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 $m\times (n\cdot l)$ 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.

After some tests and derivations, we strongly believe that the statement \begin{equation} \mathrm{rank}(\boldsymbol{W})=m \implies \mathrm{rank}\left( \boldsymbol{W} \boldsymbol{H} \right)>\frac{m\cdot (k-l)}{k}, \end{equation} is fulfilled with probability 1 (i.e., for any $\boldsymbol{H}$ except for those belonging to a set of measure 0), but we are not able to come up with a formal proof.

We have been struggling with this problem for a while and we are able to sketch a proof if $m_i = 1,\; \forall i\in\{1,2,\dots,n \}$. However, we would like to get help to prove this for any combination of $1\leq m_i \leq l$. Thanks in advance for any contribution.