Prove or provide a counterexamples: $A$ is a $m\times n$ matrix
(1) If there exists a vector $b$ such that $Ax=b$ does not have any solution, then $Ax=0$ has infinitely many solutions when $n>m$.
(2) If $Ax=0$ has infintely many solutions, then there exists a vector $b$ such that $Ax=b$ does not have any solution when $n<m$
For (1) I think it is true, if $Ax=b$ has no solution implies that $b\in R(A)$, so $rank(A)\leq n-1$, hence $N(A)$ has dimension at least 1, so there exists infinitely many solutions.
How about (2)? Thanks
For (2) consider a system of equations that will have infinite solutions, such as this augmented matrix for $Ax=0$:
$\begin{bmatrix}1 & 1 & 0\\2 & 2 & 0\\3 & 3& 0\end{bmatrix}$
Can you show/prove that if any system is overdetermined (there are more equations than unknowns) and it happens to have at least one free variable, that there exists a vector $b$ for which the overdetermined system doesn't have a solution?
(I believe the statement is true given that the premises are $Ax=0$ has infinitely many solutions (so at least one free variable) and that $A$ has less columns than rows)