Rank of sum of Galois conjugates of a matrix - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T17:39:55Z http://mathoverflow.net/feeds/question/56647 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56647/rank-of-sum-of-galois-conjugates-of-a-matrix Rank of sum of Galois conjugates of a matrix ndkrempel 2011-02-25T16:49:38Z 2011-02-25T20:33:08Z <p>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?</p> http://mathoverflow.net/questions/56647/rank-of-sum-of-galois-conjugates-of-a-matrix/56649#56649 Answer by Charles Matthews for Rank of sum of Galois conjugates of a matrix Charles Matthews 2011-02-25T17:10:20Z 2011-02-25T17:10:20Z <p>You can do this: take a normal basis for <em>K</em>, and write <em>M</em> 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.</p> http://mathoverflow.net/questions/56647/rank-of-sum-of-galois-conjugates-of-a-matrix/56676#56676 Answer by ndkrempel for Rank of sum of Galois conjugates of a matrix ndkrempel 2011-02-25T20:33:08Z 2011-02-25T20:33:08Z <p>Here is one way to get a theorem about this situation:</p> <p>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.</p>