# Rank of sum of Galois conjugates of a matrix

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?

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If $a \in K$ is non zero and has trace zero, then a diagonal matrix $M$ with $r$ ones and $n-r$ $a$'s has the property that $M'$ has rank $r$... – Laurent Berger Feb 25 '11 at 17:22

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.