Minimum of Spectrum

Question

Let $M$ be a von Neumann algebra and $e$ a projector.

Is it true that $min Spec_M(x)\leq min Spec_{eMe}(exe)$ for $x$ positive in $M$?

Thank you.

Answer 1

Yes. If $\min\sigma_M(x)=0$, there is nothing to prove. If $\min\sigma_M(x)>0$, there exists $\delta>0$ with $\min\sigma_M(x)>\delta$. suppose that $\delta<\alpha<\min\sigma_M(x)$. Then $x-\alpha\geq \alpha-\delta>0$. This gives us $$ exe-\alpha e=e(x-\alpha)e\geq(\alpha-\delta)e, $$ so $exe-\alpha e$ is invertible. As this can be done for any $\alpha\in(0,\min\sigma_M(x))$, this shows that $$ \min\sigma_{eMe}(exe)\geq\min\sigma_M(x). $$