Let $ P \in \mathbb{M}_{m\ x\ m} \space $ and $ \space A \in \mathbb{M}_{m\ x\ n} \space $ such that $ P$ is invertible. Suppose that $PA$ is a reduced row echelon form matrix with exactly one zero row (namely it has exactly one row s.t each value at that row is zero). Denote the last row of P by $[p_1\space p_2 \space ...\space p_m] \space $. Prove that $A$'s column space is equal to the solutions set of $p_1x_1 + ... + p_mx_m = 0 \space $.
I skipped that question at one of my tests since I felt it was too abstract for me. I tried to use linear transformations later then hoping the math will be more elegant but I could only figure that $rankA = m-1$ (I think) but there it becomes abstract again and I feel like I'm blind and can't proceed further. I'd like to hear some ideas please.
Here is a sketch (you can complete the details).
Let $V$ be the column space of $A$ and $W$ be the set of solutions of the equation $p_1x_1+\cdots+p_mx_m=0$. It is easy to see that $W$ is vector subspace of $\mathbb{K}^m$ (where $\mathbb{K}$ is the ground field). You can notice that the dimension of $W$ is $m-1$ which equals the rank of $A$ and hence the dimension of the column space $V$. Thus it suffices to show that $V\subseteq W$. To do so, it suffices to show that every column of $A$ belongs to $W$ (why?). Let $a=(a_1,\dots,a_m)^T$ be the $j$th a column of $A$, then the entry $(m,j)$ of $PA$ is precisely $p_1a_1+\cdots+p_ma_m$ which must be zero because the last row of $PA$ is zero. Thus the vector $a$ belongs to $W$.