About the spectral resolution theorem

Question

Let $H$ separable Hilbert space over complex numbers. $K(H)$ the space of all compact operators acting in $H$ and $x\in K(H)$ selfadjoint ($x=x^*$). Suppose $x=\sum\limits_{n} \lambda_n p_n$ is a spectral resolution, where $p_i$ one dimensional orthoprojectors. Suppose further we have one dimensional selfadjoint $y\in K(H)$ such that $y\cdot z=0$ whenever $x\cdot z=0$, where $z$ any selfadjoint operator in $K(H)$. Is it true that $y=p_{i_0}$ for some $i_0\in\mathbb N$?

Answer 1

No, it isn't. Suppose $H$ is an arbitrary Hilbert space with $\dim H\geq3$ and let $\{e_1,e_2,e_3\}$ be an orthonormal set in $H$. Consider the orthoprojectors $p_i$ onto $\text{span} \{e_i\}$ and set $x:=p_1+p_2+p_3$. Then $x$ is already given in a "spectrally resolved" form. Since the vectors $e_i$ are pairwise orthogonal, we have $$||x(v)||^2=||p_1(v)+p_2(v)+p_3(v)||^2=||p_1(v)||^2+||p_2(v)||^2+||p_3(v)||^2$$ so that $x(v)=0$ iff $v\in\{e_1,e_2,e_3\}^\perp$. Hence $x\cdot z=0$ iff $\text{ran}\,z\subseteq\{e_1,e_2,e_3\}^\perp$. Therefore, any one dimensional orthoprojector $y$ which projects to a subspace of $\{e_1,e_2,e_3\}^\perp$ satisfies your condition, however, not all of them have the form $p_{i_0}$ - pick eg $y:=p_1+p_2$, the orthoprojector onto $\text{span} \{e_1+e_2\}$.

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The idea behind my example is that the representation of $x$ as a sum of scaled orthoprojectors need not be unique if there are eigenspaces with at least 2 dimensions.