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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. :)

Berrick Caleb Fillmore
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3 Answers3

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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.

ougao
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If $A$ is a self-adjoint, compact operator on a Hilbert space $H$ and if $f:\mathbb{C}\to\mathbb{C}$ is a polynomial, then $\sigma(f(A))=f(\sigma(A))$. (This holds more generally for continuous mappings of $\sigma(A)$ to $\mathbb{C}$).

A projection is self-adjoint and compact and if you take the polynomial $f(z)=z^2-z$ you obtain that $f(p)=0$, therefore $\sigma(f(p))=\{0\}$, hence $\sigma(p)\subset f^{-1}(\{0\})=\{0,1\}$. It is easy to check that both are eigenvalues: $p$ has a kernel, therefore $0\in\sigma(p)$, and for every $v\in H$, $p(p(v))-p(v)=0$, therefore $p-I$ is not one-to-one ($p(v)\in\ker(p-I)$ and $p(v)$ is non-zero for some $v\in H$, otherwise $p=0$ and $\sigma(p)=\{0\}$).

wisefool
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    The only projections that are compact are those with finite-dimensional range. Fortunately, $\sigma(f(A))=f(\sigma(A))$ holds more generally than indicated. In fact it holds for all operators, without need for either compactness or self-adjointness, when $f$ is a polynomial. (In order to generalize to functional calculus with other types of functions, you may need to specialize the types of operators allowed. E.g., it would hold with $f$ continuous and $A$ normal, again with no need for compactness.) – Jonas Meyer May 26 '13 at 21:19
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Let $M = ran(P)$, where P is orthogonal projection. Just for convience writing $N = M^\perp$. Let z not be equal to 0 or 1. Now for any $\phi \in H$ we can uniquely write, $\phi = \phi_M + \phi_N$. That implies $$ (P-x)\phi = (1-x)\phi + (-x \phi_N) $$ Since both M and N are closed subspaces and any element of hilbert space can be uniquely written as sum of elements from these two spaces. We conclude that $(P-x)$ is bijective. Also $P(H) = Ran(P)$ and $(P-1)(H)=Ran(P)^\perp$ are bijective iff $P=I$ or $P=0$ repectively.

Hence $$\sigma(P) = \{0,1\}, \hspace{5 pt} P \neq \{0,I\}$$ whereas $\sigma(I)=1$, and $\sigma(0)={0}$.

Dharamveer Kumar
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