# 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
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## 2 Answers

You can do this: take a normal basis for K, and write M as a linear combination of the basis elements with rational matrices as coefficients. This clarifies what happens under the trace, because the normal basis elements will all end up at the same rational number. So up to a scalar you are just adding a bunch of square rational matrices and asking the rank of what you get. Obviously if those coefficients are arbitrary you have no reason to be able to say more. Some sort of "Galois module" in the background is pretty much necessary to insert more structure into the situation.

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Right. I guess the emphasis is really on "with further conditions" then. I'm relying on someone having encountered such a theorem before, since I'm not really sure what sort of "further conditions" would lead to a theorem. A similar approach to what you describe: you could also take a (non-normal) basis such that the first element has trace 1 and the rest have trace 0. Then you can use any rational matrix you like for the first one, and have all the others to tinker with just to make sure the determinant is non-zero. –  ndkrempel Feb 25 '11 at 17:26
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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.

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