Is it true that $\exists y_0$ s.t $A^Ty_0=b$?

Question

Let $A$ be a singular matrix. $\exists x_0, b \in \Bbb R^n$ s.t $Ax_0=b$ then is it true that $\exists y_0$ s.t $A^Ty_0=b$?

I was thinking that if it is true then $x_0^T A^T y_0=||b||^2$ what can I conclude from this?

Answer 1

Let $A=\begin{bmatrix} 1 & 0 \\ 2 & 0 \end{bmatrix}$ and $b=\begin{bmatrix} 1 \\ 2\end{bmatrix}$

$$A^T=\begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}$$

$b$ doesn't lie in the column space of $A^T$.