Given an invertible square matrix $M$ with entries from some number field $K$ which is Galois over $\mathbb{Q}$, sum the Galois conjugates of $M$ to form a new matrix $M' = \Sigma\_{\sigma \in \mathrm{Gal}(K/\mathbb{Q})} M^\sigma$. What can one say about the rank of $M'$? It seems like both zero and full rank can occur. Are there any theorems related to this, perhaps with further conditions on $M$, that provide some information about the rank?



Rank of sum of Galois conjugates of a matrix
Here is one way to get a theorem about this situation:


Although $M'$ may not be invertible, you can always adjust by a scalar $\lambda \in K^\times$ to ensure that $(\lambda M)'$ is invertible. This can be seen as follows: let $\{\alpha\_1, \dots, \alpha\_n\}$ be a basis for $K/\mathbb{Q}$. Write $M = \alpha\_1 M\_1 + \dots + \alpha\_n M\_n$ for rational matrices $M\_i$. Now $\mathrm{det}(x\_1 M\_1 + \dots + x\_n M\_n)$ is a polynomial in $\mathbb{Q}[x\_1,\dots,x\_n]$ which has non-zero value at the point $(\alpha\_1, \dots, \alpha\_n) \in K^n$, and so is non-zero. Since $\mathbb{Q}$ is infinite, there is also a rational point $(\beta\_1, \dots, \beta\_n) \in \mathbb{Q}^n$ at which it is non-zero. Now by the nondegeneracy of the trace form, we can pick a $\lambda \in K$ such that $\mathrm{tr}\_{K/\mathbb{Q}}(\lambda \alpha\_i) = \beta\_i$ for all $i$. Then $\mathrm{det}(\mathrm{tr}\_{K/\mathbb{Q}}(\lambda M)) \neq 0$ as desired.


