Let $A:\mathbb{R}^m \to \mathbb{R}^n$ be a linear transformation. We know that there is a unique transformation $A^*:\mathbb{R}^n \to \mathbb{R}^m$ such that $$\langle Ax,y\rangle = \langle x,A^*y \rangle, \forall x \in \mathbb{R}^m,y\in \mathbb{R}^n.$$ And we know also that the matrix of $A^*$ is $A^t$.
Having this, prove that the equation $$Ax=b$$ has a solution if, and only if, $b$ is orthogonal to all of the vectors $y \in \operatorname{ker}A^*.$ Note that this is precisely $$\operatorname{Im} A = (\operatorname{ker} A^*)^\perp.$$ To show that $\operatorname{Im}A \subset (\operatorname{ker}A^*)^\perp$ is easy. If $b \in \operatorname{Im}A$ then $\exists x\in \mathbb{R}^m$ such that $Ax=b$. Then, if $y$ is any vector in $\operatorname{ker}A^*$, we have $$\langle b,y\rangle = \langle Ax,y\rangle$$ $$=\langle x,A^*y\rangle$$ $$=\langle x,0\rangle=0$$ and, therefore, $b \in (\operatorname{ker}A^*)^\perp$, i.e. $$\operatorname{Im}A \subset (\operatorname{ker}A^*)^\perp.$$ Now, how to show that, conversely, if $b \in (\operatorname{ker}A^*)^\perp$ then $b \in \operatorname{Im}A$??
Need some help... Sorry for duplicates... (I couldn't find this question over here...)


marked as duplicate by user147263, Empty, Claude Leibovici, Strants, Najib Idrissi Sep 6 '15 at 11:48

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    $\begingroup$ You must have that $\text{Im} \ A$ is a closed space. You can easily verify that $(\text{Im} A)^\perp \subset \text{ker} A^*$, hence by taking orthogonal complements you get $( \text{ker} \ A^*)^\perp \subset( (\text{Im} \ A)^\perp)^\perp = \overline{\text{Im}\ A}$. Hence only in the case that $\text{Im} \ A$ is a closed space your claim follows, that is the equation $Ax=b$ has a solution when $\text{Im} \ A$ is closed. $\endgroup$ – Alonso Delfín Sep 3 '15 at 1:49
  • $\begingroup$ @AlonsoDelfín And $\text{Im} \, A$ will be closed, because everything here is finite dimensional. $\endgroup$ – Stephen Montgomery-Smith Sep 3 '15 at 3:01
  • $\begingroup$ To paraphrase the argument of @AlonsoDelfín, it is quite easy to prove $(\text{Im}\,A)^\perp = \text{ker} \, A^*$. But to prove what you are looking for, you have to invoke $S^{\perp \perp} = S$ for any subspace $S$. (And if you are working in infinite dimensions, you need that $S$ is closed, but that doesn't apply in your case.) $\endgroup$ – Stephen Montgomery-Smith Sep 3 '15 at 3:05
  • $\begingroup$ @StephenMontgomery-Smith you are right, I did not look that we were working on finite dimensional spaces, thus indeed $\text{Im} \ A$ is a closed set. $\endgroup$ – Alonso Delfín Sep 3 '15 at 4:44

Let $b$ be any element in $(\operatorname{ker}A^*)^\perp$.
Since $\mathbb{R}^n$ has finite dimension $$\mathbb{R}^n=\operatorname{Im}A \oplus (\operatorname{Im}A)^\perp$$ holds, and we can write $b=Au+w$, $u\in \mathbb{R}^m$, $w \in (\operatorname{Im}A)^\perp$. We shall be done if we show that $w=0$ and, therefore, $b\in \operatorname{Im}A$.
Now, for any $v \in \mathbb{R}^m$, we have $$\langle A^*w,v\rangle=\langle w,Av\rangle=0,$$ since $w\in (\operatorname{Im}A)^\perp$. This shows that $A^*w=0$ (take $v=A^*w$). So, $w \in \operatorname{ker}A^*$.
Finally, how $b\in(\operatorname{ker}A^*)^\perp$, we have $$\langle b,w\rangle=0.$$ On the other hand, $$\begin{array}{rcl} \langle b,w\rangle&=&\langle Au+w,w \rangle\\ &=& \langle Au,w\rangle+\langle w,w \rangle\\ &=&\langle w,w\rangle\end{array}$$ And hence $\langle w,w \rangle =0$ and we are done.


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