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


*

*T is oneone but not onto.

*T is onto but not one one.

*T is invertible.

*no such T exists.


could any one give me just hint?
 A: $$
T^2-T+I=0 \iff T(I-T)=I=(I-T)T,
$$
i.e. $T$ is invertible and $T^{-1}=I-T$. In particular $T$ is injective and surjective.
A: 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?
A: 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.
A: $T(T-1)= -I$ then $\det T\cdot \det (T-I) = (-1)^n$ which implies $\det (T) \neq0 \,\,.$ hence $T$ invertible
