Suppose $N\in M_{n\times n}$ invertible and $A\in M_{n\times m}$ injective. If $N$ satisfies the condition $v^{T}Nv=0\ \text{implies}\ v=0$, then can we say $\text{Im}(N^{-1}A)=\text{Im}(A)$?
This is a part of my attempt to solve the following bigger problem:
Suppose $N\in M_{n\times n}$ invertible, $M\in M_{m\times m}$ invertible, and $A\in M_{n\times m}$. $N$ satisfies the condition $v^T Nv=0$ implies $v=0$. Does it imply $Im (N^{-1}A M)=Im A$?
(I think the answer to the latter problem is true if the answer to the first one is yes, but please correct me if I'm wrong).
Any comment or suggestion is greatly appreciated. Thank you for reading.