AX = b has solution iff $\langle A, Y\rangle = 0$ implies $\langle b,Y\rangle = 0$ Let $A$ be an $m\times n$ matrix of real numbers, $b \in \mathbb{R}^n$.
Show that the system of equations $AX = b$ has a solution if and only if $A^TY = 0$ always implies $b^TY = 0$.

I'm not sure what we can use to prove this.  For one, since $A$ is not square, it can't be invertible, so I don't think we can say anything about $A^TY = 0$ right?

I can try substituting $b = AX$ into $b^TY.$  This gives $A^TY = 0 \Rightarrow (AX)^TY \Rightarrow 0 = X^TA^TY = 0$.  I'm not sure if I can use this to make any progress.

Any help would be greatly appreciated.

Thanks.
$AX=b$ have solution means $b \in \operatorname{im} A$.

$A^TY=0$ implies $b^TY=0$ means $b\in \operatorname{Ann}\ker A^T $, where $\operatorname{Ann}$ is the annihilators.

So your statement is equivalent to show $\operatorname{im} A=\operatorname{Ann}\ker A^T$, the rest is yours. (Hint: show $\operatorname{im} A\subset \operatorname{Ann}\ker A^T$ and $\operatorname{Ann} \operatorname{im} A\subset \ker A^T$)
$AX=b$ have solution means $b \in \operatorname{im} A$.

$A^TY=0$ implies $b^TY=0$ means $b\in \operatorname{Ann}\ker A^T $, where $\operatorname{Ann}$ is the annihilators.

So your statement is equivalent to show $\operatorname{im} A=\operatorname{Ann}\ker A^T$, the rest is yours. (Hint: show $\operatorname{im} A\subset \operatorname{Ann}\ker A^T$ and $\operatorname{Ann} \operatorname{im} A\subset \ker A^T$)