**Please note:** *This question has been edited after it became clear from Christian Remling's answer that the original formulation was far from what I really meant to ask.*


Let $T\ne 0$ be a self-adjoint operator on a Hilbert space $H$, with spectrum $\sigma(T)$. For any $x∈H$, denote by $μ\_x$ the spectral measure of $T$ at $x$, that is the unique Borel measure on $\sigma(T)$ such that


$$ ⟨x,f(T)x⟩ = \int\_{\sigma(T)} f(λ) d \mu\_{x}(λ) \quad \forall f \in \mathcal{C}(\sigma(T),\mathbb{C}). $$


Then, one can prove that
$$ \overline{\bigcup\_{x\in X} Supp(\mu\_x)}=\sigma(T)$$
for any orthonormal basis $X$ of $H$.


Suppose that there exists a closed subspace $H\_0$ of $H$ and a positive integer $k$ such that $T|\_{T^i H\_0}$ is injective for all $i=0,\dots,k-1$, and $$ H=H\_0 \overset{\perp}{\oplus} TH\_0 \overset{\perp}{\oplus} \dots \overset{\perp}{\oplus} T^{k-1} H\_0,\qquad T^{k}H\_0=H\_0.$$


If $k=1$, then $T$ is invertible, hence $0\in \sigma(T)$ if and only if the inverse $T^{-1}$ is unbounded. Therefore, either $0\not\in \sigma(T)$ or $\lambda=0$ is an accumulation point of $\sigma(T)$.


Suppose now that $k\ge 2$, and suppose that there is an orthonormal basis $X\_0$ of $H\_0$ such that $\mu\_x(\{0\})>0$ for all $x\in X\_0$. Then, one has $0\in \sigma(T)$. Under some additional assumptions I can actually show that $\lambda=0$ must be an isolated eigenvalue. However, I was actually wondering whether that might always be the case.


**Addendum:** This is how I would like reason. Let
$$ X:=X\_0\cup TX\_0 \cup \dots \cup T^{k-1}X\_0. $$


Let $\mathfrak{M}$ be the set of discrete positive measures $\nu$ on $X$ with the following property: $$\forall A\subseteq X,\; \nu(A)=0\implies \nu(TA)=0.$$


Then, for all non-zero $\nu\in\mathfrak{M}$, one has $\nu(X\_0)>0$. Hence $$ \int\_{X} \mu\_{x}(\{0\})d\nu(x) \ge \int\_{X\_0} \mu\_{x}(\{0\})d\nu(x)>0.$$


It seems to me that this could be enough to conclude that $\lambda=0$ must be an isolated eigenvalue. However, I can only show this with extra assumptions.


**Addendum:** It occurred to me that the case $k\ge 3$ is not that interesting because one gets
$$ 0=⟨x,T^2 x⟩ = \int\_{\sigma(T)} \lambda^2 d \mu\_{x}(λ) \implies \mu\_{x}=\delta\_0.$$
The case $k=2$ (which is actually also the one that motivated this question) is thus the only really interesting case.



Detecting isolated eigenvalues from local spectral measures
**Updated answer:**
For $k=2$ the conditions are contradictory.


We have the decomposition $H=H\_0\oplus H\_1$. You now impose the following conditions: (1) $TH\_0=H\_1$; (2) $T$ injective on $H\_0$ and $H\_1$; (3) $T^2 H\_0=H\_0$; (4) $Tu=0$ for some $u\notin H\_1$.


Since $N(T^2)=N(T)$ for the self-adjoint operator $T$, we can rephrase the last condition as: (4') $T^2u=0$ for some $u\notin H\_1$.


Let's write
$$
T= \begin{pmatrix} 0 & B \\ B^\* & C \end{pmatrix} ,
$$
and here the components refer to this decomposition of $H$; the zero in the $(1,1)$ is a consequence of (1) above.


In terms of $B,C$, your conditions become: (1') $B^\*$ surjective (onto $H\_1$); (2') $B^\*$ injective (on $H\_0$) and also if $By=Cy=0$, then $y=0$; (3') $BC=0$, $BB^\*$ surjective; (4'') $BB^\*x=$ for some $x\in H\_0$, $x\not= 0$.


Since $B^\*$ is injective, (4'') shows that $N(B)\not= 0$, but since $N(B)=R(B^\*)^{\perp}$, this contradicts (1').




---


**Comment:** The part below answers the original version of the question.


It does follow, but in a rather trivial, disappointing way. Let $S$ be the operator $T^k$, restricted to its reducing subspace $H\_0$. Since $N(T^k)=N(T)$ for a self-adjoint operator, we have $N(S)= N(T)\cap H\_0$. By assumption, if $x\in H\_0$, $x\not= 0$, then $x\notin N(T)^{\perp}$, so $x\notin N(S)^{\perp}$. Thus $S=0$ and hence $T=0$ as well on $H\_0$. But then $H=H\_0$, and your operator was the zero operator all along.


