# Some hermitian operators relations

1. Nov 7, 2011

### merkamerka

How can I formally demonstrate this relations with hermitian operators?

$$(A^{\dagger})^{\dagger}=A$$
$$(AB)^{\dagger}=B^{\dagger}A^{\dagger}$$
$$\langle x|A^{\dagger}y \rangle=\langle y|Ax \rangle ^*$$
$$If \ A \ is \ hermitian \ and \ invertible, \ then \ A^{-1} \ is \ hermitian$$

I've tried to prove them taking the definition of hermitian operator or/and considering matrices while operating, but I want something more formal.

Thanks

Last edited: Nov 7, 2011
2. Nov 8, 2011

### lanedance

$$AA^{-1}= \mathbb{I}$$