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Let $A,B\in\mathbb{R}^{n\times n}$ be two nonsingular matrices with $A\ne B$, and consider the following partitioned matrix $$ M:=\begin{bmatrix}AA^\top + BB^\top & A^\top \Delta_1 A + B^\top \Delta_2 B \\ A^\top \Delta_1 A + B^\top \Delta_2 B & A^\top A + B^\top B \end{bmatrix} $$ where $\Delta_1$ and $\Delta_2$ are diagonal matrices with $0$-$1$ diagonal entries.

It is quite easy to see that if the diagonal entries of $\Delta_1$ and $\Delta_2$ are $1$s, then $M$ is singular. However, numerical simulations with randomly generated $\Delta_i$ (whose diagonal entries take values $0$ or $1$ with probability $1/2$) suggests that, in general, it is very "unlikely" that $M$ is singular. I'd like to understand if there is a way to formalize this fact, so my question is:

Given all possibile choices of matrices $\Delta_1$ and $\Delta_2$, how is it "likely" that matrix $M$ is singular?

I'm aware that my question is not very rigorous, but I hope it's understandable.

Thanks in advance for any help.

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  • $\begingroup$ For each $n$ the answer is rational number with power of two in the denominator. Did you try to search for the sequence of numerators in OEIS? $\endgroup$
    – Vít Tuček
    Commented Apr 28, 2020 at 21:35
  • $\begingroup$ @VítTuček: No, I did not. Could you please elaborate a little more on this? $\endgroup$
    – Ludwig
    Commented Apr 28, 2020 at 21:45
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    $\begingroup$ My understanding of your question is that you have a set of $2^{n+1}$ diagonal matrices with zeros and ones on the diagonal and you ask how many there are such that $\det(M(\Delta_1, \Delta_2)) = 0$ identicaly as a polynomial in $A$ and $B$ entries. I am suggesting you calculate first few of these numbers and then look for this sequence on oeis.org If you find something, great -- it can suggest possible way of attacking the problem. If you don't find anything, well... it seems that the sequence is new and hence we don't get any clues for free. $\endgroup$
    – Vít Tuček
    Commented Apr 28, 2020 at 21:51

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