$T \in \text{Hom}V $ is nilpotent implies $I - T$ invertible

Question

Assume $T \in \text{Hom}V$ is nilpotent I.e. $T^n=0$ for some $n$. How do you show that this implies $I - T$ is invertible? The hint says the proof has something to do with the power series but I’m not sure how to proceed. Thanks

Answer 1

$T \in \text{Hom}V$

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,... $