Is my proof for this linear algebra question correct? $\newcommand{\R}{\operatorname{Ran}} \newcommand{\K}{\operatorname{Ker}}\newcommand{\b}{\mathbf}$If $A$ is a linear operator then prove that $A = A'P_{\R A^*}$ in the domain $\R A^*$; $A'$ is an invertible transformation from $\R A^* \to \R A$ such that $A' \ = A  $ on $\R A^*$ and  $P_{\R A^*}$ is orthogonal projection on $\R A^*$.

From the definition of $A'$, I see that $A\b x = A' \b x$ for all $\b x\in\R A^*$.
Since $\b x - P_{\R A^*}\b x \in (\R A^*)^\perp = \K A$,  $$A(\b x - P_{\R A^*} \b x) = \b 0 \iff A\b x = AP_{\R A^*}\b x= A'P_{\R A^*}\b x$$
Now I need to prove that $A'$ is injective and surjective. For injectivity :
Assume that $\b y \ne \b 0$ and $A \b y = \b0$ . $$A' \b y =\b 0 = A\b y  \iff\b y \in \K A = (\R A^*)^\perp$$. So $\b y \in \R A^*\cap (\R A^*)^\perp = \{\b 0\}$. Hence we get a contradiction and $A'$ is injective. 
For surjectivity I need to prove $\R A' = \R A$ : If $\b y \in \R A'$ then $\b y = A'\b x = A\b x$, so $\b y \in \R A$. 

Frankly, I don't understand the question. So please verify that my proof is what is to be proven and it is correct.
 A: In your proof of injectivity of $A'$, you started with $Ay=0$ (instead of $A'y=0$), but $Ay=0$ implies $A'y=0$ only if $y\in\operatorname{Ran}(A^\ast)$, but not for general $y\in\Bbb C^n$.
Similarly, in the proof of surjectivity, you write $Ax=A'x$, but this holds only for $x\in\operatorname{Ran}(A^\ast)$ and is false, in general, for $x\in\Bbb C^n$.

As you noticed, $\operatorname{Ran}(A^\ast)=\operatorname{Ker}(A)^\bot$, so that we have an internal direct sum decomposition
$$\Bbb C^n=\operatorname{Ker}(A)\oplus\operatorname{Ran}(A^\ast)$$
Consider $A$ as a $\Bbb C$-linear map $\Bbb C^n\to\Bbb C^n$.
Then $A$ restricted to $\operatorname{Ker}(A)$ is, clearly, the trivial morphism, while its restriction $A'$ to $\operatorname{Ran}(A^\ast)$ is an isomorphism of $\operatorname{Ran}(A^\ast)$ onto $\operatorname{Ran}(A)$.
For prove $A'$ injective, let $y\in\operatorname{Ran}(A^\ast)$ such that $A'y=0$.
Then $Ay=0$ by definition of $A'$, thus $y\in\operatorname{Ker}(A)$ hence $y=0$ because $\operatorname{Ker}(A)\cap\operatorname{Ran}(A^\ast)=\{0\}$ (recall that the sum above is direct).
For prove $A'$ surjective, let $w\in\operatorname{Ran}(A)$.
Then there exists $z\in\Bbb C^n$ such that $w=Az$.
By direct sum decomposition there exists $x\in\operatorname{Ker}(A)$ and $y\in\operatorname{Ran}(A^\ast)$ such that $z=x+y$.
Thus $w=Az=Ax+Ay=Ay=A'y$, hence $w\in\operatorname{Ran}(A')$.
Finally, we prove $A=A'P$ where $P:\Bbb C^n\to\operatorname{Ran}(A^\ast)$ denote the orthogonal projection.
Let $z\in\Bbb C^n$.
As before write $z=x+y$ for $x\in\operatorname{Ker}(A)$ and $y\in\operatorname{Ran}(A^\ast)$.
We have $Pz=y$, hence $A'Pz=A'y$.
On the other hand $Ax=0$ because $x\in\operatorname{Ker}(A)$, consequently, $Az=Ay$.
Since $y\in\operatorname{Ran}(A^\ast)$, we have $Ay=A'y$, hence we get $Az=A'Pz$.
By the arbitrariety of $z\in\Bbb C^n$ we have $A=A'P$ as wanted.
A: For this answer, I will use "range of" instead of "Ran". 
I might be missing something but the way this question is asked, projection onto the range of A* -when the domain is range of A*- is, well, just the identity.
As such, we now need to show A = A' on range of A*, but that is already given in the question.
Hopefully my answer also explains the problem to OP, though I don't know how to explain it any further.
Since the OP also asked for a critique of their answer, the steps individually look right; I don't really see their point though. 
