# Relation between spectrum of an operator and its cut down

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 $$\sigma(T)$$?

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 $$\sigma(PTP)=\{0\}\cup\sigma(T_0)=0\cup K$$.