Suppose that an $M \times 2$ complex number matrix $\boldsymbol{Q}$ can be partitioned into $\boldsymbol{Q} = [\boldsymbol{q}_1, \boldsymbol{q}_2]$ (notion: in this post, I use square bracket to indicate the matrix), where the column vectors $\boldsymbol{q}_i \in \mathbb{C}^{M},(i=1,2,M >2)$ are independent (equivalently, $rank(\boldsymbol{Q})=2$). Given $\boldsymbol{Q}' = \boldsymbol{q}_1 \boldsymbol{q}_1^H + \boldsymbol{q}_2 \boldsymbol{q}_2^H$, prove that $rank(\boldsymbol{Q}') = 2$
My understand is that, $ \boldsymbol{Q}' = [a, b]\times[a, b]^H = \boldsymbol{Q}\boldsymbol{Q}^H $, which is a linear transformation of the rank-2 matrix $\boldsymbol{Q}$, but there is no theorem telling that after linear transformation what is the rank of $\boldsymbol{Q}$ would be. Instead, there is a lemma showing that any non-singular matrix, e.g. $\boldsymbol{P}_1$, right product to $\boldsymbol{Q}$ would preserve its rank, or $rank(\boldsymbol{Q}\boldsymbol{P}_1) = rank(\boldsymbol{Q})$. Following this lemma, I partition $\boldsymbol{Q}^H $ into $M/2$ (suppose that this is an integer) sub matrix with size $2 \times 2$, $\boldsymbol{Q}^H = [\boldsymbol{P}_1, \boldsymbol{P}_2, \dots,\boldsymbol{P}_{M/2}]$, thus $\boldsymbol{Q}' = [\boldsymbol{Q}\boldsymbol{P}_1, \boldsymbol{Q}\boldsymbol{P}_2, \dots, \boldsymbol{Q}\boldsymbol{P}_{M/2}]$ , for each block $rank(\boldsymbol{Q}\boldsymbol{P}_i) = rank(\boldsymbol{Q})=2 , (i=1,\dots, M/2)$, but to this point, it seems that there is no lemma showing that what is the relationship for the rank between $[\boldsymbol{Q}\boldsymbol{P}_1, \boldsymbol{Q}\boldsymbol{P}_2, \dots, \boldsymbol{Q}\boldsymbol{P}_{M/2}]$ and its block matrix $\boldsymbol{Q}\boldsymbol{P}_i$, so I get stuck here, anybody has a good idea ?
$Q'$ has rank $2$: Its image is spanned by $q_1,q_2$
$$Q'v=q_1(q_1^*v)+q_2(q_2^*v)\in\mathrm{Span}(q_1,q_2)$$
...and $Q'q_1$, $Q'q_2$ are independent: $$0=\alpha Q'q_1+\beta Q'q_2=\alpha (q_1^*q_1)q_1+\alpha (q_2^*q_1)q_2+\beta (q_1^*q_2)q_1+\beta (q_2^*q_2)q_2$$ $$\implies \alpha\|q_1\|^2+\beta(q_1^*q_2)=0=\alpha(q_2^*q_1)+\beta\|q_2\|^2$$ since $q_1,q_2$ are independent. But this in turn implies $\alpha=0=\beta$ since $$Q^*Qx=\begin{pmatrix}\|q_1\|^2&q_1^*q_2\\q_2^*q_1&\|q_2\|^2\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$ $\det(Q^*Q)=\|q_1\|^2\|q_2\|^2-|q_1^*q_2|^2>0$ since equality for Cauchy-Schwarz only holds if $q_1,q_2$ are dependent.