Spectrum of general projection and orthogonal projection

Question

I am trying to think about this, but I seem to be stuck. Suppose $P$ is a projection on a Hilbert space $\mathcal{H}$. If I am just talking about a general projection, where I only know that $P^2=P$, can I deduce that the spectrum of $P$ can only consist of $0$ and $1$? What about if $P$ is also orthogonal, meaning $P^*=P$? Can I also say something about the type of spectrum in terms of classification to point (eigenvalues), residual, and continuous? I am relatively new to operators and functional analysis, so my toolbox might be a bit limited here. I thank all helpers for any and all help.

Answer 1

Since the formula $\sigma(P)^2=\sigma(P^2)=\sigma(P)$ holds, the spectrum of $P$ is contained in $\{0,1\}$. The condition $P=P^*$ is not necessary for this.

Answer 2

Being orthogonal makes no difference in terms of what the spectrum is. You have

• if $P=I$, then $\sigma(P)=\{1\}$;

• if $P=0$, then $\sigma(P)=\{0\}$;

• if $P$ is not $0$ nor $I$, then $\{0,1\}\subset\sigma(P)$ (because $P(I-P)=0$ shows that neither can be invertible) and for any $\lambda\in\mathbb C\setminus\{0,1\}$ one can check directly that $$ (P-\lambda I)^{-1}=\frac1{\lambda(1-\lambda)}\,P-\frac1\lambda\,I. $$

In summary, if $P^2=P$ then

• $\sigma(P)=\{1\}$ if and only if $P=I$;

• $\sigma(P)=\{0\}$ if and only if $P=0$;

• $\sigma(P)=\{0,1\}$ if and only if $P\ne 0,I$.

In all cases the spectrum agrees with the point spectrum. For if $x\in P\mathcal{H}$, then $Px=x $ and so $1$ is an eigenvalue.