Assume that $\mathcal{H}$ is a Hilbert space with inner product $\langle \cdot, \cdot \rangle$ and an orthonormal basis $\{e_j\}_{j=1}^{\infty}$ and $C: \mathcal{H} \to \mathcal{H}$ is a rank one Hilbert-Schmidt operator. I came across the statement that for such an operator it holds that for all $i<j$, $$\det \begin{bmatrix}\langle C e_{i}, e_i \rangle & \langle C e_{j}, e_i \rangle \\ \langle C e_{i}, e_j \rangle & \langle C e_{j}, e_j \rangle \end{bmatrix} = 0 $$
where $\det$ denotes the determinant. Now obviously $C$ as a rank one operator its range is in the span of some element $g \in \mathcal{H}$, i.e., $Cf = a_f g$ for some $a_f \in \mathbb{R}$ and some $g \in \mathcal{H}$, but I cannot draw the connection between this fact and the zero determinant. If someone sees it, could I get some help here? Thank you.
EDIT: After writing it down I think I got it: it just involves expanding the determinant and using the characterization of rank one operators I mentioned in my question using the orthonormal basis. But if someone wants to provide a proper answer, I would gladly accept it. Thanks.
Since $C$ is rank one, there are non-zero numbers $a_i,a_j$ such that $a_iCe_i +a_j Ce_j=0$. Then the columns of the matrix are linearly dependent, i.e., $a_1$ times the first column plus $a_2$ times the second column is zero. Hence, the determinant is zero.
This argument directly carries over to the case that the dimension of the range of $C$ is $k$, and one builds a similar matrix of dimension $n\times n$ with $n>k$.