Im studying 'Advanced Calculus' in 'Loomis-Shlomo's' book, and got stuck at a question in page 50, question 3.23:
http://www.math.harvard.edu/~shlomo/docs/Advanced_Calculus.pdf
"Suppose T is nilpotent, S commutes with T, $S^{-1}$ exists. $S,T \in Hom (V)$, show that $(S-T)^{-1}$ exists."
Ive tried to substitute I-S+T = x in the formula of question 3.22 and didn't succeed...
Could anyone give a hint?
On
One trick is to formally compute via power series. Disclaimer: inequalities all come with a grain of salt, the final result is what matters.
Try writing $$ (S-T)^{-1}=S^{-1}(I-S^{-1}T)^{-1}=\\ S^{-1}\sum_{k=0}^\infty S^{-k}T^{k} $$ which is in fact the finite sum (this is good), $$ \sum_{k=0}^nS^{-k-1}T^k $$ where $n$ is the order with $T^n=0$.
One may now check this is an inverse directly.
Well, $S-T=S(I-S^{-1}T)$, so if $I-S^{-1}T$ is invertible, then so is $S-T$.
We are given $ST=TS$. Therefore $S^{-1}T=TS^{-1}$. We are also given $T$ nilpotent, so $T^n=0$ for some $n$. What can we deduce from these about $S^{-1}T$? About $I-S^{-1}T$?