Characterising minors of diagonal matrices

Let $$k,d$$ be positive integers, $$1. Let $$\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.

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, i.e. non-top minors of a square matrix.

• I guess you mean $\lambda_{i_1, \dots, i_k} = \sigma_{i_1} \cdot ... \cdot \sigma_{i_k}$? – Severin Schraven Jan 27 at 12:51
• Yes, thank you! corrected now. – 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.
• I think we have $k$ fixed. – Severin Schraven Jan 27 at 13:06
• 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? – Severin Schraven Jan 27 at 14:57
• @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). – Asaf Shachar Jan 27 at 15:08