$T \in \text{Hom}V $ is nilpotent implies $I - T$ invertible [duplicate]
$T \in \text{Hom}V$
And $T$ is nilpotent operator with $deg_{null} (T) =n$
Claim: $(I-T)$ is Invertible.
$G_0(T) =Null\{T^k : k\in \Bbb{N}\}$
As, $T^n=0$, $G_0(T)= V$
Hence, $spec(T) =\{0\}$
Now, eigen value of $(I-T) = 1-\text{ eigenvalue of } $T$.$
Now, $0\notin spec(I-T) $
$0$ is not an eigen value of $(I-T) $ implies $(I-T) $ is Invertible.
And check \begin{align}(I-T) ^{-1}&=I+T+T^2 +... +T^{{deg_{null}}-1}\end{align}
$\frac{I}{I-T} =\sum_{k=0}^{\infty} T^k =\sum_{k=0}^{n-1} T^k$
As, $ T^k=0 \space , \forall k=n, n+1, n+2,... $