What is the resolvent and spectrum of the projection operator?

Question

Well, i am reading about this subject and for one projection operator $P:X\to X$ ($P^2=P$) the excercise ask me to find the resolvent and spectrum. (The excersice do not say who is $X$. I suppose that $X$ is a Banach space)

My attempt. I note that If $\lambda x=Px$ implies that $P(\lambda x)=P^2(x)=P(x)$ then $\lambda=1$. Therefore, for me the spectrum is $\{1\}$ and the resolvent is $\mathbb C-\{1\}.$

Am i fine or there is other elements? Please somebody can to help me? Thank you.

Best

If $P$ is onto, it is the identity and $1$ is the spectrum. But if $P$ is not onto, then $\lambda=0$ is also part of the spectrum. — plop, Aug 26 '22 at 20:05
$\lambda=1$ or $\lambda=0$ satisfy $\lambda^2=\lambda$. So the spectrum is a non-empty subset of ${0,1}$. — Disintegrating By Parts, Aug 26 '22 at 20:05

Martin Argerami · Answer 1 · 2022-08-26T20:13:03.380

4

Most projections will have a kernel, so $0$ is also an eigenvalue. So $$\sigma(P)\subset \{0,1\}.$$ All possibilities can occur, though. If $P=0$, then $\sigma(P)=\{0\}$, if $P=I_X$ then $\sigma(P)=\{1\}$, and if $P$ is any other projection then $\sigma(P)=\{0,1\}$.

But for this you need to show that no other $\lambda$ is in the spectrum: , if $\lambda\not\in \{0,1\}$, you can check explicitly that $P-\lambda I$ is invertible by noting that $$ (P-\lambda I_X)^{-1}=\frac1{\lambda(1-\lambda)}P-\frac1\lambda \,I_X. $$

edited Aug 26 '22 at 20:13

answered Aug 26 '22 at 20:06

Martin Argerami

205,756

2

I prefer to write $$\lambda I-P= \lambda (I-P)+(\lambda -1)P$$ Then it is clear that the inverse is equal $$\lambda^{-1}(I-P)+(\lambda-1)^{-1}P$$ provided $\lambda \neq 0,1.$ – Ryszard Szwarc Aug 26 '22 at 20:37
Very nice! $ \ $ – Martin Argerami Aug 26 '22 at 20:41

Disintegrating By Parts · Accepted Answer · 2022-08-27T10:29:42.383

If $P^2=P$, then \begin{align} \lambda(\lambda-1)I&=\lambda(\lambda-1)I-P(P-I) \\ &=\lambda^2I-\lambda I-P^2+P \\ &=(\lambda^2 I-P^2)-(\lambda I-P) \\ &=(\lambda I-P)(\lambda I+P)-(\lambda I-P) \\ &=(\lambda I-P)((\lambda-1)I+P) \end{align} Therefore, the resolvent of $P$ is defined for $\lambda\notin\{0,1\}$ by \begin{align} (\lambda I-P)^{-1}=\frac{1}{\lambda(\lambda-1)}((\lambda-1)I+P) \end{align} The spectrum can be $\{0\},\{1\},\{0,1\}$.

NOTE: This technique works for any operator $P$ such that $p(P)=0$ for some polynomial $p$, which includes all $n\times n$ matrices because the characteristic polynomial is such a polynomial. But it also covers operators in infinite dimensions that are annihilated by a polynomial $p(\lambda)$, such as a projection. The technique for finding the resolvent $(\lambda I-A)^{-1}$ amounts to looking at the annihilating polynomial $p$ and writing $$ p(\lambda)-p(\mu)=(\lambda-\mu)q(\lambda,\mu) $$ and substituting $\mu=A$, and using $p(A)=0$: $$ p(\lambda)I = (\lambda I-A)q(\lambda,A) \\ I=(\lambda I-A)\left[\frac{1}{p(\lambda)}q(\lambda,A)\right]=\left[\frac{1}{p(\lambda)}q(\lambda,A)\right](\lambda I-A) $$ This works for all $\lambda$ for which $p(\lambda)\ne 0$ and gives the two-sided inverse: $$ (\lambda I-A)^{-1}=\frac{1}{p(\lambda)}q(\lambda,A). $$ This also works for all matrices over $\mathbb{C}$ by setting $p$ equal to the characteristic or minimal polynomial.

Amazing, thank you anybody, all is very useful for me!!! – weymar andres Aug 27 '22 at 14:53 — weymar andres, Aug 27 '22 at 14:53

What is the resolvent and spectrum of the projection operator?

2 Answers2