Division theorem for vector-valued distributions

Question

The classical division theorem for scalar distributions on $\mathbb R^n$ can be formulated as follows. Let $T$ be a tempered distribution on $\mathbb R^n$ and let $P$ be a non-zero polynomial of $n$ variables. Then there exists a tempered distribution $S$ such that $ T=PS. $ The above result was proven by Lojasiewicz (MR0096120) in the more general case where $P$ is an analytic function and by Hörmander (MR0124734) for the polynomial case.

My question: is there a vector-valued version ? For instance let us consider a tempered distribution $T\in \mathscr S'(\mathbb R^n; \mathbb R^n)$ and let $P$ be a $n\times n$ matrix with polynomial entries and whose determinant is not identically 0. I would like to find $S\in \mathscr S'(\mathbb R^n; \mathbb R^n)$ such that $T=PS$.

Answer 1

I think I have an answer to my own question : let us consider $Q$ the transposed of the comatrix of $P$. The determinant of $P$ is a polynomial and by the Lojasiewicz-Hörmander theorem, we can find a tempered distribution $\tilde T$ such that $$ T=(\det T) \tilde T=P Q\tilde T= PS, \quad S=Q\tilde T. $$