Invertible Linear Operator to a Power I am trying to solve the following question. I know that this can be solved by a more simple method but I am trying to gain insight on my attempted proof, if I can proceed or if this method will not yield a result.
Suppose $T$ $\in$ $L(V)$ and there exists a positive integer n such that $T^n=0$. 
Prove that $I-T$ is invertible and that $$(I-T)^{-1}=I+T+...+T^{n-1}$$
I have tried to argue that since $T$ and $I$ are operators then $I-T$ must be an operator. To then show that $I-T$ is invertible we need only show that it is surjective i.e $Null(I-T)=0$. Then consider a $u\in$ $Null(I-T)$ $\Rightarrow$ $Tu=u$ so $u$ is an eigenvector corresponding to the eigenvalue $\lambda=1$. Is there a way to force $u=0$ here to show that $I-T$ is injective hence invertible?
 A: You can show the inverse by definition. Simply expand:
\begin{align*}
(I - T)(I + T + \ldots + T^{n-1}) &= I(I + T + \ldots + T^{n-1}) - T(I + T + \ldots + T^{n-1}) \\
&= I + T + \ldots + T^{n-1} - T - T^2 - \ldots - T^{n-1}-T^n \\
&= I - T^n = I.
\end{align*}
Similarly,
$$(I + T + \ldots + T^{n-1})(I - T) = I - T^n = I,$$
so the formula for the inverse is proven by definition of the inverse.
A: Take $T^{n-1}$ on both sides of $Tu=u$ to get $T^nu=T^{n-1}u$. Since $T^n=0$ the left hand side must be $0$, and hence $T^{n-1}u=0$. Now take $T^{n-2}$ on both sides of $Tu=u$ to get $T^{n-1}u=T^{n-2}u$. We already know that the left hand side is $0$, hence $T^{n-2}u=0$. Continue this process by induction, and finally you will get $Tu=0$. The equation $Tu=u$ then implies $u=0$. 
A: Note that if $u \ne 0$ is an eigenvector with eigenvalue $\mu$, that is,
$Tu = \mu u, \tag 1$
then
$T^2 u = T(Tu) = T(\mu u) = \mu Tu = \mu(\mu u) = \mu^2 u; \tag 2$
taking this as the basis, we may argue via induction as follows:
$T^k u = \mu^k u$
$\Longrightarrow T^{k + 1}u = TT^k u = T(\mu^k u) = \mu^k Tu = \mu^k (\mu u) = \mu^{k + 1}u, \tag 3$
from which we conclude that
$\forall m \in \Bbb N, \; T^m u = \mu^m u; \tag 4$
with $m = n$ we find that
$\mu^n u = T^n u = 0u = 0, \tag 5$
so if
$\mu \ne 0, \tag 6$
$u = 0, \tag 7$
contrary to assumption; therefore we must have
$\mu = 0, \tag 8$
i.e. the only eigenvalue of $T$ is $0$; but this precludes
$Tu = u = 1u = Iu, \tag 9$
for non-zero $u$, and thus 
$\ker (I - T) = \{0\}, \tag{10}$
as per request.
A perhaps more direct route to the invertibility of $I - T$ is the observaion that, for any $k \in \Bbb N$,
$(I - T)(T^{k - 1} + T^{k - 2} + \ldots + T + I) = I - T^k; \tag{11}$
with $k = n$ this yields
$(I - T)(T^{n - 1} + T^{n - 2} + \ldots + T + I) = I, \tag{12}$
showing that $I - T$ is invertible and
$(I - T)^{-1} = T^{n - 1} + T^{n - 2} + \ldots + T + I. \tag{13}$
(11) is easily seen via the observation that
$(I - T)(T^{k - 1} + T^{k - 2} + \ldots + T + I)$
$= T^{k - 1} + T^{k - 2} + \ldots + T + I - T^k - T^{k - 1} - \ldots - T^2 - T$
$= I - T^k, \tag{14}$
and this complete or argument.
