Let $k,d$ be positive integers, $1<k<d$. Let $\lambda_I=\lambda_{i_1,\ldots,i_k}$ be real numbers, indexed by multi-indices $I=(i_1,\ldots,i_k)$, where $1\le i_1<\ldots<i_k \le d$.

Are there necessary and sufficient conditions on $\lambda_{i_1,\ldots,i_k}$ which are equivalent to the existence of $\sigma_1,\ldots,\sigma_d \in \mathbb{R}$ such that $\lambda_{i_1,\ldots,i_k}=\sigma_{i_1}\cdot \ldots\cdot\sigma_{i_k}$ holds for every multi-index $I$?

In other words, I am asking whether we can characterise which sequences of real numbers can arise as the $k$-minors of diagonal $d \times d$ matrices?

I am interested mainly in the case where all the $\lambda_{i_1,\ldots,i_k}$ are non-zero.

I have heard that the general problem of recognizing $k$-minors of arbitrary square matrices is open, but I am hoping that for diagonal matrices, the situation maybe better understood.

I guess this should be easier by starting over $\mathbb{C}$. What is known about that case?

Commnet: If I understand correctly, the Plucker relations only describe the minors of top-degree of a non-square matrix. Here I am talking about the minors of degree $k$, when $1<k<d$, i.e. non-top minors of a square matrix.

  • $\begingroup$ I guess you mean $\lambda_{i_1, \dots, i_k} = \sigma_{i_1} \cdot ... \cdot \sigma_{i_k}$? $\endgroup$ – Severin Schraven Jan 27 at 12:51
  • $\begingroup$ Yes, thank you! corrected now. $\endgroup$ – Asaf Shachar Jan 27 at 13:03

Let me give an answer over complex numbers. The conditions are the following: for each pair $i \ne j$ and for each pair of subsets $I,J \subset \{1,\dots,d\} \setminus \{i,j\}$ of cardinality $k - 1$ one has a relation $$ \lambda_{I \sqcup \{i\}} \cdot \lambda_{J \sqcup \{j\}} = \lambda_{I \sqcup \{j\}} \cdot \lambda_{J \sqcup \{i\}}. $$ Clearly these relations are necessary.

Let us also show that they are sufficient. In fact, let me just explain how $\sigma_i$ can be reconstructed. Take any set $J \subset \{1,\dots,d-1\}$ of cardinality $k$. Then set $$ \sigma_d = \sqrt[k]{\frac{\prod_{j \in J} \lambda_{J \setminus \{j\} \sqcup \{d\}}}{\lambda_J^{k-1}}}. $$ After that for each $i \in \{1,\dots,d-1\}$ choose $J \subset \{1,\dots,d-1\} \setminus \{i\}$ of cardinality $k-1$ and set $$ \sigma_i = \frac{\prod_{j \in J} \lambda_{J \setminus \{j\} \sqcup \{i,d\}}}{\lambda_{J \sqcup \{i\}}^{k-2}\sigma_d^{k-1}}. $$ One can check that this solves the required equations.

Over real numbers the only extra problem is the existence of the root.

  • $\begingroup$ I think we have $k$ fixed. $\endgroup$ – Severin Schraven Jan 27 at 13:06
  • $\begingroup$ @SeverinSchraven: You are right, I will change the answer. $\endgroup$ – Sasha Jan 27 at 13:50
  • $\begingroup$ @AsafShachar: Thanks for correction! $\endgroup$ – Sasha Jan 27 at 14:53
  • $\begingroup$ If $k$ is even, don't we have to be careful with the choice of the sign of $\sigma_d$? How do you know that you can pick the positive? Respectively, that you can take the kth root of the expression? $\endgroup$ – Severin Schraven Jan 27 at 14:57
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    $\begingroup$ @SeverinSchraven I will just note that when you work over the complex numbers, you don't really have a problem: Instead of possible non-existence of roots (in the real case), you now have too many of them. However, it does not matter "which one you choose", since choosing a different $k$-th root only amounts to multiplying all the $\sigma_i$ by some specific $k$-th root of unity, which is the only ambiguity we have here. (This was to be expected in advance, as I will elaborate in the question). $\endgroup$ – Asaf Shachar Jan 27 at 15:08

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