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Not exactly sure how to approach this: my idea was to try to show that $S-1$ is injective and since $S-1$ is a linear map from V to V, it follows that $S-1$ also surjective by the rank-nullity theorem (on the condition that V was finite-dimensional), and hence $S-1$ is bijective. I am still stuck. Any help is appreciated.

Saran Wrap
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    Hint $\frac{1}{1-x} = 1 + x + x^2 + \dots$. – Magdiragdag Nov 08 '19 at 21:16
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    do you know the geometric series formula? that and the assumption that there exist $n$ such that $S^n=0$ are very useful. you'll find that you can explicitly write what the inverse is as a polynomial in $S$. – peek-a-boo Nov 08 '19 at 21:18
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    Dupe of https://math.stackexchange.com/questions/119904/units-and-nilpotents and https://math.stackexchange.com/questions/325318/prove-that-if-matrix-a-is-nilpotent-then-ia-is-invertible and https://math.stackexchange.com/questions/483173/relation-between-nilpotent-matrix-and-identity-matrix – Hanno Nov 08 '19 at 22:10

2 Answers2

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Hint:

For a scalar $\;\alpha\; ,\;\; S-\alpha I\; $ is an isomorphism iff $\;\alpha\;$ is not an eigenvalue of $\;S\;$ . But it is given $\;S\;$ is nilpotent, thus its only eigenvalue is...

DonAntonio
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An eigenvalue argument is not required.

Given that

$S^n = 0, \tag 1$

we observe that

$(I - S) \left ( \displaystyle \sum_0^{n - 1} S^k \right ) = (I - S)(I + S + S^2 + \ldots + S^{n - 1} ) = I - S^n = I, \tag 2$

which shows that $I - S$ is invertible and that

$(I - S)^{-1} = \displaystyle \sum_0^{n - 1} S^k. \tag 3$

Note that this result binds no matter what the dimension of $V$ might be; in particular, $\dim V$ need not be finite.

Robert Lewis
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