Proving that a symmetric real matrix with a specific structure is positive definite.

Question

Let $H$ be an $N\times N$ symmetric matrix with the following structure: \begin{equation} [H]_{n,m}\triangleq\begin{cases} {\left|x_{n}\right|}, &\text{if}\ n=m,\\ -\mathsf{Re}{\left(\frac{x_{n}\left|x_{m}\right|}{x_{m}\left|x_{n}\right|}\sum_{k=1}^{K}w_{k,n}w_{k,m}^{*}\right)}, \ &\text{otherwise}, \end{cases} \end{equation} where $a_{k}$, $w_{k,n}$ are complex numbers for all $1\leq k\leq K$ and $1\leq n\leq N$. Furthermore, $$x_{n}\triangleq\sum_{k=1}^{K}w_{k,n}^{*}\left(a_{k}+\sum_{\substack{i=1 \\ i\neq n}}^{N}w_{k,i}\cfrac{x_{i}}{\left|x_{i}\right|}\right).$$ I want to prove that $H$ is positive definite. (Numerical results show that for over 10000 random realizations of the coefficients, $H$ exhibits positive definiteness).

Answer 1

This is not true. Consider the case where $K=N$, $x_1=\cdots=x_N=\epsilon>0$ and $W$ be any real invertible matrix such that $W^TW$ is not a diagonal matrix. Since $W$ is invertible, the condition $$ x_{n}=\sum_{k=1}^{K}w_{k,n}^{*}\left(a_{k}+\sum_{\substack{i=1 \\ i\neq n}}^{N}w_{k,i}\cfrac{x_{i}}{\left|x_{i}\right|}\right) $$ is always satisfied by some $a_1,a_2,\ldots,a_N$. Now let $F$ be the off-diagonal part of $H$. As $W^TW$ is a real non-diagonal matrix, $F$ is nonzero. Also, by definition, $F$ depends on the phases but not the moduli of the $x_i$s. Since the $x_i$s in our example are real numbers, $F$ remains constant when $\epsilon\to0$. Therefore $\lim_{\epsilon\to0}H=F$, which is indefinite because it is a nonzero real symmetric hollow matrix. It follows that $H$ is indefinite when $\epsilon$ is small.

E.g., when $K=N=2$, $x_1=x_2=1$, $(a_1,a_2)=(-1,0)$ and $W=\pmatrix{2&1\\ 1&1}$, $$ H=\pmatrix{1&-3\\ -3&1} $$ is indefinite.