Minimizing the determinant of a Hermitian matrix subject to certain constraints

Question

We have an $n \times n$ Hermitian and positive definite matrix $\mathbf{A}$ with diagonal elements equal to $1$ and off-diagonal elements of the (polar) form $r_{ij} e^{j \theta_{ij}}$, where the magnitudes satisfy $|r_{ij}| < 0.01$ and the phases $\theta_{ij} \in [0, 2\pi)$. Note that $n$ is usually big, e.g., $10^3$.

The matrix $\mathbf{A}$ can be written as: $$ \mathbf{A} = \begin{bmatrix} 1 & r_{12} e^{j \theta_{12}} & \cdots & r_{1n} e^{j \theta_{1n}} \\ r_{12} e^{-j \theta_{12}} & 1 & \cdots & r_{2n} e^{j \theta_{2n}} \\ \vdots & \vdots & \ddots & \vdots \\ r_{1n} e^{-j \theta_{1n}} & r_{2n} e^{-j \theta_{2n}} & \cdots & 1 \end{bmatrix}. $$ which is in fact a Gram matrix corresponding to a normalized basis$^\color{magenta}{\dagger}$. The optimization problem is: $$ \min_{\{r_{ij}, \, \theta_{ij}\}} \det(\mathbf{\mathbf{I}+\mathbf{A}}) $$ subject to the following constraints$^\color{blue}{\dagger}$: $$ \begin{aligned} & \mathbf{A} = \mathbf{A}^H \quad (\text{Hermitian condition}), \\ & \mathbf{A} \succ 0 \quad (\text{positive definiteness}), \\ & |r_{ij}| < 0.01, \quad \forall \, 1 \le i < j \le n, \\ & \theta_{ij} \in [0, 2\pi), \quad \forall \, 1 \le i < j \le n, \\ & a_{ii} = 1, \quad \forall \, i = 1, \dots, n. \end{aligned} $$

I want to know the eigenvalues of the optimal $\mathbf{A}$. However, I have difficulty in using the condition $|r_{ij}| < 0.01$ to solve the problem. I cannot explain how $|r_{ij}| < 0.01$ is related to the eigenvalues of $\mathbf{\mathbf{I}+\mathbf{A}}$ or how it constrains them. Could you give me some insights or any intuitive thinking if you have learned things related?

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Numerical experiment

To get some intuition, I tried to do this in a numerical way. I use a $4\times4$ (yes, I know I said $n$ is usually big. I use $4$ here to make it easier for the solver) matrix and set $|r_{ij}| < a$.

If I use $a=0.1$ (yes, I know it is $0.01$ in the original optimization problem. I use bigger $a$ here because I will make $(n-1)a>1$ later as it should be in the optimization problem above), here are the eigenvalues of the optimal $\mathbf{\mathbf{I}+\mathbf{A}}$:

    1.7000
    2.1000
    2.1000
    2.1000

If we use $a=0.3$, we have the

    1.1000
    2.3000
    2.3000
    2.3000

We can see it is "taking" the value $(n-1)\times a$ from the smallest eigenvalue and distributing it to the others from the two examples above. And the value one of the other eigenvalues receives is $a$. If we use $a=0.5$, things will be different. Now, we already have $(n-1)a>1$ as in the original optimization problem.The corresponding eigenvalues would be

    1.0000
    1.2865
    2.7500
    2.9635

It is different and I cannot see the rules behind it.

Here is the Matlab code for numerical optimization:

