What is the resolvent and spectrum of the projection operator?
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
Best Answer
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.