Given,$A$ be a $n\times n $ real matrix. $(1)$ $Ax=b$ , where $b\in \mathbb{R}^{n\times 1 }$ $(2)$ $A'Ax=A'b $ , where $A'$ is transpose of $A$
(a) both $(1),(2)$ may not have solution. (b) $(1)$ may not have solution, but $(2)$ always have solution.
For $(1)$ it's clear that whenever $b$ is not in the column space of $A$ ,then $Ax=b$ have no solution.
For $(2)$ , is it always possible that $A'b$ is in the column space of $A'A$?
Please help me here.
$$\mathcal{C}(A'A)\subseteq\mathcal{C}(A')$$ Now $Ax=0\implies A'Ax=0$, so null space of $A$ is a subset of nullspace of $A'A$. Also $$A'Ax=0\implies x'A'Ax=0\implies\|Ax\|=0\implies Ax=0.$$ Hence we also have the reverse inclusion, that the null space of $A'A$ is contained in the nullspace of $A$. Hence we have $\mathrm{nullity}(A)=\mathrm{nullity}(A'A)$, which implies $$\mathrm{rank}(A'A)=\mathrm{rank}(A)=\mathrm{rank}(A')$$
Thus we must have $\mathcal{C}(A'A)=\mathcal{C}(A')$. For any $b$, $A'b\in \mathcal{C}(A')=\mathcal{C}(A'A)$.