Range perpendicular to Nullspace I'm stuck in this Linear Algebra problem:

Let $A\in M_n(\mathbb{C})$ with $\mathrm{rank}(A)=k$. Prove that the following are equivalent:
a) $R(A) \bot N(A)$
b) $N(A)=N(A^*)$
c) $R(A)=R(A^*)$

for a) implies b) I should prove double contention, i.e. $N(A)\subset N(A^*)$ and $N(A^*)\subset N(A)$. So for the first contention, I took $x \in R(A)$ so $\exists \ u \in \mathbb{C}^n$ such that $Au=x$, then I took $y \in N(A)$.  And by hypothesis (I mean $R(A) \bot N(A)$ ), $(Au)^*y=0$ this is the same as $u^*A^*y=0$. I had to prove that $y$ is in $N(A^*)$ i.e. $A^*y=0$ but i've got this upset $u^*$, how do I can improve that? Or from here can I conclude that $y\in N(A^*)$? Thanks in advance.
 A: If $A^*y = u \neq 0$ then $(y,Au)=(A^*y,u)>0$, where $(.,.)$ is the inner product. However $Au \in R(A)$ and $y \in N(A)$ by assumption so it must be the case that $(y,Au)=0$, a contradiction.
A: We actually have the equivalent statements
a') $R(A)\perp N(A)$
b') $N(A)\subset N(A^*)$
c') $R(A)\subset R(A^*)$
Indeed, it all depends on the equality (proven at the end)
$$\tag{1}
N(A)=R(A^*)^\perp.
$$
For a')$\implies$b'), by hypothesis, $N(A)\subset R(A)^\perp=N(A^*)$. 
For b')$\implies$c'), from $N(A)\subset N(A^*)$ we get
$$
R(A)=N(A^*)^\perp\subset N(A)^\perp=R(A^*). 
$$
And for c'$\implies$a'), from $R(A)\subset R(A^*)$ we get $$R(A)\subset R(A^*)=N(A)^\perp,$$ so $$N(A)\perp R(A). $$

Now, a posteriori, we can show that if the equivalent conditions above are satisfied, then $N(A)=N(A^*)$ and $R(A)=R(A^*)$. This is an easy consequence of the Rank-Nullity Theorem. We have 
$$
n=\text{rank}(A^*)+\text{null}(A^*)\geq\text{rank}(A)+\text{null}(A)=n.
$$
Since $\text{null}(A)\leq\text{null}(A^*)$ and $\text{rank}(A)\leq\text{rank}(A^*)$, we have $\text{null}(A)=\text{null}(A^*)$ and $\text{rank}(A)=\text{rank}(A^*)$, and so $N(A)=N(A^*)$ and $R(A)=R(A^*)$. 

To establish the equality
$$
N(A^*)=R(A)^\perp,
$$
if $A^*x=0$ and $y\in \mathbb C^n$, then 
$$
\langle x,Ay\rangle=\langle A^*x,y\rangle=0,
$$
so $x\in R(A)^\perp$. That is, $N(A^*)\subset R(A)^\perp$. Conversely, if $y\in R(A)^\perp$, then for any $z$
$$
\langle A^*y,z\rangle=\langle y,Az\rangle=0,
$$
As we can do this for all $z$ (in particular for $z=A^*y$) we get that $A^*y=0$; that is $R(A)^\perp\subset N(A^*)$. 

