$Ax=b\Leftrightarrow b\in\left(\ker A^*\right)^\perp$

Question

Let $A:\mathbb{R}^m\to\mathbb{R}^n$ be a linear map and $A^*:\mathbb{R}^n\to\mathbb{R}^m$ be the adjoint of $A$ (that's $\langle Ax,y\rangle=\langle x,A^*y\rangle$ for all $x\in\mathbb{R}^m,y\in\mathbb{R}^n$). Given $b\in\mathbb{R}^n$, are equivalent:

(i) there exists $x\in\mathbb{R}^n$ such that $Ax=b$;
(ii) $\langle b,z\rangle=0$ for all $z\in\ker A^* $.

I don't know how to prove (ii)$\Rightarrow$(i). Can someone help me?

Thanks.

Answer 1

\begin{align*} w \in \ker A^* &\iff A^* w = 0 \\ &\iff \langle v, A^* w\rangle = 0 \quad \forall v \in \mathbb R^m \\ &\iff \langle Av, w\rangle = 0 \quad \forall v \in \mathbb R^m \\ &\iff w \in \left(\operatorname{im}A\right)^\perp \end{align*}

Hence $\ker A^* = \left(\operatorname{im}A\right)^\perp$.

Since $\left(V^\perp\right)^\perp = V$ for any subspace $V \subset \mathbb R^n$, it follows that $\left(\ker A^*\right)^\perp = \left(\left(\operatorname{im}A\right)^\perp\right)^\perp = \operatorname{im}A$ as desired.

Answer 2

Suppose $\langle b,z\rangle=0$ for all $z\in\ker A^\ast$. Note $z\in\ker A^\ast$ if and only if $A^\ast(z)=0$ if and only if $\langle Ax,z\rangle=0$ for all $x\in\mathbb{R}^m$, that is $z\in\ker A^\ast$ if and only if $z$ is orthogonal to $\operatorname{im}A$. So we are supposing $b$ is orthogonal to the orthogonal complement of $\operatorname{im}A$, i.e. $b\in\operatorname{im}A.$