"Conjugacy rank" of two matrices over field extension - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T20:26:15Z http://mathoverflow.net/feeds/question/9162 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9162/conjugacy-rank-of-two-matrices-over-field-extension "Conjugacy rank" of two matrices over field extension darij grinberg 2009-12-17T08:36:36Z 2009-12-18T17:46:16Z <p>EDIT: ARGH! I've got to go and I have no idea how to do the damn subscripts right.</p> <p>I have posted this <a href="http://www.mathlinks.ro/viewtopic.php?t=198806" rel="nofollow">elsewhere</a> and got only a partial reply. I don't know whether this qualifies the question for an open-problem tag; if it does, please anyone insert it.</p> <p>Let $L$ be a field, and $K$ a subfield of $L$. For any two $n\times n$ matrices $A$ and $B$ from <code>$\mathrm{M}_n\left(K\right)$</code>, and any field $S$ containing $K$, we denote <code>$$\rho_{S}\left( A,B\right) = \max\left\{\mathrm{Rank}Q\mid Q\in\mathrm{M}_{n}\left( S\right) ;\ AQ = QB\right\}.$$</code> We can call $\rho_{S}\left( A,B\right)$ the "conjugacy rank" of the matrices $A$ and $B$ over the field $S$ (noting that $\rho_{S}\left( A,B\right) = n$ if and only if the matrices $A$ and $B$ are conjugate to each other in $\mathrm{M}_{n}\left( S\right)$).</p> <p>My question is: Do we have $\rho_{K}\left( A,B\right) = \rho_{L}\left( A,B\right)$ for any two matrices $A$ and $B$ from $\mathrm{M}_{n}\left( K\right)$ ?</p> <p>This can be shown in the case of $n\leq\left\vert K\right\vert$ by a "polynomials which vanish everywhere must be identically $0$" argument. Besides, in the case of $\rho_{L}\left( A,B\right) = n$, it can be shown using the rational canonical form. I am interested in the most general case of the problem - neither restricting $\left\vert K\right\vert$, nor $\rho_{L}\left( A,B\right)$ -, but there may even be counterexamples.</p> <p>What also might be of help: For any field $S$ containing $K$, the space <code>$$R_{S}\left( A,B\right) = \left\{ Q\in\mathrm{M}_{n}\left( S\right) \mid AQ = QB\right\}$$</code> is a subspace of the vector space <code>$\mathrm{M}_{n}\left(S\right)$</code>. Besides, every basis of the space $R_{K}\left( A,B\right)$ is also a basis of the space $R_{S}\left( A,B\right)$ for every field $S$ containing $K$. However, <em>this alone is not enough</em>; you can easily construct a subspace of <code>$\mathrm{M}_{n}\left(\mathbb{F}_p\right)$</code> that consists of singular matrices only but loses this property when extended into <code>$\mathrm{M}_{n}\left(\mathbb{F}_{p^2}\right)$</code>.</p> http://mathoverflow.net/questions/9162/conjugacy-rank-of-two-matrices-over-field-extension/9187#9187 Answer by t3suji for "Conjugacy rank" of two matrices over field extension t3suji 2009-12-17T17:05:34Z 2009-12-17T17:05:34Z <p>I think this is true, and can be proved by brute force: write an explicit formula for conjugacy rank. I'll prefer to restate the problem in terms of modules.</p> <p>To an $n\times n$ matrix $A$ over a field $K$, associate the $K[x]$-module $M$ that is $K^n$ as a vector space, while $x$ acts as $A$. Everywhere below, all $K[x]$-modules are finite-dimensional as $K$-vector spaces. Then your definition becomes as follows:</p> <p>Let $M$ and $N$ be two $K[x]$-modules. Define their conjugacy rank $\rho(M,N)$ to be the maximal dimension (over $K$) of a $K[x]$-module that is simultaneously isomorphic to a submodule of $M$ and a quotient-module of $N$. We aim to prove that $\rho(M,N)$ is stable under field extensions of $K$.</p> <p>By structure theorem for modules over PID, we can write $M\simeq\bigoplus K[x]/f_i$, where invariant factors $f_i=f_i(M)\in K[x]$ satisfy $f_{i+1}|f_i$. It is easy to check the following claim:</p> <p>Lemma: $M'$ is isomorphic to a quotient of $M$ if and only $f_i(M')|f_i(M)$. The same criterion holds for $M'$ being isomorphic to a submodule of $M$.</p> <p>Corollary: There is unique up to isomorphism maximal-dimensional module $M'$ that is simultaneously isomorphic to a submodule of $M$ and a quotient-module of $N$; its invariant factors are given by $f_i(M')=gcd(f_i(M),f_i(N))$.</p> <p>Since the formula for $M'$ is stable under field extensions of $K$, the claim follows.