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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 $a, bin mathbb F$ 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.

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

$q(X) = -I - X in Bbb F[X]. tag{32}$

Do it by induction (i.e. that $p(T)$, where $p$ is of degree $n$, can be written as $aI + bT$).

Note that for all $n$, $T^{n} = -T^{n-1}-T^{n-2}$.

Suppose $p$ is of degree $1$, then we are done automatically.

Otherwise, suppose that $p$ is of degree $n$, and $p(T) = sum_{k=0}^n a_nT^n$. Note that $a_nT^n = a_n(-T^{n-1}-T^{n-2})$, so that we can rewrite $p(T) = sum_{k=0}^{n-3} a_kT^k + (a_{n-2} - a_n) T^{n-2} + (a_{n-1} - a_n)T^{n-1}$, which is now of degree at most $n-1$ in $T$. By the induction hypothesis, $p(T) = aI+bT$ for some $a$ and $b$. Hence, the induction is complete.

$T$ is invertible, since $-T^2 -T = I$, which factorizes as $(-I-T)T = T(-I-T) = I$. Hence, $(-I-T)$ is the inverse of $T$, and therefore $q(T) = -T-I$ is the inverse of $T$ and a polynomial in $T$.