How to find the spectrum $\sigma_p(P)$

Question

How to find the spectrum $\sigma_p(P)$:

Let $P:H\rightarrow H$ be an orthoprojection, $P\neq 0, P\neq I$.

could you please help

Answer 1

Let $H_0=\mathrm{Im}(P)$ and $H_1=\mathrm{Ker}(P)$. Since $P\neq 0, I$, so $$ \mathrm{Ker}(P)=H_1\neq\{0\}\qquad\mathrm{Ker}(P-\lambda 1_H)=H_0\neq\{0\} $$ which gives $\{0,1\}\subset\sigma_p(P)$. If $\lambda\notin\{0,1\}$, then consider $$ R=(1-\lambda)^{-1}P-\lambda^{-1}(1_H-P) $$ We have $$ \begin{align} (P-\lambda 1_H)R&=(P-\lambda 1_H)((1-\lambda)^{-1}P-\lambda^{-1}(1_H-P))\\ &=(1-\lambda)^{-1}P^2-\lambda P(1_H-P)-\lambda(1-\lambda)^{-1}P+(1_H-P)\\ &=(1-\lambda)^{-1}P-\lambda P+\lambda P^2-\lambda(1-\lambda)^{-1}P+(1_H-P)\\ &=(1-\lambda)^{-1}P-\lambda(1-\lambda)^{-1}P-\lambda P+\lambda P+(1_H-P)\\ &=1_H\\ \end{align} $$ $$ \begin{align} R(P-\lambda 1_H)&=((1-\lambda)^{-1}P-\lambda^{-1}(1_H-P))(P-\lambda 1_H)\\ &=(1-\lambda)^{-1}P^2-\lambda (1_H-P)P-\lambda(1-\lambda)^{-1}P+(1_H-P)\\ &=(1-\lambda)^{-1}P-\lambda P+\lambda P^2-\lambda(1-\lambda)^{-1}P+(1_H-P)\\ &=(1-\lambda)^{-1}P-\lambda(1-\lambda)^{-1}P-\lambda P+\lambda P+(1_H-P)\\ &=1_H\\ \end{align} $$ This means that for $\lambda\notin\{0,1\}$ operator $P-\lambda 1_H$ is invertible. Thus $\sigma(P)=\sigma_p(P)=\{0,1\}$