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Fix $d\in\mathbb{N}$ and consider $e_{i,j}\in\mathbb{C}$ for $i=1,\dots,d+3$ and $j=1,\dots,d-1$. Suppose to have the following matrices $$N_{i,1}=\begin{pmatrix} 1 & 0 \\ e_{i,1} & 1 \end{pmatrix}$$ $$N_{i,2}=\begin{pmatrix} (-1)^de_{d-1} & (-1)^{d-1}e_{d-2} & \cdots & (-1)^2e_1 \\ 0 & (-1)^de_{d-1} & \cdots & (-1)^3e_2 \end{pmatrix}$$ $$A_{i}=\begin{pmatrix} I_{d-1} & {^t}(N_{i,1}N_{i,2}) \end{pmatrix}$$ $$B_{i}=\begin{pmatrix} -N_{i,1}N_{i,2} & I_2 \end{pmatrix}$$ $$ S_i=\begin{pmatrix} 0 & 0 & \cdots & 0 & (-1)^d\\ (-1)^{d}e_{i,d-1} & 0 & \cdots & 0 & e_{i,1}\\ 0 & 0 & \cdots & (-1)^{d-1} & 0\\ (-1)^{d-1}e_{i,d-2} & 0 & \cdots & e_{i,1} & (-1)^d\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & (-1)^3 & \dots & 0 & 0 \\ (-1)^3e_{i,2} & e_{i,1} & \cdots & 0 & 0\\ 1 & 0 & \dots & 0 & 0\\ 2e_{i,1} & (-1)^3 & \dots & 0 & 0 \end{pmatrix}$$ and consider the following matrix $$M=\begin{pmatrix} A_1\otimes B_1 & S_1 & 0 & 0 & \cdots & 0 & 0\\ A_2\otimes B_2 & 0 & S_2 & 0 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ A_{d+3}\otimes B_{d+3} & 0 & 0 & 0 & \cdots & 0 & S_{d+3}\\ \end{pmatrix}$$

I would like to prove that $M$ has full rank for the generic choice of the $e_{i,j}$'s (in particular it would be enough to prove it for a specific choice of the values). I know that the rank of the Kronecker product is the product of the rank of the matrices so that any $A_i\otimes B_i$ has full rank for the generic choice of the parameters, but I do not know how to say something about the rank of the whole matrix.

To be precise and honest, I can prove "by hand" that the whole matrix has full rank, for $d=3,4,5$, but the procedure is quite long and tedious, and it would be impossible to generalise it to the generic case. So, what I am looking for, is an argument more "intrinsic" and generalisable: I tried in many ways, but I couldn't find a smart way, could someone please help me? I would also appreciate a lot any kind of reference helpful for this problem.

Since the matrices $S_i$ are a little bit hard to understand, I write them here in the cases $d=3,4,5,6$.

If $d=3$, then $$\begin{pmatrix} 0 & -1 \\ -e_{i,2} & e_{1,1}\\ 1 & 0 \\ 2e_{1,1} & -1 \end{pmatrix}.$$

If $d=4$, then $$\begin{pmatrix}0 & 0 & 1\\ e_{i,3} & 0 & e_{i,1}\\ 0 & -1 & 0\\ -e_{i,2} & e_{i,1} & 1\\ 1 & 0 & 0 \\ 2e_{i,1} & -1 & 0 \end{pmatrix}.$$

If $d=5$, then $$\begin{pmatrix} 0 & 0 & 0 & -1\\ -e_{i,4} & 0 & 0 & e_{i,1}\\ 0 & 0 & 1 & 0\\ e_{i,3} & 0 & e_{i,1} & -1\\ 0 & -1 & 0 & 0\\ -e_{i,2} & e_{i,1} & 1 & 0\\ 1 & 0 & 0 & 0 \\ 2e_{i,1} & -1 & 0 & 0 \end{pmatrix}$$

If $d=6$ then $$\begin{pmatrix} 0 & 0 & 0 & 0 & 1\\ e_{i,5} & 0 & 0 & 0 & e_{i,1}\\ 0 & 0 & 0 & -1 & 0\\ -e_{i,4} & 0 & 0 & e_{i,1} & 1\\ 0 & 0 & 1 & 0 & 0\\ e_{i,3} & 0 & e_{i,1} & -1 & 0\\ 0 & -1 & 0 & 0 & 0\\ -e_{i,2} & e_{i,1} & 1 & 0 & 0\\ 1 & 0 & 0 & 0 & 0\\ 2e_{i,1} & -1 & 0 & 0 & 0 \end{pmatrix}$$

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    $\begingroup$ The tedious work for these particular matrices can be delegated to a computer algebra system. What is the generic problem? $\endgroup$
    – Max Alekseyev
    Commented Jul 11 at 11:10
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    $\begingroup$ The generic problem is obtained considering not $e_{i,2},e_{i,3},e_{i,4}$ but $e_{i,j}$ for $j=2,\dots, d-1$ with the matrices changing as a consequence. And then you have to consider not just 8 blocks but $d+3$. Do you think that a computer algebra system could solve the problem for a generic d? $\endgroup$
    – Stefano
    Commented Jul 11 at 11:31
  • $\begingroup$ This does not seem clear, for $A_i$ and $B_i$, say $j=2$ , is $A_i=\begin{pmatrix} -e_{i,2} & 0 & 1 & 0\\ 0&-e_{i,2} & 0 & 1 \end{pmatrix}$ ? If yes maybe $B_i$ expressions follows but the expression of $S_i$ is not very symmetrical, please recheck, giving examples in lower dimensions ($j=2$ or $3$) also helps. $\endgroup$
    – Toni Mhax
    Commented Jul 15 at 6:48
  • $\begingroup$ @ToniMhax I edited the text and now the problem is written in all its generality. If you could give me some suggestion, I would be very thankful $\endgroup$
    – Stefano
    Commented Jul 18 at 14:52
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    $\begingroup$ @ToniMhax done, I added them in the cased d=4,5,6, I hope now it is clearer, thank you! $\endgroup$
    – Stefano
    Commented Jul 18 at 18:02

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This is long comment not provide an anwser to your question

Let us consider the structure of the block matrix $M$ and the properties of the Kronecker product. First, note that $B_i \otimes A_i$ has full rank 8 for generic choices of $(e_{i,2}, e_{i,3}, e_{i,4})$ because the rank of a Kronecker product is the product of the ranks of the individual matrices. Since each $A_i$ is a $2 \times 6$ matrix of rank 2 and each $B_i$ is a $4 \times 6$ matrix of rank 4, their Kronecker product $B_i \otimes A_i$ will have rank $2 \cdot 4 = 8$.

Now, consider the block matrix $M$: $$ M=\begin{pmatrix} B_1 \otimes A_1 & S_1 & 0_{8,4} & \cdots & 0_{8,4}\\ B_2 \otimes A_2 & 0_{8,4} & S_2 & \cdots & 0_{8,4}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ B_8 \otimes A_8 & 0_{8,4} & 0_{8,4} & \cdots & S_8\\ \end{pmatrix} $$

By the structure of $M$, each $S_i$ provides additional rows that are linearly independent of the rows in $B_i \otimes A_i$. Hence, the combined structure of $M$ ensures that the full rank condition propagates through the entire block matrix. Therefore, for a generic choice of the parameters $(e_{i,2}, e_{i,3}, e_{i,4})$, the matrix $M$ will indeed have full rank. This argument can be generalized to larger matrices with similar structures.

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    $\begingroup$ Some of your answer is a copy of the OP, did you have computer assistance in formulating this answer? $\endgroup$
    – David Roberts ♦
    Commented Jul 16 at 6:02

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