Is it possible to find the determinant of an $n\times n$- matrix, only given the determinant of all $p\times p$ sub-matrices in it? Here $p\leq n$ is fixed. This is obviously true if $p=1,n$. But what happens in other cases?



Find the determinant of a matrix given the determinant of all $p\times p$ sub-matrices?
As [mentioned](https://mathoverflow.net/questions/416659/find-the-determinant-of-a-matrix-given-the-determinant-of-all-p-times-p-sub-ma#comment1069136_416659) by Will Sawin, a necessary condition is that $p$ divides $n$. Thus let us assume that $n=pk$. Denoting $e\_1,\dotsc,e\_n$ the canonical basis, the knowledge of the $p\times p$ minors is the knowledge of the $p$-vectors $$(Ae\_{i\_1})\wedge\cdots\wedge(Ae\_{i\_p})\in\Lambda^p(K^n),$$
where $K$ is the field of scalars (e.g. $\mathbb C$).


Splitting
$$(Ae\_1)\wedge\cdots\wedge (Ae\_n) = [(Ae\_1)\wedge\cdots\wedge (Ae\_p)]\wedge\cdots\wedge[(Ae\_{n-p+1})\wedge\cdots\wedge (Ae\_n)],
$$
we see that $(Ae\_1)\wedge\cdots\wedge (Ae\_n)$ is a polynomial function in the $p\times p$ minors. Since $(Ae\_1)\wedge\dotsb\wedge (Ae\_n)=(\det A)e\_1\wedge\dotsb\wedge e\_n$, we deduce the value of $\det A$.


Let me describe how it works when $n=4$ and $p=2$. The minors are denoted
$$A\binom{i\alpha}{j\beta}=a\_{i\alpha}a\_{j\beta}-a\_{i\beta}a\_{j\alpha}.$$
Then
$$(Ae\_\alpha)\wedge(Ae\_\beta)=\sum\_{i<j}A\binom{i\alpha}{j\beta}e\_i\wedge e\_j.$$
We thus obtain
$$\det A=\sum\_{\substack{\rho\in{\frak S}\_4 \\ \rho(1)<\rho(2),\rho(3)<\rho(4)}}A\binom{\rho(1)1}{\rho(2)2}A\binom{\rho(3)3}{\rho(4)4}.$$


