Let $A$ be an $m \times n$ matrix. If $Ax=b$ has a solution for every $m$-vector $b$, then does $A^T y = c$ also have a solution for every $n$-vector $c$?
My teacher quickly mentioned this property, but there has no explanation or proof. Could someone provide some intuition behind this?
I understand that a pivot in every row means that Ax=b has a solution for all b, but does that apply to this?
On
The statement is not necessarily true. Put
$$ A = \left(\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 3 \end{matrix}\right) $$
Then $Ax = y$ is solvable for any $y = (\begin{matrix} y_1 & y_2 & y_3 \end{matrix})^T \in \mathbb{R}^3 $ (Just put $x = (\begin{matrix} y_1 & y_2 & y_3 & 0 \end{matrix})^T$). However, that is not the case for $A^Ty = c$, to see this consider $$ c = \left( \begin{matrix} 1 \\ 1 \\1 \\ 1 \end{matrix} \right) $$
We use that $$Ax=b$$ has a solution if and only if $A$ has the same rank as $A'$ which is $A$ concated with $b$. If a solution exist for every $b$, we can conclude that $A$ has full rank. $A^T$ also has full rank and concenating another column cannot change it. Hence $$A^Ty=c$$ is solvable for every $c$.