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The Schur-Horn theorem says that there exists a Hermitian matrix with diagonal entries $d_1,d_2,\ldots,d_n$ and eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_n$ if and only if $(\lambda_1,\lambda_2,\ldots,\lambda_n) \succeq (d_1,d_2,\ldots,d_n)$, where ``$\succeq$'' refers to majorization.

I am interested in how this generalizes to principal minors. In particular, is the following true?

Conjecture. Let $1 \leq k \leq n$ and let $\vec{d},\vec{\lambda} \in \mathbb{R}^r$, where $r = \binom{n}{k}$. There exists an $n \times n$ Hermitian matrix with its $k \times k$ principal minors equal to the entries of $\vec{d}$, and $k$-fold products of its eigenvalues equal to the entries of $\vec{\lambda}$, if and only if $\vec{\lambda} \succeq \vec{d}$.

Some notes and examples:

  • If $k = 1$ then this conjecture is just the usual Schur-Horn theorem, and is thus true.
  • If $k = n$ then this conjecture just says that if $d,\lambda \in \mathbb{R}$ then there exists a Hermitian matrix with its $n \times n$ principal minor (i.e., its determinant) equal to $d$, and the product of all $n$ of its eigenvalues equal to $\lambda$, then $d = \lambda$ (which is trivially true).
  • If $n = 3$ and $k = 2$, for example, the conjecture states that there is a matrix with its three $2 \times 2$ principal minors equal to $d_1$, $d_2$, $d_3$, and its three eigenvalues equal to $\lambda_1$, $\lambda_2$, $\lambda_3$, if and only if $(\lambda_1\lambda_2, \lambda_1\lambda_3, \lambda_2\lambda_3) \succeq (d_1,d_2,d_3)$.

One implication of the conjecture is not too difficult to prove: for every Hermitian matrix $A$, the majorization specified by the conjecture indeed holds. To see this, just construct the $k$-th compound matrix of $A$, whose eigenvalues are the entries of $\vec{\lambda}$ and whose diagonal entries are the entries $\vec{d}$, and apply the Schur-Horn theorem to it.

However, the converse (i.e., the fact that the majorization implies existence of such a matrix) seems much trickier. Is it known?

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    $\begingroup$ This can't work for dimension reasons. There are only $n^2$ real parameters to give an $n \times n$ matrix, and there are $\binom{n}{k}$ principal minors. We often have $\binom{n}{k} > n^2$, for example $\binom{8}{4} = 70 > 8^2 = 64$, so many $\binom{n}{k}$-tuples are not the principal minors of any matrix. $\endgroup$
    – David E Speyer
    Commented Feb 1, 2022 at 1:02
  • $\begingroup$ Thanks David. I actually had a similar thought, but I second-guessed it because it’s not clear to me that we can always find a $\vec{\lambda}$ that majorizes $\vec{d}$ either. So could it be that when a tuple does not describe principal minors of any matrix, it’s also not majorized by any vector of $k$-fold products of numbers? $\endgroup$
    – Nathaniel Johnston
    Commented Feb 1, 2022 at 1:24
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    $\begingroup$ Thanks for pointing that out, I hadn't thought about that, but no, I don't think that can save you. Start with a Hermitian matrix and let $\vec{\lambda}$ and $\vec{d}$ be the vectors of eigenvalues and minors, as you describe. Assume that all the majorization inequalities are strict (which will generally hold). Then there is a small open ball around $\vec{d}$, inside the hyperplane where $\sum d_i$ is fixed, so that all the majorization inequalities still hold in that ball. And the dimension of the ball is $\binom{n}{k}-1$, so we still have the problem. $\endgroup$
    – David E Speyer
    Commented Feb 1, 2022 at 2:12
  • $\begingroup$ @DavidESpeyer - Thanks very much; I'm convinced. If you write your comments as an answer, I'll accept it. $\endgroup$
    – Nathaniel Johnston
    Commented Feb 1, 2022 at 12:30

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