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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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up vote 1 down vote accepted

I was looking at a problem similar to this, and these answers are great but I needed a bit more to make it click. Here is a step by step with the rules used.

$T^2 − T + I = 0$

$TT = T - I$

$TT = T - TT^{-1}$ , because $I = TT^{-1}$

$TT = IT - TT^{-1}$ , because $IT = T$

$TT = T(I-T^{-1})$ , factor out the T

$T = I - T^{-1}$ , remove T from both sides

In my case we had to prove $T^{-1} = 2I - T$ given almost the same equation (that 2 being the difference). I know this break down is probably too basic for most, but this what helped me understand how to apply those rules.

I was lead here by this duplicate.

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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? – Un Chien Andalou Nov 9 '12 at 16:38
I know about minimal polynomial, it has distinct roots over the field $\mathbb{C}$ – Un Chien Andalou 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

$T(T-1)= -I$ 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

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