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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.

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  • $\begingroup$ the image of $A$ and the kernel of $B$ are linear subspaces of $R^{10} \; . \;$ What can you say about the dimension of each? $\endgroup$
    – Will Jagy
    Aug 30, 2018 at 1:30
  • $\begingroup$ @WillJagy Well, the image of $A$ has dimension $5$, as the columns of $A$ are linearly independent. I think the kernel of $B$ has dimension $\leq6$, because the max rank of $B$ is $4$, so the max nullity of $B$ should be $6$ $\endgroup$
    – Atsina
    Aug 30, 2018 at 1:34
  • $\begingroup$ Which, if it is $=6$, then that implies that the column space of $A$ and the null space of $B$ intersect non-trivially, implying the desired result $\endgroup$
    – Atsina
    Aug 30, 2018 at 1:36
  • $\begingroup$ Alright...if $B$ is all zeros, what is the dimension of the kernel? $\endgroup$
    – Will Jagy
    Aug 30, 2018 at 1:37
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    $\begingroup$ @WillJagy Oh, I got it backwards. The max rank of $B$ being $4$ means the $\textit{min}$ nullity of $B$ is 6 $\endgroup$
    – Atsina
    Aug 30, 2018 at 1:39

1 Answer 1

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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.

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