Find the inverse of $\lambda I-T$ for $\lambda\neq 0,1$.

Question

Let $T$ be a bounded operator on a normed space $X$ such that $T^2=T$ .

Find the inverse of $\lambda I-T$ for $\lambda\neq 0,1$.

I am hardly getting any idea how to do it. Should I use some trial and error method for finding a polynomial in $T$ say $f(T)$ such that $(\lambda I-T)f(T)=I$

But how should I do it?Any hints would suffice.

Another question:Can anyone suggest how to solve these type of problems.

Answer 1

Hint: Just try $2I - T$ first, and multiply it by, say, $I+T$ (which is not the right answer). When you compute the product, how can you simplify? You use $T^2 = T$, of course! Now try multiplying by $aI + bT$, and ask which $a$ and $b$ make things work out. Pretty soon you'll have an answer, and will be able to generalize.

Answer 2

The standard resolvent expansion for $|\lambda| > \|T\|$ is $$ (\lambda I-T)^{-1}=\lambda^{-1}(I-\frac{1}{\lambda}T)^{-1}=\sum_{n=0}^{\infty}\frac{1}{\lambda^{n+1}}T^{n} $$ In your case that works out to be $$ \frac{1}{\lambda}I+\sum_{n=1}^{\infty}\frac{1}{\lambda^{n+1}}T = \frac{1}{\lambda}I+\frac{1}{\lambda^2}\frac{1}{1-1/\lambda}T = \frac{1}{\lambda}I+\frac{1}{\lambda(\lambda-1)}T. $$ You can check that the right side works for all $\lambda\notin\{0,1\}$, and this has to be true because the final expression is holomorphic everywhere except at $\lambda=0,1$, and it equals $R(\lambda)=(\lambda I-T)^{-1}$ for $|\lambda| > \|T\|$. You're finished at this point.

It is customary to separate the singularities, which can be done by using the partial fraction decomposition $$ \frac{1}{\lambda(\lambda-1)}=\frac{1}{\lambda-1}-\frac{1}{\lambda}. $$ The resulting expression is $$ (\lambda I-T)^{-1}=\frac{1}{\lambda}(I-T)+\frac{1}{\lambda-1}T. $$