Relationship between linear transformations Suppose that for any $N\times 1$ vector $t$ that satisfies $At=0$ for some $2\times N$ non-zero matrix $A$, that same vector necessarily satisfies $Bt=0$ for some $1\times N$ non-zero matrix $B$. What is the relationship between the matrix $A$ and the matrix $B$? (i.e., if the null space of $A$ is a subset of the null space of $B$, how must $A$ be related to $B$?)
 A: $Bt=0\implies t$ is orthogonal to $B$.  As for $\operatorname{ker}A$, we have $\operatorname{ker}A\subset \operatorname{ker}B=B^{\perp}$.
A: If $t$ and $B$ are non-zero, write $B \cdot t$ as $\langle v, t \rangle$, then it is clear that any vector $v$ in the $N-1$ dimensional space perpendicular to t, i.e. $t^\perp$, will satisfy $\langle v, t \rangle = 0$.
Suppose $N \gt 3$. Since you can choose the lines of A to be any vector in $t^\perp$, and $t^\perp$ is $N-1$ dimensional, there is no necessary relation between the null spaces of $A$ and $B$, i.e., the null space of $A$ may or may not contain the null space of $B$.
if $N = 3$, then either the two lines of $A$ are linearly independent, or they are not. If they are linearly independent, they span $t^\perp$, so the null space of $A$ contains the null space of $B$. If they are not linearly independent, they either the null space of $A$ is identical to the null space of $B$, or they are different.
Finally, if $N = 2$, then $t^\perp$ is one dimensional, and both lines of $A$ must be proportional to $v$, so the null space of both $A$ and $B$ must be the same.
