Let us assume that we have a full-rank $(n\cdot l)\times k$ matrix, $\boldsymbol{H}$, with no specific structure (e.g., a realization of a Gaussian i.i.d. random matrix), and an $m\times (n\cdot l)$ full-rank block diagonal matrix $\boldsymbol{W}$ such that \begin{equation} \boldsymbol{W}= \begin{bmatrix} \boldsymbol{W}_1 & \boldsymbol{0}_{m_1\times l} & \dots & \boldsymbol{0}_{m_1\times l}\\ \boldsymbol{0}_{m_2\times l} & \boldsymbol{W}_2 & \dots & \boldsymbol{0}_{m_2\times l}\\ \vdots & \vdots & \ddots & \vdots \\ \boldsymbol{0}_{m_n\times l} & \boldsymbol{0}_{m_n\times l} & \dots & \boldsymbol{W}_n \end{bmatrix}, \end{equation} where $\boldsymbol{W}_i$ are $m_i \times l$ matrices with $\sum_{i=1}^nm_i=m$, $1\leq m_i \leq l$, and $\boldsymbol{0}_{m_i\times l}$ denotes the $m_i\times l$ all-zero matrix. A new $\boldsymbol{W}$ can be selected for every possible $\boldsymbol{H}$
What is the minimum rank of the matrix product $\boldsymbol{WH}$?
In the case where $m_i = 1,\; \forall i\in\{1,2,\dots,n \}$, we have been able to roughly demonstrate that \begin{equation} \mathrm{rank}\left( \boldsymbol{W} \boldsymbol{H} \right)>\frac{m\cdot (k-l)}{k}, \end{equation} but we still lack an elegant proof.
