Proof verification: Null space of $A$ = $($Column space of $A^*)^\perp$ where $A\in M_{m\times n} (F)$

Question

Want to show : Null space of $A$ = $($Column space of $A^*)^\perp$ where $A\in M_{m\times n} (F)$

$^*$ is conjugate transpose

Note that both are subspaces of $F^n$

Let $x\in N(A)$ and $y\in Col(A^*)$

$\Rightarrow Ax=0$ and $\exists z\in F^m $ such that $A^* z = y$

Consider $\langle x,y\rangle =\langle x,A^*z\rangle = \langle Ax, z\rangle=0$

Since $x$ and $y$ were arbitrary elements of $N(A)$ and $col(A^*)$ respectively, so it is reasonable to conclude $N(A)=col(A^*)^\perp $? or do I need to explicitly do element chasing in order to prove the result?

Answer 1

Your proved that $N(A) \subseteq \left(\textrm{col}(A^*)\right)^\perp$. However, I don't perceive in what you wrote a clean proof of reverse inclusion, namely $\left(\textrm{col}(A^*)\right)^\perp \subseteq N(A)$.

Let's prove this reverse inclusion. $x \in \left(\textrm{col}(A^*)\right)^\perp$ means that for all $z \in F^m$ we have $\langle A^*(z) , x \rangle = \langle z , A(x) \rangle = 0$. Therefore $x=0$ (because the bilinear form $(x,y) \mapsto \langle x,y \rangle$ is not degenerated).

A comment: I don't understand why you note the base field $F$. It should be $\mathbb C$ as you speak of conjugate.