# Reconstructing a (unitary) matrix from the determinant of its sub-matrices

I want to find the unitary $N \times N$ matrix U from the following data. Let $M$ be an integer $(1< M<N-1)$ and let $\mathcal S$ be the space of all the possible subsets of $\{1,2,\dots, N\}$ with exactly $M$ elements.

Let $U(s,s')$ be the $M\times M$ submatrix of U whose rows are specified by the elements of $s\in\mathcal S$ and whose columns are specified by the elements of $s'\in\mathcal S$.

My data is a set of $I(s,s')\in\mathbb Q$ such that $$\det U(s,s') = e^{i \pi I(s,s')}.$$

Can I find the full matrix $U$? I would be interested also in understand why this problem might not have a solution. However, if the solution exists (for certain values of $M,N$), than I would like to find the algorithm to obtain $U$.

-
Homework assignment...? –  Wlodek Kuperberg Feb 3 '14 at 17:10
No, I can solve a research problem if there is a solution to this "homework" :-/ –  benkj Feb 3 '14 at 17:21
Do you know that the adjoint is the case when M=N-1? I suspect Cayley or Sylvester handled cases for other M. Gerhard "Does More Solitary Than Unitary" Paseman, 2014.02.03 –  Gerhard Paseman Feb 3 '14 at 17:31
You are right, thanks... but unfortunately when M=1 or M=N-1 the original problem is quite trivial –  benkj Feb 3 '14 at 17:38
Evidently, I misinterpreted the question. Sorry. –  Wlodek Kuperberg Feb 3 '14 at 20:22

$\newcommand{\bC}{\mathbb{C}}$ View the unitary matrix as a unitary operator $U:\bC^N\to\bC^N$. For any $1\leq M\leq N$ it canonically induces a unitary map

$$\Lambda^MU: \Lambda^M\bC^N\to\Lambda^M\bC^N,$$

where $\Lambda^M$ denotes the $M$-th exterior power of a vector space. Your problem can be rephrased as follows: how much of $U$ can one recover given that we have complete knowledge of $\Lambda^M U$.

Immediately one ses that we can recover quite a bit. If $\lambda_1,\dotsc,\lambda_N$ are the eigenvalues of $U$ then the eigenvalues of $\Lambda^M U$ are

$$\lambda(S)=\prod_{k\in S}\lambda_k,$$

where $S$ runs through the subsets of $\{1,\dotsc, N\}$ of cardinality $M$. Observe that if $M< N$, then the collection of eigenvalues

$$\{ \lambda(S);\;\;|S|=M\}$$

determines the collection of ratios

$$\frac{\lambda_i}{\lambda_j},\;\; 1\leq i<j\leq N.$$

With this extra knowledge we can compute the powers $\lambda_1^M,\dotsc, \lambda_N^M$, and thus the eigenvalues of $U$ up to an overall multiplicative ambiguity $\rho$ which is an $M$-th root of $1$. You cannot do better than this because

$$\Lambda^M U=\Lambda^M(\rho U).$$

-
Thanks for your answer. Do you think that one can reconstruct also the entire matrix from the conditions? For example, let us consider the first non-trivial case: M=2 (or M=N-2). The constraints that I have are N(N-1)/2, a number which is smaller than the number of independent coefficients of U. So I hope that, at least in this case, there is a way for determining U (not in a unique way) –  benkj Feb 4 '14 at 8:23
The real dimension of $U$ is $N^2$ and if you are given $N(N-1)/2$ complex constraints, this indicates that the space of solutions of $\Lambda^2 U=T$ is a real variety of dimension $N$. –  Liviu Nicolaescu Feb 4 '14 at 13:29
Cannot edit my previous comment. You actually have $\binom{N}{2}^2$ constraints so my above comment is misleading. One case that seems hard to decide is when $U$ is an involution and $\dim \ker(1\pm U)>1$. –  Liviu Nicolaescu Feb 4 '14 at 13:41
Here is an even more challenging case. Assume that the spectrum of $U$ as a set consists of all the $2k$-th roots of $1$ for some $k\geq 1$ and each eigenvalue has multiplicity $>1$. It seems difficult to recover the the eigenspaces of $U$ from those of $\Lambda^2 U$. –  Liviu Nicolaescu Feb 4 '14 at 13:50