$f(x)=\|A^* x \|_2^2 - \| Ax \|_2^2=0$

Question

given $f(x)=\|A^* x \|_2^2 - \| Ax \|_2^2=0$ , $\forall x \in \mathbb{C}$ ,show that $ g(x)= \| (AA^*-A^*A)x \|_2$ $=0, \forall x\in \mathbb{C}$.

It feels like, I have to to differentiate $f(x)$ in some way to get $g(x)$. But since $f'(x)h=<A^*x,A^*h>-<Ax,Ah>$ I think I am stuck. Can some give me a hint how to start here?

Answer 1

Hint: Rewrite $$ \frac 12 f'(x)h = \langle AA^*x,h \rangle - \langle A^*A x, h \rangle = \langle (AA^* - A^*A)x,h \rangle $$

Answer 2

Also note that $f(x) = \langle x, (A A^* - A^* A) x \rangle$. Since $(A A^* - A^* A) = (A A^* - A^* A)^*$, it has a basis of eigenvectors and so for each eigenvalue, eigenvector pair $\lambda, v$ of $A A^* - A^* A$ we have $f(v) = \lambda \|v\|^2= 0$. From this we can conclude that $(A A^* - A^* A) = 0$.