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.

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.