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Let $V$ be a finite dimensional vector space. Let $T$ be a linear transformation such that $(T-I)^k=0$ for some $k\in\mathbb{N}$. Then prove that $T\in\text{SL}(V)$.

Any ideas ?

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    $\begingroup$ To get the idea for $k=2$: $$0=(T-I)^2=T^2-2T+I,$$ hence $$T(2I-T)=I.$$ $\endgroup$ Nov 27, 2016 at 14:06

3 Answers 3

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$(T-I)^k=0$ implies that the minimal polynomial of $T$ is a divisor of $(x-1)^k$. This implies that the only eigenvalue of $T$ is $1$. This implies that the determinant of $T$ (which is the product of all eigenvalues) is $1$.

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  • $\begingroup$ We are using the fact that minimal polynomial and the characteristic polynomial have the same roots (possibly with different multiplicities). Am I right ? $\endgroup$ Nov 27, 2016 at 14:24
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    $\begingroup$ well, you can proof "roots of min-poly = eigenvalues" without using characteristic polynomials at all. $\endgroup$
    – Simon
    Nov 27, 2016 at 14:26
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We don't even need that $V$ is finite-dimensional. Let $\displaystyle S := \sum_{n=0}^{k-1}(I-T)^n$. Then $\displaystyle TS = (I-(I-T))S = S - (I-T)S = \sum_{n=0}^{k-1}(I-T)^n-\sum_{n=1}^k(I-T)^n = (I-T)^0 - (I-T)^k = I$.

Similarly we get $ST = I$

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Hint: For $k = 2$ the equation is $0 = (T - I)^2 = T^2 - 2T + I = T(T - 2I) + I$, i.e. $T(2I - T) = I$.

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