$N(A^T) \subset N(B^T)$ and the system $Bx=b$ has at least one solution, then the system $Ax=b$ has at least one solution

Question

Let $A$ be $m\times n$ matrix and $B$ be a $m \times k$ matrix such that $N(A^T) \subset N(B^T)$ and the system $Bx=b$ has at least one solution, then the system $Ax=b$ has at least one solution, where $b \in \Bbb R^m$.

Facing difficulty to prove this, hints required.

Answer 1

You can use the fundamental theorem of linear algebra (see also this wonderful article by Gilbert Strang). In particular, note that $$ N(C^T)=R(C)^\perp $$ for any matrix $C$, the assumption $N(A^T)\subset N(B^T)$ tells you that $$ R(A)^\perp\subset R(B)^\perp $$ and thus $$ R(B)\subset R(A) $$ which is exactly your conclusion.

Note that the notation $N(C)$ corresponds to $\ker (C)$ and $R(C)$ is $\text{im}(C)$ in the Wikipedia page. Also, saying that "$Ax=b$ has (at least) one solution" is exactly the same as saying that $b$ is in the range of $A$, i.e., $b\in R(A)$.

Answer 2

Hint: $Ker(A) = Im^\perp(A^T)$ and $V \subseteq W \implies W^\perp \subseteq V^\perp$.