The Quillen-Suslin theorem states that projective modules over $\mathbb{Q}[x_1, \dots , x_n]$ are free. (This also holds over more general fields or rings.)
I have a square $m\times m$ matrix $M$ over $\mathbb{Q}[x_1, \dots , x_n]$. I am interested in the space spanned by its columns. Is this space projective?
Personally, I don't think this is projective in general, but I would still like to apply the Quillen-Suslin theorem partially to my setting, but I have dificulty interpreting the theorem. Therefore, I looked for a formulation of the result on the level of matrices. Eventually, I found the article by Youla and Pickel named 'The Quillen-Suslin Theorem and the Structure of $n$-Dimensional Elementary Polynomial Matrices'. They phrased the result in the following may:
If the $m\times r$ matrix $A$ with $r\leq m$ is zero prime (ZP), i.e. its $r\times r$ minors have no common zero, then it is projectively free (PJF), i.e. it is the first $r$ columns of some $m\times m$ matrix $V$ with constant determinant. As a consequence, $V^{-1}A = (1_{r\times r}, 0_{r\times (m-r)})^T$.
In my setting, I have a $m\times m$ matrix $M$ and a number $r$, such that its $r\times r$ minors have no common zero. I would like to conclude that there exist $V_1$ and $V_2$ with constant determinant such that $$V_1^{-1}M V_2 = \begin{pmatrix}1_{r\times r} & \ast \\ \ast & \ast \end{pmatrix} $$ or some similar form.
If there would exist transformations of the form $Q_1 M Q_2 $ where $Q_1$ and $Q_2$ have constant determinants such that the first $r$ columns of $Q_1 M Q_2 $ satisfy ZP, then we would be done. But I have no idea whether this would be achievable. Maybe a direct argument is also possible.
I guess there is also a connection with Am I interpreting Artin Theorem 14.9.1 correctly?, but I'm not familiar with the formulation there.