Find the inverse of $\lambda I-T$ for $\lambda\neq 0,1$. 
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.
 A: 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
$$
      (\lambda I-T)^{-1}=\frac{1}{\lambda}(I-T)+\frac{1}{\lambda-1}T.
$$
A: 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. 
