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Let $T\in\mathcal L(V)$. Show that if $T^n=0$, for some $n\in\Bbb N_{>0}$, then $I-T$ is invertible.

I can show that the above is true whenever $\dim V<\infty$, but I dont know how to prove it if $\dim V=\infty$. I dont know where start because nearly all my knowledge of linear algebra is based on finite dimensional spaces.

I though a sketch for a proof by contradiction, assuming that $I-T$ is not invertible but $T^{n-1}\neq 0$ and $T^n=0$ for some $n\in\Bbb N_{>0}$, but I dont know what to do from here because $I-T$ not being invertible doesnt necessarily means that $I-T$ is not bijective (I was thinking to try to reduce the problem to show that $I-T$ is bijective), it can be the case that $I-T$ is bijective but discontinuous, hence not invertible.

Maybe the exercise would need to add that $V$ is finite dimensional, but this is not stated so I assumed that I must show that the statement is true also when $\dim V=\infty$.

Some help will be appreciated, thank you.

Masacroso
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2 Answers2

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$(I+T+\cdots+ T^{n-1})(I-T) = I$.

copper.hat
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$$ (I-T)(I+T+T^2+ \dotsb + T^{n-1}) = I-T+T-T^2+T^2+\dotsb+T^{n-1}-T^n = I-T^n = I, $$ so $\sum_{k=0}^{n-1} T^k$ is an left-inverse for $I-T$. We can show it's a right-inverse in exactly the same way.

Chappers
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