%% minimize_determinant_script.m
% Clear workspace and command window
clear; clc;
r_d = 0;
r_u = 2/3;
%% 1. Set problem parameters
n = 4;                       % Matrix dimension
m = n(n-1)/2;               % Number of off-diagonal elements in the upper triangle
dim = 2m;                   % Each off-diagonal position has two variables: r and phi
% Initial guess: set all r = 0.005 (intermediate value), phi strictly within (0, 2pi) (e.g., 0.1)
x0 = zeros(dim,1);
for k = 1:m
    x0(2k-1) = rand * r_u;       % Initial value for r (within [0, 0.01])
    x0(2k)   = rand  2 * pi;    % Initial value for phi (avoiding exactly 0)
end
%% 2. Set variable bounds
lb = zeros(dim,1);           % Lower bounds
ub = zeros(dim,1);           % Upper bounds
for k = 1:m
    lb(2k-1) = 0;           % Lower bound for r
    ub(2k-1) = r_u;         % Upper bound for r
    lb(2k)   = 0;           % Lower bound for phi
    ub(2k)   = 2 * pi;      % Upper bound for phi
end
%% 3. Set optimization options and nonlinear constraints
options = optimoptions('fmincon', 'Algorithm', 'interior-point', ...
                       'Display', 'iter', 'FiniteDifferenceStepSize', 1e-6);
% Nonlinear constraint: ensure the constructed matrix A is positive definite,
% i.e., the minimum eigenvalue lambda_min >= epsilon
nonlcon = @(x) pdConstraint(x, n);
%% 4. Call fmincon to solve the optimization problem for minimizing the determinant
[x_opt, fval, exitflag, output] = fmincon(@(x)objFun(x, n), x0, [], [], [], [], lb, ub, nonlcon, options);
%% 5. Display results
% fprintf('Optimal objective function value (determinant):\n');
% disp(fval);
A_opt = constructA(x_opt, n);
% fprintf('Optimally constructed matrix A:\n');
% disp(A_opt);
fprintf('Eigenvalues of matrix A:\n');
disp(eig(eye(size(A_opt)) + A_opt));
%% --- Local Functions ---
% Objective function: compute the determinant of matrix A 
% (taking the real part to avoid numerical errors causing complex results)
function f = objFun(x, n)
    A = constructA(x, n);
    f = real(det(eye(size(A)) + A));
end
% Construct Hermitian matrix A from the variable vector x
function A = constructA(x, n)
    A = eye(n);  % Diagonal elements are fixed to 1
    idx = 1;
    for i = 1:n
        for j = i+1:n
            r = x(idx);
            phi = x(idx+1);
            A(i,j) = r * exp(1i * phi);
            A(j,i) = r * exp(-1i * phi);
            idx = idx + 2;
        end
    end
end
% Nonlinear constraint function: ensure A is positive definite,
% i.e., the minimum eigenvalue is not less than epsilon
function [c, ceq] = pdConstraint(x, n)
    A = constructA(x, n);
    epsilon = 1e-7;
    lambda_min = min(eig(A));
    % Inequality constraint form: epsilon - lambda_min <= 0  => lambda_min >= epsilon
    c = -lambda_min;
    ceq = [];
end
3

Any advice would be appreciated!

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$\color{magenta}{\dagger}$ Thanks for Rodrigo de Azevedo's question. I call it a correlation matrix since it originally comes from correlation between vectors. Not a correlation matrix in the probability sense.

$\color{blue}{\dagger}$ Thanks for Rodrigo de Azevedo's question. Here, the problem comes from the Shannon capacity formula and $\mathbf{A}$ is $\mathbf{H}^H\mathbf{H}$, where $\mathbf{H}$ is a (tall) channel matrix with size $m \times n$ and $m > n$.

Answer 1

Too long for a comment.

As $r_{i,j}\le \epsilon$ the hermitian matrix $I+A$ is strongly diagonal dominant, and consequently positive definite. Regarding the constraints $r_{i,j}\le \epsilon$ we can handle an equivalent unrestricted approach as

$$ r_{i,j}= \frac{\epsilon}{2}\left(1+\tanh\left(\alpha u_{i,j}\right)\right) $$

with $u_{i,j}$ irrestricted. Considering the phases, as they are convertible $\mod 2\pi$, they can be considered unrestricted as well. Follows a MATHEMATICA script which handles the optimization procedure. Any way, the great challenge is the dimensional curse.

sigma[x_] := s (1 + Tanh[alpha x])/2;
a[i_, j_] := If[i == j, 1, If[j > i, sigma[u[i, j]] Exp[I v[i, j]], sigma[u[j, i]] Exp[-I v[j, i]]]]
n = 10;
s0 = 0.01;
alpha = 1/20;
A = Table[a[i, j], {i, 1, n}, {j, 1, n}];
obj = Det[IdentityMatrix[n] + A];
U = Flatten[Table[u[i, j], {i, 1, n}, {j, i + 1, n}]];
V = Flatten[Table[v[i, j], {i, 1, n}, {j, i + 1, n}]];
sol = NMinimize[Re[obj /. {s -> s0}], Join[U, V], Method -> "DifferentialEvolution"]
IA0 = (IdentityMatrix[n] + A) /. {s -> s0} /. sol[[2]]
Eigenvalues[IA0]