Sum of determinants of block submatrices

Solution 1:

As no one has answered let me offer something between a comment and a non-answer.

You have the formula $$ \sum_{S \subseteq [n]} M_{S,S} = \det(M + I) $$ expressing the sum of the principal minors of an $n\times n$ matrix $M$ in terms of $\det(M+I)$. We can generalise this a bit, replacing $I$ by the diagonal matrix $\Lambda:=\text{diag}(\lambda_1,\dots,\lambda_n)$. We will get $$ \sum_{S \subseteq [n]} M_{S,S}\lambda^{S'} = \det(M + \Lambda), $$ where $S'$ is the complementary set of indices, and $\lambda^{S'}$ is the monomial $\prod_{i\in S'}\lambda_i$.

Now suppose we are working with block matrices, with $M$ being $mn\times mn$, the blocks being of size $m\times m$. [Note, I choose to do the more general case (i) to illustrate the general technique; (ii) more importantly, to make clear that the line I am pursuing probably complicates matters, although (luckily?) for $m=2$ it does not.]

We want to use our formula for $\det(M+\Lambda)$; the 'problem' is how to exclude the minors which don't arise from the given block structure. What we can do is to put $\Lambda:=\Gamma(\gamma_1,\dots,\gamma_n)$ where $\Gamma(\gamma_1,\dots,\gamma_n)$ is the block-diagonal matrix $\text{diag}(\gamma_1 I_m,\dots, \gamma_n I_m)$.

The terms we want to retain in our expansion of $\det(M+\Lambda)$ are those where either all or none of the indices $1,2\dots,m$ occur, and all or none of the indices $m+1, m+2,\dots ,2m$ occur, and so on. In other words we must look only at the terms where the power of each $\gamma_i$ is divisible by $m$.

That is we must "section" the power series (actually polynomial) $\det(M+\Gamma(\gamma_1,\dots,\gamma_n))$. This we can do in the old-fashioned way. (Is this baby Fourier theory?) Let $\omega:=\text{e}^{2\pi\text{i}/n}$. Then the terms of interest are got by averaging the various $\det(M+\Gamma(\gamma_1 \omega^{r_1},\dots,\gamma_n \omega^{r_n}))$.

To be precise we have that the sum of the interesting principal minors is $$ \frac{1}{m^n} \sum_{r_1=0}^{m-1}\dots\sum_{r_n=0}^{m-1} \det(M+\Gamma(\omega^{r_1},\dots,\omega^{r_n})). $$

For $m=1$ this just recovers the original formula. For $m=2$ it gives an expression in terms of $2^n$ determinants, so it is no improvement. For larger $m$ it does seem to make things worse, although the expression seems to me of theoretical interest.