I was hoping someone could help me with the following proof question:
"Prove that if $A$ and $B$ are two matrices with $m$ rows, and $N(A^T) \subset N(B^T)$, then $R(B) \subset R(A)$. Hint: Use the fact that for any matrix $M$ with $m$ rows, we have dim $R(M) +$ dim $N(M^T) = m$
Lemma #1: Suppose $W_1,W_2$ are subspaces of an inner product space $V$ such that $W_1 \subseteq W_2$. Then we have that $W_2^{\perp}\subseteq W_1^{\perp}$.
Proof of Lemma #1: Choose $v\in W_2^{\perp}$ and $w_1\in W_1$ arbitrarily. Since $W_1\subseteq W_2$ we have $w_1\in W_2$ and so $v\cdot w_1=0$. But this means $v\in W_1^{\perp}$ since $w_1\in W_1$ was chosen arbitrarily and containment follows.
Lemma #2: Suppose $U$ is a subspace of an inner product space $V$. Then $\Big(U^{\perp}\Big)^{\perp}=U$
Proof of Lemma #2: Choose $v\in \Big(U^{\perp}\Big)^{\perp}$ arbitrarily and write $v=v_{||}+v_{\perp}$ where $v_{||}=\text{Proj}_{U}(v)\in U$ and $v_{\perp}=v-v_{||}\in U^{\perp}$. Since $v$ is orthogonal to any element of $U^{\perp}$ we have that $$0=v\cdot v_{\perp}=(v_{||}+v_{\perp})\cdot v_{\perp}=v_{||}\cdot v_{\perp}+||v_{\perp}||^2=||v_{\perp}||^2$$ This implies $v_{\perp}=0$ and so $v=v_{||}\in U$ which shows $\Big(U^{\perp}\Big)^{\perp}\subseteq U$. On the other hand, if $u\in U$ is arbitrary, then by definition of $U^{\perp}$ we have that $u \cdot v=0$ for any $v\in U^{\perp}$ meaning $u\in \Big(U^{\perp}\Big)^{\perp}$ and the result follows.
To prove your result let's use the fact that $N(M^T)=R(M)^{\perp}$ for any matrix $M$. We get $$N(A^T)\subseteq N(B^T) \iff R(A)^{\perp} \subseteq R(B)^{\perp} \implies \Big(R(B)^{\perp}\Big)^{\perp} \subseteq \Big(R(A)^{\perp}\Big)^{\perp} \iff R(B)\subseteq R(A)$$
