# Spectral subspace is nontrivial iff it has a non-trivial intersection with an invariant closed subspace

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 $$1_E(A_0)$$.
• I'm not familiar with reducing subspaces of operators, but I will think about it/look in to it. Can you explain how this isn't a counterexample though: Consider $L^2([-1,1],m) \oplus L^2([-1,1],m)$ with $m$ Lebesgue measure. Let $A = X \oplus I$ where $X,I$ are bounded self adjoint on $L^2([-1,1],m)$ with $Xf(x) := xf(x)$ and $If = f$. Then let $W := L^2([-1,1],m) \times \{0\}$. Then consider the spectral subspace associated with $\{1\}$. This should contain precisely the functions $(0,f)$ I believe. But this has a trivial intersection with $W$. Commented Apr 16, 2020 at 18:37
• The set $E=\{1\}$ has measure zero. So $1_E=0$. "Non trivial" for a set $E$, in this context, means "positive measure". Commented Apr 16, 2020 at 18:42
• Just because the set has measure zero doesn't mean the spectral subspace is the 0 subspace. Even though $E$ is a null set, in this case $V_E$ is distinctly non-trivial, right? Commented Apr 16, 2020 at 18:44