Nilpotent Operators over Vector Spaces

Question

Exercise Prove that if $S:V \to V$ (a vector space) is nilpotent, then $I_V-S$ in invertible.

An operator $S: V \to V$ is nilpotent if there some some natural number $k$ such that $S^k = \underbrace{S \circ S \circ \dots \circ S}_{k \, times}$ is the zero map.

$I_V$ simply refers to the identity matrix/mapping with respect to $V$

The author provides a hint by asking us to consider the product of $I_v - S$ and $(I_V +S + S^2 + \dots + S^{k-1})$

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Proof.

Suppose $S: V \to V$ is nilpotent. Then

$$\exists \, k \in \mathbb{N}, S^k(v) = 0 \hspace{0.3cm} \forall v \in V$$

We now consider the product

$$\Big[ I_v - S \Big] \cdot \Big[ I_v + S+ S^2 + \dots + S_{k-1}\Big]$$

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I'm not exactly seeing here how considering this product is helpful. Does anyone have any advice on where to go from here? Is there some property of matrices that would be helpful?

Answer 1

Expanding the brackets shows that $$\Big[ I_V - S \Big] \cdot \Big[ I_V + S+ S^2 + \dots + S^{k-1}\Big]=I_v-S^k=I_V,$$ which shows that $I_V-S$ is invertible.

Answer 2

Expanding the product yields $$ \Big[ I_v - S \Big] \cdot \Big[ I_v + S+ S^2 + \dots + S^{k-1}\Big] = \\ \Big[ I_v + S+ S^2 + \dots + S^{k-1}\Big] - \Big[ S+ S^2 + \dots + S^{k-1} + S^k\Big]. $$ Note that this sum can be simplified.