# "Conjugacy rank" of two matrices over field extension

I have posted this elsewhere 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.

Let $$L$$ be a field, and $$K$$ a subfield of $$L$$. Let $$n$$ and $$m$$ be two nonnegative integers.

For any $$n\times n$$ matrix $$A \in \operatorname{M}_n\left(K\right)$$ and any $$m\times m$$-matrix $$B \in \operatorname{M}_m\left(K\right)$$, and any field $$S$$ containing $$K$$, we define \begin{align} \rho_{S}\left( A,B\right) := \max\left\{\operatorname{Rank} Q \mid Q\in\operatorname{M}_{n,m}\left( S\right) ; \ AQ = QB\right\}. \end{align} We can call this number $$\rho_{S}\left( A,B\right)$$ the "conjugacy rank" of the matrices $$A$$ and $$B$$ over the field $$S$$.

(Note that if $$n = m$$, then this conjugacy rank is directly connected with conjugacy -- i.e., similarity -- of matrices: Namely, in this case, we have $$\rho_{S}\left( A,B\right) = n$$ if and only if the matrices $$A$$ and $$B$$ are conjugate to each other in the ring $$\operatorname{M}_{n}\left( S\right)$$.)

My question is: Do we have $$\rho_{K}\left( A,B\right) = \rho_{L}\left( A,B\right)$$ for any two matrices $$A \in \operatorname{M}_n\left(K\right)$$ and $$B \in \operatorname{M}_m\left(K\right)$$ ?

This can be shown in the case of $$n = m \leq \left\vert K\right\vert$$ by a "polynomials which vanish everywhere must be identically $$0$$" argument. Besides, in the case of $$\rho_{L}\left( A,B\right) = n = m$$, 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 $$\rho_{L}\left( A,B\right)$$, nor requiring $$n = m$$.

What also might be of help: For any field $$S$$ containing $$K$$, the space $$R_{S}\left( A,B\right) = \left\{ Q\in\operatorname{M}_{n,m}\left( S\right) \mid AQ = QB\right\}$$ is a subspace of the vector space $$\operatorname{M}_{n,m}\left(S\right)$$. 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, this alone is not enough; you can easily construct a subspace of $$\operatorname{M}_{n}\left(\mathbb{F}_p\right)$$ that consists of singular matrices only but loses this property when extended into $$\operatorname{M}_{n}\left(\mathbb{F}_{p^2}\right)$$.

• I fixed all of your formulas. Dec 17, 2009 at 8:45
• Thanks! Though honestly I still have no idea when I have to use the grave accent and when I don't - seems perfectly random to me. But using it is a safe way, at least. Dec 17, 2009 at 11:05

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.

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:

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 $$K$$.

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 $$f_{i+1}|f_i$$. (We set $$f_i = 1$$ when $$i$$ is larger than the number of invariant factors.) It is easy to check the following claim:

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 $$M$$.

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 $$f_i(M')=gcd(f_i(M),f_i(N))$$.

Since the formula for $$M'$$ is stable under field extensions of $$K$$, the claim follows.

• Great! I knew about invariant factors, but I didn't make the observation that they are invariant under base change (i. e., that the $f_i$ are the same over $K$ and over $L$). Of course, this is trivial, but one doesn't think of it if one has always been thinking of primary decomposition instead of invariant factors. Dec 18, 2009 at 15:35
• Another thing your argument shows: $\rho_K\left(A,B\right)=\rho_K\left(B,A\right)$. Okay, this actually follows from $A$ being similar to $A^T$ and $B$ to $B^T$, but this is another consequence of your argument. Dec 20, 2009 at 0:06

Edit: As pointed out in comment, there's a flaw in this argument.

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

$(I \otimes A - B^T \otimes I)Q = 0$

and we are to find the maximal rank of the solution $Q$, where the coefficients of $Q$ may lie in a larger field.

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 $D(FQ) = 0$.

Since $F$ is invertible, $Q'=FQ$ has the same rank as $Q$.

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.

• $Q'$ does not necessarily have the same rank as $Q$: in the formula $Q'=FQ$, you view $Q$ as a vector that is acted upon by an invertible $n^2\times n^2$ matrix $F$. For instance, $F$ could be the transformation $\begin{pmatrix} a&b\\c&d\end{pmatrix}\mapsto\begin{pmatrix} a&b\\d&c\end{pmatrix}$, and $Q$ could be the identity, then $Q'$ is singular. Dec 18, 2009 at 17:34

Suppose that the field extension $$L/K$$ is separable and that $$K$$ is infinite.

Let $$A\in M_n(K)$$, $$B\in M_m(K)$$ and suppose there exists a matrix $$Q\in M_{n,m}(L)$$ is such that $$QA=BQ$$ and which has rank $$Q=r$$. We want to show there is a matrix $$Q'\in M_{n,m}(K)$$ such that $$Q'A=BQ'$$ and which has rank at least $$r$$.

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 transcendental extension of $$K$$ contained in $$L$$ as an intermediate step, we see that it is enough to consider separately the cases in which (i) $$L/K$$ is purely transcendental or (ii) $$L/K$$ is finite.

In case (i), let $$S$$ be a transcendence 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 $$Q'\in M_{n,m}(K)$$ (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.

Let us now consider case (ii). 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 $$g_1=1_G,g_2,\dots,g_j$$, and consider the polynomial $$f(X_1,\dots,X_j)=\det_M\left(\sum_{i=1}^j g_i(Q)X_i\right)\in L[X_1,\dots,X_j]$$; 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 not the zero polynomial, because the coefficient of $$X_1^r$$ is precisely $$\det_M Q\neq0$$.

Since $$L$$ is infinite and the elements of $$G$$ are algebraically independent (Lang, Algebra, VI, \S12, Theorem 12.2), the map $$x \in L \mapsto f(g_1(x),\dots,g_j(x))\in L$$ is not identically identically zero. It follows that there exists a $$\xi\in L$$ such that the matrix $$Q'=\sum_{i=1}^j g_i(\xi)g_i(Q)$$ has $$\det_M Q'\neq0$$; 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 $$Q'\in M_{n,m}(K)$$. Finally, since the matrices $$A$$ and $$B$$ have their coefficients in $$K$$, $$Q'A=BQ'$$.

• In both of your cases you are assuming that $K$ is an infinite field, and as grinberg pointed out if $|K|>n$ the question is easily settled. Dec 18, 2009 at 1:21