# linear operator on a vector space V such that $T^2 -T +I=0$

let T be a linear operator on a vector space V such that $T^2 -T +I=0$.Then

1. T is oneone but not onto.
2. T is onto but not one one.
3. T is invertible.
4. no such T exists.

could any one give me just hint?

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$$T^2-T+I=0 \iff T(I-T)=I=(I-T)T,$$ i.e. $T$ is invertible and $T^{-1}=I-T$. In particular $T$ is injective and surjective.

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Let $\mathbb{x}$ be any vector in the nullspace. Then $T\mathbb{x} = \mathbb{0}$. Using your equation $T^2 - T + I = 0$, what can you conclude about $\mathbb{x}$?

Alternatively if you know about minimal polynomials: How does your polynomial split?

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$T^2(x)-T(x)=0$ and so $I(x)=0$ so $x=0$ so $T$ is injective. am I right? –  Bunuelian Trick Nov 9 '12 at 16:38
I know about minimal polynomial, it has distinct roots over the field $\mathbb{C}$ –  Bunuelian Trick Nov 9 '12 at 16:40
Your mapping is injective. It is not necessarily surjective unless $V$ is finite-dimensional. –  EuYu Nov 9 '12 at 16:52
It is surjective: $T(-Tx + x) = x$ –  Robert Israel Nov 9 '12 at 16:54
Oh, well that's that. Thank you for catching my mistake @RobertIsrael –  EuYu Nov 9 '12 at 16:54
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$T(T-1)= -I$ then $\det T\cdot \det (T-I) = (-1)^n$ which implies $\det (T) \neq0 \,\,.$ hence $T$ invertible

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Is it true for a vector space with not a finit dimension ? –  Ricky Bobby Nov 9 '12 at 16:40