</p> http://mathoverflow.net/questions/9162/conjugacy-rank-of-two-matrices-over-field-extension/9214#9214 Answer by Mariano Suárez-Alvarez for "Conjugacy rank" of two matrices over field extension Mariano Suárez-Alvarez 2009-12-18T00:13:28Z 2009-12-18T02:02:49Z <p><em>Suppose that the field extension $L/K$ is separable and that $K$ is infinite.</em> </p> <p>Let $A$, $B\in M_n(K)$ and suppose there exists a matrix $Q\in M_n(L)$ is such that $QA=BQ$ and which has rank $Q=r$. We want to show there is a matrix $Q'\in M_n(K)$ such that $Q'A=BQ'$ and which has rank at least $r$.</p> <p>First, by replacing $L$ by the subfield of $L$ generated over $K$ by the coefficients of $Q$ if we need to, we can suppose that $L/K$ is a finitely generated extension. By using a maximal purely trascendental extension of $K$ contained in $L$ as an intermediate step, we see that it is enough to consider separately the cases in which <em>(i)</em> $L/K$ is purely trascendental or <em>(ii)</em> $L/K$ is finite.</p> <p>In case <em>(i)</em>, let $S$ be a trascendence basis of $L/K$. Since the matrix $Q$ has rank $r$, it has an $r\times r$ minor $M$ with non-zero determinant. As the entries of $Q$ are finitely many rational functions in a finite number of elements of the indeterminates $S$, and since the field $K$ is infinite, we assign values from $K$ to the indeterminates which appear in $Q$ in such a way that we obtain a matrix $Q'\in M_n(K)$ (ie, we avoid zeroes in denominators) and such that the minor of $Q'$ corresponding to $M$ still has non-zero determinant. It is clear that $Q'A=BQ'$ and that the rank of $Q'$ is at least $r$, so we are done in this case.</p> <p>Let us now consider case <em>(ii)</em>. Up to enlarging $L$, we can assume that $L/K$ is Galois, with Galois group $G$. As before, the matrix $Q$ has an $r\times r$ minor $M$ with non-zero determinant. Suppose the elements of $G$ are $g_1=1_G,g_2,\dots,g_n$, and consider the polynomial $f(X_1,\dots,X_n)=\det_M\left(\sum_{i=1}^ng_i(Q)X_i\right)\in L[X_1,\dots,X_n]$; here the elements of $G$ act on the matrix $Q$ in the obvious way, and $\det_M$ denotes the determinant of the minor of its argument corresponding to $M$. Notice that $f$ is <em>not</em> the zero polynomial, because the coefficient of $X_1^r$ is precisely $\det_MQ\neq0$.</p> <p>Since $L$ is infinite and the elements of $G$ are algebraically independent (Lang, Algebra, VI, \S12, Theorem 12.2), the map $$x \in L \mapsto f(g_1(x),\dots,g_n(x))\in L$$ is not identically identically zero. It follows that there exists a $\xi\in L$ such that the matrix $Q'=\sum_{i=1}^ng_i(\xi)g_i(Q)$ has $\det_MQ'\neq0$; in particular, the rank of $Q'$ is at least $r$. Since the extension $L/K$ is Galois and $Q'$ is fixed by all elements in $G$, we see that $Q'\in M_n(K)$. Finally, since the matrices $A$ and $B$ have their coefficients in $K$, $Q'A=BQ'$.</p> http://mathoverflow.net/questions/9162/conjugacy-rank-of-two-matrices-over-field-extension/9291#9291 Answer by Ho Chung Siu for "Conjugacy rank" of two matrices over field extension Ho Chung Siu 2009-12-18T17:24:49Z 2009-12-18T17:46:16Z <p>Edit: As pointed out in comment, there's a flaw in this argument.</p> <p>Rewrite the equation $AQ-QB = 0$ using Kronecker's product (i.e. regard Q as a column vector), then we are to solve the equation </p> <p>$(I \otimes A - B^T \otimes I)Q = 0$</p> <p>and we are to find the maximal rank of the solution $Q$, where the coefficients of $Q$ may lie in a larger field.</p> <p>Let $M = I \otimes A - B^T \otimes I \in M_{n^2} (K)$, and perform Gaussian elimination to get invertible matrices $E,F$ in $M_{n^2}(K)$ such that $M = EDF$, with $D = diag(1,...,1,0,..,0)$ (r 1s). Now the maximal rank of the solution of $MQ = 0$, is obviously the same as the maximal rank of the solution $EDFQ = 0$, or the rank of the solution $D(FQ) = 0$.</p> <p>Since $F$ is invertible, $Q'=FQ$ has the same rank as $Q$.</p> <p>Yet the maximal rank of $Q'$ satisfying $DQ' = 0$ is clearly only dependent on $r$ - this is obviously independent of the field where $Q'$ lies in - so the conjugacy rank has to be independent of which extension of $K$ you are considering.</p>