Sum of determinants of block submatrices

Question

I have a $2n \times 2n$ matrix, $M$. I view it a block matrix, of $n^2$ blocks, each of shape $2\times 2$. Computing the determinant of $M$ is easy by conventional methods. I could also look at diagonal block submatrices: given a set $S \subseteq [n]$, I take the submatrix of all the rows in $S$ and columns in $S$.

For instance, if $n=3$, I have 9 blocks (each of 4 entries), $A_{1,1}, A_{1,2} \dots A_{3,3}$. $S$ could be ${1,3}$, in which case I take the 4x4 submatrix

$$\begin{bmatrix} A_{1,1} & A_{1,3} \\ A_{3,1} & A_{3,3}\end{bmatrix}$$

There are $2^n$ such submatrices, and each has its own determinant. I would like to compute the sum of those determinants as efficiently as possible. (The trivial case, where $S = \emptyset$, naturally has a determinant of 1.) This might sound like a bold hope, but consider that if this were not a block matrix, then you can very easily compute the sum of the determinants of the diagonal submatrices:

$$ \sum_{S \subseteq [n]} M_{S,S} = \det(M + I) $$

So I'm hoping that there's some way to make a good "block identity matrix" that I can carry out operations with to compute this sum. Honestly, I'd be happy with anything in polynomial time.

Proofs of hardness would also be interesting. For instance, I could believe that the problem of computing the permanent might somehow be reduced this problem, in which case it would be NP-Hard. I have several fairly different approaches that take $2^n$ time, so anything in even $1.9^n$ would be stimulating.

Answer 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.