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 $n^2 - r$ - 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.

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.

Rewrite the equation $AQ-QB = 0$ using Kronecker's product (i.e. regard Q as a row 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 $n^2 - r$ - 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.

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 $n^2 - r$ - 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.