How to show that a null potent linear transformation is invertible

Question

V is a K vector space and $ψ : V → V$ is a null potent linear transformation i.e. $ψ^N = 0$ for a certain $N ∈ N$. Prove that $Id_V − ψ$ est an invertible element in the ring $L(V, V )$. My assistant told me to consider the product of $Id_V − ψ$ and $(Id_V + ψ + ψ^2 + · · · + ψ^{N−1})$. But why or how would i know to consider the product of those two? How would i know that i would need to consider the product of those 2 things. What is the significance of $(Id_V + ψ + ψ^2 + · · · + ψ^{N−1})$?

Answer 1

Hint: use the geometric series identity, writing $\frac{1}{\textrm{Id}_V - \psi}$ instead of $\frac{1}{1-x}$.

Answer 2

I will use $I$ for the identity transformation. Note that $I\in Z(L(V,V))$ is in the center of the algebra, so that it commutes with everything. But then

$$(I-\psi)(I+\psi+\psi^2+\ldots + \psi^{N-1})$$ $$=I+\not\psi+\not\psi^2+\ldots + \not\psi^{N-1}$$ $$\qquad\qquad\quad -(\not\psi+\not \psi^2+\ldots + \not\psi^{N-1}+\psi^{N})$$

and since $\psi^N=0$ this is just $I$, hence $(I+\psi+\psi^2+\ldots + \psi^{N-1})$ is a right inverse for $I-\psi$, but the same computation with the factors reversed shows it is also a left inverse.