Relation between spectrum of an operator and its cut down

Question

Let $T$ be a self-adjoint operator in $\mathcal{H}$ with spectrum $\sigma(T)$, Let $P$ be a projection in the commutant of $\text{vN}\{T\}$, the von Neumann algebra generated by $T$, question what is the spectrum of $PTP$? What are the relation betweeen $\sigma(PTP)$ and $\sigma(T)$?

Answer 1

There is no relation. This already shows in $M_2(\mathbb C)$. Take any $t\in[0,1]$, $$ T=\begin{bmatrix} t &\sqrt{t-t^2}\\ \sqrt{t-t^2}&1-t\end{bmatrix},\ \ P=\begin{bmatrix} 1&0\\0&0\end{bmatrix} . $$ Then $\sigma(T)=\{0,1\}$, $\sigma(PTP)=\{0,t\}$.

And this idea can be used even more brutally. Let $K$ be any compact subset of $[0,1]$. Construct a selfadjoint operator $T_0$ with spectrum $K$. Now, on $H\oplus H$, form $$ T=\begin{bmatrix} T_0 & (T_0-T_0^2)^{1/2} \\ (T_0-T_0^2)^{1/2} & I-T_0\end{bmatrix}, \ \ P=\begin{bmatrix} I&0\\0&0\end{bmatrix}. $$ Then $T$ is a projection, so $\sigma(T)=\{0,1\}$, while $\sigma(PTP)=\{0\}\cup\sigma(T_0)=0\cup K$.