I want to show that $ \sigma(p) = \{ 0,1 \} $ for any orthogonal projection operator $ p \notin \{ 0,I \} $ on a Hilbert space $ \mathcal{H} $. Recall that an orthogonal projection operator $ p $ on $ \mathcal{H} $ is a bounded linear operator such that $ p = p^{*} = p^{2} $. What should I do to prove this?
Suppose that $ \alpha \in \sigma(p) $. Then $ p – \alpha I $ is not invertible, but what next? I can’t imagine how to come up with $ \alpha \in \{ 0,1 \} $. Thank you. 🙂
Best Answer
For any $\alpha\in \mathbb{C}$ and $\alpha\neq 0, 1$, it is easy to check that $(\alpha-P)^{-1}=\frac{1}{\alpha}(I+\frac{P}{\alpha-1})$. And it is obvious that $P, I-P$ are both projections not equal to $I$, so neither of them are invertible, so we are done.