Linear endomorphism such that $T^2 + 2T + Id_V = 0$ is invertible 
Let $ T: V \rightarrow V$ be an endomorphism such that $T^2 + 2T + Id_V = 0$. Show that $T$ is invertible.

Any hint?
 A: Hint: What does $T(T+2\operatorname{Id}_V)$ equal to?
A: From
$T^2 + 2T + I = 0 \tag 1$
we can of course write
$T(T + 2I) = -I, \tag 2$
or
$T(-(T + 2I)) = I, \tag 3$
which of course shows that
$T^{-1} = -(T + 2I), \tag 4$
in accord with the answer given by user10354138; indeed, any time an operator such as $T$ satisfies a polynomial with non-vanishing constant term,
$\displaystyle \sum_0^n \beta_i T^i = 0, \; \beta_0 \ne 0, \tag 5$
we may write
$T\displaystyle \sum_1^n \beta_i T^{i -1} = -\beta_0I, \tag 6$
from which we infer that
$T^{-1} = -\beta_0^{-1} \displaystyle \sum_1^n \beta_i T^{i - 1}. \tag 7$
Another path to establishing the existence of $T^{-1}$ is via the observation that
$(T + I)^2 = T^2 + 2T + I = 0; \tag 8$
$T + I$ is nilpotent.  Now for any nilpotent operator $N$, 
$N^m = 0, \; m \ge 2, \tag 9$
we have the identity
$I = I - N^m = (I - N)\displaystyle \sum_0^{m - 1} N^i; \tag{10}$
which shows that
$(I - N)^{-1} = \displaystyle \sum_0^{m - 1} N^i; \tag{11}$
and so $I -N$ is invertible.  Taking
$N = T + I, \tag{12}$
we see that
$-T = I - (I + T) \tag{13}$
is invertible, hence so is $T$.  $OE\Delta$.
A: The only zero of the annihilating polynomial $p(x)=x^2+2x+1=(x+1)^2$ is $-1$. This suggests that the only eigenvalue of $T$ is $-1$. Thus, $T$ is non-singular.
