If $\|Ax\|=\|A^* x\|$ for all $ x\in E$, why $A$ is normal?

Question

Let $E$ be an infinite-dimensional complex Hilbert space.

Let $A\in {\cal{L}}(E)$. If $\|Ax\|=\|A^* x\|$ for all $ x\in E$, why $A$ is normal (i.e. $A^*A=AA^*$)?

Answer 1

Note that $$\| Ax\|^2=(Ax,Ax)=(A^*Ax,x),\quad \| A^*x\|^2=(A^*x,A^*x)=(AA^*x,x).$$ So $$((A^*A-AA^*)x,x)=0,$$ for all $x$. Hence by (1) $A^*A=AA^*$.

Answer 2

For any $x \in X$ we have:

$$\langle A^*A - AA^*x, x\rangle = \langle A^*Ax, x\rangle - \langle AA^*x, x\rangle = \langle Ax, Ax\rangle - \langle A^*x, A^*x\rangle = \|Ax\|^2 - \|A^*x\|^2 = 0$$

Now notice that $A^*A - AA^*$ is a self-adjoint map:

$$(A^*A - AA^*)^* = A^*A - AA^*$$

and recall the formula for the norm of a self-adjoint map $B$:

$$\|B\| = \sup_{\|x\| = 1}|\langle Bx, x\rangle|$$

Therefore,

$$\|A^*A - AA^*\| = \sup_{\|x\| = 1}|\langle A^*A - AA^*x, x\rangle| = 0 \implies A^*A - AA^* = 0$$