galois theory - "Conjugacy rank" of two matrices over field extension

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

Let $A$, $Bin M_n(K)$ and suppose there exists a matrix $Qin
M_n(L)$ 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 $r$.

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 (i) $L/K$ is purely trascendental or
(ii) $L/K$ is finite.

In case (i), let $S$ be a trascendence basis of $L/K$. Since the matrix $Q$
has rank $r$, it has an $rtimes r$ 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.

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
$rtimes r$ 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
$f(X_1,dots,X_n)=det_Mleft(sum_{i=1}^ng_i(Q)X_iright)in
L[X_1,dots,X_n]$; 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 not the zero polynomial, because the coefficient of $X_1^r$ is precisely $det_MQneq0$.

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 $xiin L$ 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
$K$, $Q'A=BQ'$.