k}$ integers. We can multiply elements $\alpha = \sum_{i=1}^{n} a_i \omega_i$ and $\beta = \sum_{j=1}^{n} b_j \omega_j$ by $$ \alpha \beta = \sum_{i=1}^{n} \sum_{j=1}^{n} (a_i b_j) (\omega_i \omega_j) $$ Multiplication by $\alpha$ is still a linear map and the matrix $M_{\alpha}$ is formed the usual way (the columns are the coefficients of $\alpha \omega_j$). The norm $\| \cdot \|_2$ extends to $V$ with the same formula.

Now $\| \cdot \|_1$ and $\| \cdot \|_2$ are two norms on $V$, which is a finite-dimensional vector space over $\mathbb{R}$. So they must be equivalent. Thus they are equivalent over the subset $K$.