Let $\mathcal{H}$ a Hilbert space, $A$ a bounded self-adjoint operator on the space, $W \subseteq \mathcal{H}$ a closed vector subspace that it $A$-invariant, i.e. $AW \subseteq W$. Then there is a result (which we will assume here), that $\sigma(A|_W) \subseteq \sigma(A)$, where $\sigma(B)$ denotes the spectrum of $B$, and that $f(A|_W) = f(A)|_W$, where we are using the respective functional calculi associated to $A|_W$ and $A$ resp.
I have the following conjecture: For each Borel $E \subseteq \sigma(A|_W)$, the spectral subspace $V_E$ corresponding to $E$ (which can be defined as the image of the projection $1_E(A)$) is nontrivial iff it has a nontrivial intersection with $W$.
I don't know if this is true, but seems to be used as a step of a proof I'm trying to work through (which I asked about here, with little luck given the complexity; though if I get a proof of this proposition I can figure out the linked problem). If you want, you can assume that $\mathcal{H}$ is separable. Note that in the case that $W=0$, the spectrum is empty and the statement holds trivially.
Note: I first asked about if the subspace if nontrivial iff it has a nontrivial intersection with $W$. I then changed it to "nontrivial iff not orthogonal to $W$". I now realize that these are the same, as if $x \in V_E$ and $(x,w) \neq 0$ for some $w \in W$, then $1_E(A)w \neq 0$ and $1_E(A)|_Ww = 1_E(A|_W)w$, so $1_E(A)w \in V_E \cap W$.
Edit: I have accepted Martin Argerami's answer, though look at the comments as things were clarified there.
Because $A$ is selfadjoint, the subspace $W$ is not only invariant, but reducing. That is, you can write $$ A=\begin{bmatrix}A_0&0\\0&B\end{bmatrix}, $$ where $A_0=A|_W$. From this you can see that $$ 1_E(A)=\begin{bmatrix}1_E(A_0)&0\\0&1_E(B)\end{bmatrix}. $$ So, if $V_E$ is non-trivial, this means that $E$ isn't. Then $1_E(A_0)\ne0$. And so $W\cap V_E$ contains the range of $1_E(A_0)$.