let ${G}_{j-1}$ be an $N\times d_{j-1}$ matrix, the columns of ${G}_{j-1}$ are independent. let $P$ be an $N\times s$ matrix, the columns of $P$ are independent. if $rank(P^{H}G_{j-1}) <s$, then we must have $x^{H}G_{j-1}=0^H$ for some nonzero $x=Py$.
I don't know why we must have $x^{H}G_{j-1}=0^H$ for some nonzero $x=Py$.
Thanks for any helps.
First note, that since all columns of $G_{j-1}$ are independent the rank of $G_{j-1}$ is $d_{j-1}$ and similarly rank of $P$ is $s$. Now lets denote rows of $G_{j-1}$ by $g_1,g_2...g_n$ and cols of $P$ by $p_1,p_2...p_s$. Now since no. of rows in $P^HG_{j-1}$ are s and we have rank $<s$, the rows of $P^HG_{j-1}$ are linearly dependent. And so there exist $a_1,a_2...a_s$ such that,$$a_1\sum_{j=1}^{N} \bar{p_{1j}}g_j +a_2\sum_{j=1}^{N} \bar{p_{2j}}g_j+...+a_s\sum_{j=1}^{N} \bar{p_{sj}}g_j=0$$ $$\sum_{i=1}^{s}a_i\sum_{j=1}^{N} \bar{p_{ij}}g_j= \sum_{j=1}^{N}(\sum_{i=1}^{s} a_i\bar{p_{ij}})g_j = 0$$ Which gives us the desired result with $y$ being vector $a^H$.