Determinants and $2\times 2$ minors

Question

I just proved that

$$0=\left|\begin{array}{cccc} a_0 & a_1 & a_2 & a_3 \\ b_0 & b_1 & b_2 & b_3 \\ a_0 & a_1 & a_2 & a_3 \\ b_0 & b_1 & b_2 & b_3 \end{array}\right|=2(\Delta_{01}\Delta_{23}-\Delta_{02}\Delta_{13}+\Delta_{03}\Delta_{12})$$ where $$\Delta_{ij}=\left|\begin{array}{cc} a_i & a_j \\ b_i & b_j \end{array}\right|$$ Is there any way of expressing the following determinant in a similar way? That is, is there any way of expanding the following determinant as a sum of products of minors of order $2\times 2$ of the form $\Delta_{ij}$?

$$0=\left|\begin{array}{cccc} a_0 & a_1 & \cdots & a_n \\ b_0 & b_1 & \cdots & b_n \\ \vdots & \vdots & \cdots & \vdots \\ a_0 & a_1 & \cdots & a_n \\ b_0 & b_1 & \cdots & b_n \end{array}\right|$$ (of course, one must take into account the parity of $n$). I need it in order to find the set of zeroes that defines a projective algebraic variety.

Answer 1

I'm sorry, I now realise I changed the notation so that my indices run $1$ to $n$ and not $0$ to $n$.

What you have written down is essentially the Laplace expansion of your matrix using $2\times 2$ minors from the top two rows against the complementary $(n-2)\times (n-2)$ minors. With $n=4$ and your symmetry it collapses the way you say.

If $n=2k$ we can expand by the minors of the first two rows against their complements, and then repeat the process. I think we get the rather horrid

$$ \sum\epsilon(i_1,j_1,\dots,i_k,j_k)\ \Delta_{i_1,j_1}\cdot\Delta_{i_2,j_2}\cdot\dots\cdot\Delta_{i_k,j_k} $$ where the sum is over all $$\begin{align} i_1&<j_1\\ i_2&<j_2 \ \text{and}\ i_2,j_2\in\{1,\dots,2k\}\setminus\{i_1,j_1\}\\ i_3&<j_3 \ \text{and}\ i_3,j_3\in\{1,\dots,2k\}\setminus\{i_1,j_1, i_2,j_2\}\\ \text{and so on} \end{align} $$ and $\epsilon(i_1,j_1,\dots,i_k,j_k)$ is the sign of the permutation.

In fact the distinct terms will each turn up $k!$ times but to write this down I'd need much better notation.