a linear variety {x : Ax = b} contains all the lines that pass through any of its two points. Prove the converse of this statement. That is; if a set contains all the lines that pass through any of its two points, then it is a linear variety.
Can anybody help me on this question?
Presumably you're talking about a subset of $\Bbb R^n$. Say $S\subset\Bbb R^n$ has this property. Fix $b\in S$ and let $$V=S-b=\{x-b:x\in S\}.$$
Then $V$ has also contains the line connecting any two points, and in addition $0\in V$. You can use this to show that $V$ is a linear subspace of $\Bbb R^n$. Now show that any subspace is the nullspace of some matrix and you're done.
Hint for that last bit: If $Y\subset\Bbb R^n$ let $Y^\perp$ be the set of all $x$ such that $x\cdot y=0$ for all $y\in Y$. If $F$ is finite it's clear that $F^\perp$ is the nullspace of some matrix. So if $W$ is a subspace then $W$ is the nullspace of a matrix, since $W^\perp=B^\perp$ if $B$ is a basis for $W$. Now given that $V$ is a subspace, let $W=\dots$; then $V=W^\perp$.