Proof that restriction of hermitian operator to its invariant subspace is also hermitian

Question

Proof that restriction of hermitian operator to its invariant subspace is also hermitian

What would be the most elegant way to prove this?

Answer 1

Let $L$ be invariant subspace of $A$, then $$ \begin{align} A\text{ is Hermitian}&\Longleftrightarrow \forall x,y\in H\quad \langle Ax,y\rangle=\langle x,Ay\rangle\\ &\Longrightarrow \forall x,y\in L\quad \langle Ax,y\rangle=\langle x,Ay\rangle\\ &\Longleftrightarrow A|_L \text{ is Hermitian} \end{align} $$

Answer 2

Let $A$ an Hermitian operator ($A=A^\dagger$) and $P$ a projection onto the subspace $V$. If we define the restriction of $A$ to the subspace $V$ as $A_V=PAP$, then

$$ (PAP)^\dagger=P^\dagger A^\dagger P^\dagger= PAP $$

Properties used:

1. $(AB)^\dagger=B^\dagger A^\dagger$
2. projections are hermitian operator, since $P=\sum_i |i\rangle \langle i|$ and $(|\psi\rangle\langle \phi |)^\dagger=|\phi\rangle \langle \psi |$, therefore, thanks to the linearity in the second argument of the inner product, we have $P^\dagger=P$