Rank-two matrix formulation of the 4M-4 conjecture

Let $\phi_1,\ldots,\phi_{4M-5}\in\mathbb{C}^M$, and let $L_{\Phi}$ be the Hermitian matrix subspace

$$L_{\Phi}=\{Q\in\mathbb{C}^{M\times M}_{\rm Herm}:\phi_n^*Q\phi_n=0\text{ for }n=1,\ldots,4M-5\}.$$

Equivalently, $L_{\Phi}$ is the orthogonal complement of the span of $\phi_n\phi_n^*$ for $n=1,\ldots,4M-5$. The 4M-4 rank-two formulation. The space $L_{\Phi}$ always contains a matrix of rank at most $2$. Equivalently, if $d=(M-2)^2+1$ Hermitian matrices $A_1,\ldots,A_d$ span $L_{\Phi}$, then some linear combination $x_1A_1+\cdots+x_dA_d$ has rank two.

This is an equivalent reformulation of the open lower-bound part of the 4M-4 conjecture. The source notes that the first open case is $M=4$ and proves the claim in the special case $M=2^k+1$.