Recovering eigenvalues of a matrix from its $p$th compound matrix

Question

This question was motivated by Find the determinant of a matrix given the determinant of all $p\times p$ sub-matrices?. Let $A$ be an $n\times n$ matrix over a field. Suppose we are given the $p$th compound matrix $C_p(A)$, together with the canonical indexing of its rows and columns by $p$-element subsets of $\{1,\dotsc,n\}$. (The entries of $C_p(A)$ are the determinants of the $p\times p$ submatrices of $A$.) Assume that $C_p(A)$ is nonsingular. Can we recover from this information the $p$th powers of the eigenvalues of $A$? The eigenvalues of $C_p(A)$ are products of $p$ of the eigenvalues of $A$. If the eigenvalues of $A$ are $\theta_1,\dotsc,\theta_n$ and we know for each eigenvalue $\psi$ of $C_p(A)$ which $p$ elements of $\{\theta_1,\dotsc,\theta_n\}$ $\psi$ is a product of, then my comment at the link above explains that the answer is yes. In this situation we don't need to know $C_p(A)$ itself but only its eigenvalues. But what if we are not given this information about $\psi$ but instead know $C_p(A)$ (and the indexing of its rows and columns)?

Answer 1

We'll consider only the case $C_p(A)$ ( and so $A$) nonsingular.

The $\binom{n}{p}$ minors of size $p\times p$ obtained from $p$ rows of the matrix determine the span of those $p$ rows ( Plucker coordinates). Hence we can determine the span of any $p$ rows, and by intersection, any row up to proportionality. Similarly for columns. So now we know the rows of $A$ up to proportionality, and also the columns. That is we have

$$a_{ij} = \lambda_i \cdot b_{ij} \\ a_{ij} = \mu_j \cdot c_{ij}$$

with $(b_{ij})$, $(c_{ij})$ known. To find $(a_{ij})$ (up to proportionality):

$$\frac{b_{ij}}{c_{ij}} = \frac{\mu_j}{\lambda_i}$$ that is, a matrix of rank $1$. So find $\mu_j$, $\lambda_i$ up to proportionality, and hence $(a_{ij})$.

Conclusion: $A$ is determined up to proportionality ( a $p$-th root of $1$).

Notes:

1. In the generic case, $A$ does not need to be square.

2. The case $p=n-1$ is simple since $C_{n-1} ( C_{n-1} (A)) \simeq A$ ( proportional)

3. One can recover powers of minors larger than $q$ by using the formula for the determinant of a compound matrix.

4. Using condensation formulas one can recover a fixed Hadamard power of $A$ ( up to proportionality).

$bf{Added:}$ If $p$ and $n$ -- the size of $A$, are relatively prime, then we can get a formula for $A$ (up to proportionality. Here is the idea:

1. evey minor of size $k p$ can be determined inductively ( use Laplace expansion.

2. Get every minor of size $q p$, where $n = q \cdot p + r$, is the division of $n$ by $p$. Now, get the minors of the adjoint of $A$ of size $r$. Repeat, using Euclid's algorithm. In end we get all the minors of size $d = \operatorname{gcd}(n,p)$ of $A$ or $\operatorname{adj}(A)$. If $d=1$ we are done.

Answer 2

Converted from a comment by another user:

Knowing the successive compound matrices of $A$ without knowing their eigenvalues one can recover the elementary symmetric polynomials in the eigenvalues of $A$. Example 5.6 in Prells, U.; Friswell, M. I.; Garvey, S. D.: Use of geometric algebra: compound matrices and the determinant of the sum of two matrices, Proceedings of the Royal Society of London A: Mathematical, Physical and Engineering Sciences 459 (2003): 273-285 (over $\mathbb{C}$, says that $${\rm det}(A-\lambda I) = (-1)^n \lambda^n + \sum_{j=1}^{(n-1)^j} \lambda^j {\rm tr}C_{n-j}(A) + {\rm det}A.$$