Semi-simple matrices over fields of finite characteristic - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T10:26:36Z http://mathoverflow.net/feeds/question/51944 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51944/semi-simple-matrices-over-fields-of-finite-characteristic Semi-simple matrices over fields of finite characteristic Andreas Thom 2011-01-13T11:06:19Z 2011-01-14T19:17:06Z <p>Well-known and useful facts are: </p> <ul> <li>any symmetric matrix over $\mathbb R$ is semi-simple (i.e. diagonalizable), and</li> <li>any hermitean matrix over $\mathbb C$ is semi-simple.</li> </ul> <p>I will loosely speak about the shape of a matrix and mean the existence of some (linear) relations between matrix-entries (or functions of the matrix-entries).</p> <blockquote> <p><strong>Question:</strong> Let $k$ be an algebraically closed field of characteristic $p$. Is there any result whatsoever, which says that a rich class of matrices of a given shape consists only of semi-simple matrices.</p> </blockquote> <p>Since I am more interested in positive results, the notion of shape is kept flexible. However, if it could be proved that semi-simplicity is not implied by any shape in some reasonable class of shapes, this would be interesting as well.</p> http://mathoverflow.net/questions/51944/semi-simple-matrices-over-fields-of-finite-characteristic/51972#51972 Answer by ndkrempel for Semi-simple matrices over fields of finite characteristic ndkrempel 2011-01-13T15:48:55Z 2011-01-13T15:48:55Z <p>Well, distinct eigenvalues over the algebraic closure is enough to ensure semisimplicity, so the discriminant of the characteristic polynomial is a polynomial in the matrix entries whose non-vanishing ensures semisimplicity. If you prefer a closed set, you could require it to have a specific non-0 value, say 1. Whether this qualifies as a "shape" is another matter.</p> http://mathoverflow.net/questions/51944/semi-simple-matrices-over-fields-of-finite-characteristic/51980#51980 Answer by David Speyer for Semi-simple matrices over fields of finite characteristic David Speyer 2011-01-13T16:46:55Z 2011-01-13T16:46:55Z <p>A basic observation: If $K$ is a field where $0$ is a sum of nonzero squares, say $0=\sum_{i=1}^n x_i^2$, then $\left( x_i x_j \right)_{1 \leq i,j \leq n}$ is a symmetric, nonzero, matrix with square $0$. Such a matrix cannot be semisimple.</p> <p>So the implication "symmetric implies semisimple" only works over <a href="http://en.wikipedia.org/wiki/Formally_real" rel="nofollow">formally real fields</a>.</p> http://mathoverflow.net/questions/51944/semi-simple-matrices-over-fields-of-finite-characteristic/51994#51994 Answer by Anatoly Kochubei for Semi-simple matrices over fields of finite characteristic Anatoly Kochubei 2011-01-13T19:17:12Z 2011-01-13T19:17:12Z <p>There is a kind of spectral theorem describing a class of linear operators on Banach spaces over non-Archimedean fields possessing orthogonal (in the non-Archimedean sense) spectral decompositions. See A. N. Kochubei, Non-Archimedean normal operators, J. Math. Phys. 51 (2010), article 023526 (or ArXiv: 0908.4381).</p> http://mathoverflow.net/questions/51944/semi-simple-matrices-over-fields-of-finite-characteristic/52100#52100 Answer by BS for Semi-simple matrices over fields of finite characteristic BS 2011-01-14T18:32:35Z 2011-01-14T19:17:06Z <p>This is only a hint, not an answer.</p> <p>There is a simple characterization of semisimple matrices over finite fields. Namely, if $A\in M_n(F_q)$, its eigenvalues lie in $F_{q^m}$, $m=lcm(2,\dots,n)$, and there is $P\in GL_n(F_{q^m})$ such that $P^{-1}AP$ is a diagonal of Jordan blocks $\lambda_i I + N_i$, $i=1,\dots,s$. But it is easy to see that $(\lambda_i I+N_i)^{q^m}=\lambda_i I$ (note that $q^m \gt n$), so that $A$ is semisimple if and only if $A^{q^m}=A$.</p> <p>Now you might want to start to study the possibilities for vector spaces $V\subset M_n(F_q)$ (or other subvarieties) such that $A^{q^m}=A$ for all $A\in V$. </p>