I need to show that $\operatorname{ker}(A^T) = (\operatorname{Im} A)^\perp$

Question

Let $V=\mathbb{R}^n$ with an inner product. I need to show that $\operatorname{ker}(A^T) = (\operatorname{Im} A)^\perp$. I showed that $\operatorname{Im} A^T \subset \operatorname{ker}(A^T)$ but somehow I'm not able to show the other way $\operatorname{ker}(A^T) \subset \operatorname{Im} A^T$. Can someone give me just the idea how to show the other way (no full proof!)

Thank you very much! ;)

I think you are confused . You have to prove they are orthogonal and together with rank nullity the proof follows — Cloud JR K, Jun 23 '20 at 20:00
Note that $x\in\ker A\iff A^tx=0 \iff\langle A^tx,y\rangle=0\text{ for all }y\iff \langle x, Ay\rangle =\langle A^tx,y\rangle=0\text{ for all }y\iff x\in\big(\text{Im }A\big)^\perp.$ — Sumanta, Jun 23 '20 at 20:11
Related question: https://math.stackexchange.com/questions/21144/intuitive-explanation-of-the-fundamental-theorem-of-linear-algebra — littleO, Jun 23 '20 at 20:11
Does this answer your question? The range of $T^*$ is the orthogonal complement of $\ker(T)$ — PinkyWay, Jun 23 '20 at 22:07

score 2 · Answer 1 · answered Jun 23 '20 at 20:04

Let $A:\Bbb{R}^n \rightarrow \Bbb{R}^m$. You want to prove that $\ker(A^T)=\text{Im}(A)^{\perp}$. Since $\text{Im}(A)=\text{Col}(A)$, column space of $A$. By defintion: \begin{align*} \text{Im}(A)^{\perp} & =\{x \in \Bbb{R}^n \, | \, x \cdot b=0 \, \forall \, b \in \text{Col}(A)\}\\ &=\{x \in \Bbb{R}^n \, | \, x \cdot b=0 \, \forall \, b \in \color{red}{\text{Row}(A^T)}\} && (\because \text{Col}(A)=\text{Row}(A^T))\\ &=\ker(A^T). \end{align*} The last equality follows from the fact that if $b \perp$ to every row of $A^T$, then $A^Tb=0$, so $b \in \ker(A^T)$.

I need to show that $\operatorname{ker}(A^T) = (\operatorname{Im} A)^\perp$

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