Let $A$ and $B$ be $n\times n$ matrices. Prove that null space of $A$ is a subset of null space of $B$ if and only if, row space of $B$ is a subset of the row space of $A$.
I have proved that "if row space of $B$ is a subset of the row space of $A$ then null space of $A$ is a subset of null space of $B$", but I am unable to prove the converse.
Any help will be appreciated.
Let's think about a single matrix $A$ which we are going to allow to be $m$ by $n$. We have a pairing between $n$ dimensional row vectors $a$ and $n$ dimensional column vectors $x$ by matrix multiplication; write the pairing as $(a, x)$. The null space $N(A)$ of $A$ is the set (vector space, actually) of column vectors $x$ such that $(a, x) = 0$ for all row vectors $a$ in the row space $R(A)$ of $A$.
Let's ask the question what is the "null space of the null space". That is, what are all the row vectors $b$ such that $(b, x) = 0$ for all $x \in N(A)$? Let's call this $N(N(A))$.
Proof. It is obvious that $R(A) \subseteq N(N(A))$.
We are going to apply the rank-nullity theorem, which says that $$\dim R(A) + \dim N(A) = n$$ to get the opposite inclusion. Take a row vector $b$ in $N(N(A))$. We have to show that $b \in R(A)$.
Augment $A$ by adding $b$ as an extra row; call the augmented matrix $A'$. On the one hand, the row space of $A'$ evidently contains the row space of $A$, $R(A') \supseteq R(A)$. On the other hand, since $b \in N(N(A))$, it follows that $N(A') = N(A)$; the null space hasn't gotten any smaller. But apply the rank nullity theorem to $A'$ and discover that $\dim R(A') = \dim R(A)$, which tell us that $R(A') = R(A)$; but that means that our extra row vector $b$ is in $R(A)$.
I'm going to stop here, because the proposition gives you the extra tool you need to complete the problem, but it still could be challenging to work it out.