Selfadjoint operators I have a self adjoint operator $A$ with domain $D(A)$; I consider the restriction of $A$ in a subset of $D(A)$. Is this operator still self adjoint?
 A: The restriction of a self-adjoint operator to a subspace of its domain cannot
be self-adjoint.
I don't know if it is necessary, but just in case I first remark the
difference between symmetric and self-adjoint unbounded operators:
The domain of the adjoint operator is
$$ \mathcal{D} ( A^{\ast}) := \left\{ y \in \mathcal{H}: \exists y^{\ast} \in
   \mathcal{H}
   \text{ such that } 
   \langle y, A x \rangle = \langle y^{\ast}, x \rangle 
   \text{ for all } x \in
   \mathcal{D} ( A) \right\}, $$
and for such elements one defines $A^{\ast} y := y^{\ast}$. (This defines
$A^{\ast}$ uniquely if the domain of $A$ is dense).
An operator is symmetric if $\langle y, A x \rangle = \langle A y, x \rangle$
for all $x, y \in \mathcal{D} ( A)$, that is, if $A \subset A^{\ast}$ (meaning
that $\mathcal{D} ( A) \subset \mathcal{D} ( A^{\ast})$ and $A x = A^{\ast} x$
for all $x \in \mathcal{D} ( A)$). An operator is self-adjoint if $A =
A^{\ast}$ (the domains must coincide in this case).
Being symmetric is therefore weaker than being self-adjoint. For bounded
operators both are the same, because the domain is the whole space.
From the definition of the adjoint operator it follows that if $A, B$ are
densely defined operators such that $B \subset A$ (in the above sense), then
$A^{\ast} \subset B^{\ast}$.
So, if you start with a self-adjoint operator $A = A^{\ast}$ and consider its
restriction $B$ to a smaller subspace you will get $B \subsetneq A = A^{\ast}
\subseteq B^{\ast}$. Therefore, $B \subsetneq B^{\ast}$, so the restriction is
still symmetric but it is not self-adjoint.
A: If $A$ is self-adjoint and $\lambda \notin \sigma(A)$ (in particular if $\lambda$ is not real), $A - \lambda I: D(A) \to \cal H$ must be one-to-one and onto.  But if $B$ is the restriction of $A$ to any proper subspace of $D(A)$,
$B - \lambda I$ will no longer be onto.  So $B$ can't be self-adjoint.
A: For example in $M_2(\mathbb{C})$, take $A=Id$ and $V=\mathbb{C}(1,0)$. 
Then
$$Id:V\longrightarrow \mathbb{C}^2$$ 
is not self-adjoint while 
$$Id:V\longrightarrow V$$ 
is self-adjoint.
