Is every bounded operator normal?

Question

Let $A: {\scr H} \to {\scr H}$ be a linear bounded endomorphism on a Hilbert space $\scr H$, in brief: $A \in {\frak B} ({\scr H})$.

If it's true that $\ \forall A \in {\frak B} ({\scr H}) \quad A^*A \in {\frak B} ({\scr H}) $ and $A^*A \geq 0 $, which implies $A^*A$ self-adjoint (every positive bounded operator is self-adjoint), and hence $A^*A=A A^*$ ($^*$ is an antilinear involution)

Then $\ \forall A \in {\frak B} ({\scr H}) \quad A^*A=AA^*$, which is precisely the definition of normal operator.

So every bounded linear operator is normal. It can't be true... yet I don't see the error.

Answer 1

Note that $A^*A$ is self-adjoint, not $A$ itself. So we can conclude $$(A^* A)^*(A^* A) = (A^* A)(A A^*)^*$$ for any $A \in \mathscr{B}(\mathscr{H})$, which is an entirely more reasonable proposition.

Answer 2

In general, $ A^*A=AA^*$ is not true !

For example let ${\scr H}= \ell^2$ and $A$ given by $A(x_1,x_2,x_3,...)=(0,x_1,x_2,x_3,...).$