$u^3 + u^2-id=0\text{ and } tr(u) \in \mathbb{Q}$ show that the dimension of the vector space is a multiple of 3

Question

$\text{let E a vector space} \:dimE=n \\ \text{let}\:\: u \in L(E,E) \text{such as} \: u^3 + u^2-id=0\text{ and } tr(u) \in \mathbb{Q}\\ \text{show that n is a multiple of 3}$

Answer 1

$X^3+X^2-1$ has 3 distinct roots $a_1,a_2,a_3$ in $\overline{\Bbb{Q}}$.

$E$ is a finite dimensional $K$-vector space where $K$ contains $\Bbb{Q}$, let $E' = E\otimes_K \overline{K}$ and $u':E'\to E'$. Then

$$E' = \ker(u'-a_1)\oplus \ker(u'-a_2)\oplus \ker(u'-a_3)$$

$$tr(u') = \sum_{j=1}^3 a_j \dim(\ker(u'-a_1))$$ (dimension as $\overline{K}$ vector space)

That $tr(u')\in \Bbb{Q}$ implies that $$\dim(\ker(u'-a_1))=\dim(\ker(u'-a_2))=\dim(\ker(u'-a_3))$$ ie. $$n=\dim(E')=3 \dim(\ker(u'-a_1))$$

Answer 2

the eigenvalues are distinct, two are complex conjugates. The minimal polynomial is the given $x^3 + x^2 -1.$ The degree of each eigenvalue term $(x- \lambda)$ in the minimal polynomial is the size of the largest Jordan block. Thus the Jordan form is diagonal; the trace is $n_1 \lambda + n_2 \mu + n_3 \bar{\mu}$ where the real eigenvalue is $\lambda.$ For the trace to be real, we know $n_2 = n_3$

We also know that the sum of the eigenvalues is $-1.$ If $n_1 \neq n_2,$ write $t = n_1 - n_2$ so that the trace is now $$ t \lambda + n_2 \lambda + n_2 \mu + n_2 \bar{\mu}= t \lambda - n_2 $$ If $t \neq 0$ this is irrational.

So, $n_1 = n_2 = n_3$ and the dimension is $3 n_1$