Express determinant as a determinant of smaller order

I want to know whether the determinant (as a polynomial) can always be expressed as a determinant of some polynomial matrix of any lower dimension. To be precise, can the polynomial $$p(x_{11},...,x_{nn}) = \mathrm{det}\Big( (x_{ij})_{i,j=1}^n \Big)$$ be written in the form $$p(x_{11},...,x_{nn}) = \mathrm{det}(A), \; \; A \in \mathbb{C}[x_{11},...,x_{nn}]^{m \times m},$$ an $(m \times m)$ determinant of a polynomial matrix, for all $1 \le m \le n$?

It is clearly true when $m = 1.$

A nontrivial case is when $n = 4$ and $m = 2$ where we can use the block matrix determinant $$\mathrm{det} \begin{pmatrix} A & B \\ C & D \end{pmatrix} = \mathrm{det}(A) \mathrm{det}(D - CA^{-1}B) = \mathrm{det}(AD - ACA^{-1}B)$$ which I believe can be made to work even when $A$ is singular by using the adjugate rather than the inverse. I'm not sure about the details but I think there is a solution.

When trying to write the $n=3$ determinant as an $m=2$ determinant I ran into situations where I needed to take square roots of variables. I'm not convinced this can be done but I would be interested in a proof either way.

• After writing this it occured to me that the question is silly as stated since we can always take $A$ to be the diagonal matrix $(\mathrm{det}(x_{ij}),1,...,1).$ Maybe the question is about a natural choice for $A$, but more than likely this should just be closed. Dec 21, 2016 at 23:42

You can have a look at this formula.

https://en.m.wikipedia.org/wiki/Dodgson_condensation

• I think this is exactly what I'm looking for, thanks! Dec 21, 2016 at 23:54