Existence of scalars in polynomial of operator Let $V$ a vector space over a field $\mathbb F$. Let $p(X) \in \mathbb F[X]$ a polynomial and $T:V \rightarrow V$ an operator with $T^2+T+I=0$ s.t. $I$ is the identity operator over $V$.
How to prove that there exists $a, b\in \mathbb F$ with $p(T)=aI+bT$.
and is it possible that $T$ is invertible ? If it is, How can I find a polynomial $q(X) \in \mathbb F$ such that $q(T)$ is the inverse operator of $T$?
I have tried to represent $T$ as $T=-T^2-I$ and then putting it in the polynomial given, but it didn't given me anything.
 A: While I was staring at the equation
$T^2 + T + I = 0, \tag{1}$
and wondering what to do, I was struck by a curious similarity to the equation
$\omega^2 + \omega + 1 = 0, \tag{2}$
which arises when $\omega - 1$ is factored out of
$\omega^3 - 1 = 0, \tag{3}$
the equation satisfied by the (complex) cube roots of unity.  Therefore I multiplied (1) by $T - I$,
$(T - I)(T^2 + T + 1) = T^3 + T^2 + T - T^2 - T - I = T^3 - I, \tag{4}$
and discovered that
$T^3 - I = (T - I)(T^2 + T + 1) = (T - I)(0) = 0, \tag{5}$
that is,
$T^3 - I = 0 \tag{6}$
or
$T^3 = I.  \tag{7}$
Equation (7) implies there is a certain cyclicity to the powers of $T$:
$T^0 = I, \tag{8}$
$T^1 = T, \tag{9}$
$T^2 = -T - I, \tag{10}$
$T^3 = I, \tag{11}$
$T^4 = T^3 T = IT = T, \tag{12}$
$T^5 = T^3T^2 = IT^2 = T^2 = -T - I, \tag{13}$
$T^6 = (T^3)^2 = I^2 = I, \tag{14}$
$T^7 = T^6T = IT = T, \tag{15}$
and so forth.  In general, for $0 \le m \in \Bbb Z$ we have, by the division algorithm,
$m = 3q + r, \tag{16}$
with
$q \ge 0 \tag{17}$
and
$0 \le r \le 2; \tag{18}$
thus,
$T^m = T^{3q + r} = T^{3q}T^4 = (T^3)^qT^r = I^qT^r = IT^r = T^r, \tag{19}$
$r$ being the remainder when $m$ is divided by $3$.
Now for $p(X) \in \Bbb F[X]$ we have
$p(X) = \sum_0^n p_iX^i, \tag{20}$
where $p_i \in \Bbb F$ and $n = \deg p$.  We can break the sum (20) down into groups of $3$ terms each by again using the division algorithm to write
$n = 3q + r, 0 \le r \le 2, \tag{21}$
so that
$p(X) = \sum_0^q (p_{3i}X^{3i} + p_{3i + 1}X^{3i + 1} + p_{3i + 2}X^{3i + 2})$
$= \sum_0^q (p_{3i}X^0 + p_{3i + 1}X^1 + p_{3i + 2}X^2)X^{3i}, \tag{22}$
where, when $i = q$, we have chosen the $p_{3i + r} = p_{3q + r}$, $0 \le r \le 2$, to be zero as necessary to preserve $\deg p = n$.  With $p(X)$ as in (22), we have
$p(T) = \sum_0^q (p_{3i}T^0 + p_{3i + 1}T^1 + p_{3i + 2}T^2)T^{3i}$
$= \sum_0^q (p_{3i}T^0 + p_{3i + 1}T^1 + p_{3i + 2}T^2)(T^3)^i = \sum_0^q (p_{3i}I + p_{3i + 1}T + p_{3i + 2}(-I - T)), \tag{23}$
since
$(T^3)^i = I^i = I \tag{24}$
for $0 \le i \le q$, and we have made use of (10).  We re-arrange (23) in order to group terms in $I$ and $T$:
$p(T) = \sum_0^q ((p_{3i} - p_{3i + 2})I + (p_{3i + 1} - p_{3i + 2})T). \tag{25}$
From (25) we immediately see that we may write
$p(T) = a + bT \tag{26}$
with
$a = \sum_0^q (p_{3i} - p_{3i + 2}) \tag{27}$
and
$b = \sum_0^q  (p_{3i + 1} - p_{3i + 2}); \tag{28}$
not only have we shown the existence of the requisite $a$ and $b$, but we have obtained explicit formulas for them as well.
Finally, $T$ is most definitely invertible.  By (1) we have
$T(I + T) = T + T^2 = -I, \tag{29}$
whence
$T(-(I + T))  = I, \tag{30}$
so that we must indeed have
$T^{-1} = -(I + T).  \tag{31}$
We then take
$q(X) = -I - X \in \Bbb F[X].  \tag{32}$
