This thread is Q&A.
Problem
Given Hilbert spaces $\mathcal{H}$ and $\mathcal{K}$.
Consider an operator: $$A:\mathcal{D}A\subseteq\mathcal{H}\to\mathcal{K}$$
Then for the kernel: $$\mathcal{N}A=\mathcal{N}(A^*A)$$
How can I prove this?
Reference
This is a lemma for: Polar Decomposition, Partial Isometries
We must assume ${\mathcal D}A$ is dense, otherwise $A^*$ is not well defined.
If $x \in {\mathcal D}A$ and $Ax = 0$ then of course $A^*A = 0$.
For the converse, note that $y \in {\mathcal D} A^*$ with $A^* y = z$ iff for every $w \in {\mathcal D} A$, $\langle A w, y\rangle = \langle w, z\rangle$. Now $x \in {\mathcal N} A^* A$ iff $x \in {\mathcal D} A$ and $Ax \in {\mathcal D} A^*$ with $A^*(Ax) = 0$, i.e. for every $w \in {\mathcal D}A$, $\langle Aw, Ax\rangle = \langle w, 0\rangle = 0$. In particular this must be true for $w = x$, which says $Ax = 0$.
On the one hand: $$\varphi\in\mathcal{N}A\implies A\varphi=0\in\mathcal{D}A^*\implies\varphi\in\mathcal{N}(A^*A)$$
On the other hand: $$\varphi\in\mathcal{N}(A^*A)\implies\|A\varphi\|^2=\langle A^*A\varphi,\varphi\rangle=0\implies\varphi\in\mathcal{N}A$$
Concluding the assertion.
