Let $A$ be a $10\times5$ matrix whose columns are linearly independent. Let $B$ be a $4\times10$ matrix. Show that there exists a nonzero vector $y\in\mathbb{R}^{10}$ such that $Ax=y$ has a solution and $By=0$.
It's clear that there exists a non-zero $y$ such that $Ax=y$ has a solution. Indexing the columns of $A$ by $a_1,a_2,\cdots,a_{10}$ and the elements of $x$ by $x_1,x_2,\cdots,x_{10}$ gives that $y=a_1x_1+a_2x_2+\cdots+a_{10}x_{10}$ (and obviously $x=0$ is not a solution). My question is how to show that $By=0$ for some $y$ of this form. I can write $By=B(a_1x_1+\cdots+a_{10}x_{10})=b_1a_1x_1+\cdots+b_{10}a_{10}x_{10}$, but it's unclear to me how to utilize this. Certainly it has to do with the fact that the $a_i$'s are linearly independent. And since the $x_i$'s are not all $0$, the important fact must come from the $b_i$'s, but I don't see it. Maybe a better approach is to select some element of the nullspace of $B$ and then work backwards, but I'm also getting stuck trying that.
Best Answer
Because the columns of $A$ are linearly independent, the rank of $A$ is 5. Then, considering the dimension of $B$, the maximum rank of $B$ is $4$, implying that, by the rank-nullity theorem, the minimum nullity of $B$ is $6$. Hence, the column space of $A$ and the null space of $B$ intersect non-trivially. The desired result follows.