Q:

The dimension of a linear space is a multiple of 3 if it has an endomorphism of a certain feature

I am trying to solve the following problem:

Let $K$ be a field of characteristic zero and $V$ be a finite-dimensional $K$-vector space. Assume that there exists a linear endomorphism $T$ over $V$ such that $T^3=\mathrm{id}$ and $\mathrm{Tr}\ T = 0$. Prove that the dimension of $V$ is a multiple of 3.

If $K=\mathbb C$, things are much simplified, because the eigenvalues of $T$ are among 1, $\omega$ and $\omega^2$, where $\omega$ is the primitive cubic root of 1.
But for general $V$ that is not necessarily algebraically closed, I do not how to use the condition for the trace.
I would be very grateful  if you could give a clue. (If possible, I prefer an elementary solution that does not use field theory.)

A:

For the case $K=\Bbb C$ you would use the fact that the only $\Bbb Q$-linear dependency between the cubic roots of unity $1,\omega,\omega^2$ is (up to a scalar) $1+\omega+\omega^2=0$, so that $m_1.1+m_2\omega+m_3\omega^2=0$ with $m_1,m_2,m_3\in\Bbb N$ implies $m_1=m_2=m_3$ and hence $m_1+m_2+m_3\equiv0\pmod3$.
The same argument works whenever $\omega\in K$, so assume that this is not the case, in other words that $X^3-1=(X-1)(X^2+X+1)$ is a decomposition into irreducible factors over$~K$.
The condition $T^3=\operatorname{id}$ means that the minimal polynomial$~\mu$ of$~T$ divides $X^3-1$; it is a product of a subset of those factors. Without using field extensions, every irreducible factor of the characteristic polynomial$~\chi$ of$~T$ is also a factor of$~\mu$, so $\chi=(X-1)^{m_1}(X^2+X+1)^{m_2}$ for some $m_1,m_2\in\Bbb N$. Then simple computation shows that
$$\chi= X^n+(m_2-m_1)X^{n-1}+\text{lower order terms} \qquad\text{where $n=m_1+2m_2$,}$$
and the subdominant coefficient $m_2-m_1$ equals $-\operatorname{Tr} T=0$, so $m_1=m_2$ and $n=3m_1$.
Of course the latter is just a watered-down version of the argument used when $\omega\in K$, in other words one can avoid case distinctions be reasoning in an extension field where $X^3-1$ does split into linear factors. But I think you asked for a proof that does not use (the existence of) such extensions, and as you can see this can be done. 
Added. I'll add a bit about the "whenever $\omega\in K$" above. First of all this should really say "whenever $K$ contains a root of $X^2+X+1$, which we may call$~\omega$". So supposing that, why should $1+\omega+\omega^2=0$ be up to a scalar the only $\Bbb Q$-linear dependency between $1,\omega,\omega^2$? Basically because $X^2+X+1$ is irreducible over$~\Bbb Q$. One has in general (apply it for $F=\Bbb Q$):
Lemma. Let $K$ be a field containing a subfield$~F$ and a root$~\alpha$ of an irreducible monic polynomial$~P=p_0+p_1X+\cdots+p_{n-1}X^{n-1}+X^n$ in $F[X]$. Then the elements $1,\alpha,\alpha^2,\ldots,\alpha^{n-1}$ of$~K$ are $F$-linearly independent, and the unique $F$-linear dependency between $1,\alpha,\alpha^2,\ldots,\alpha^n$ involving $\alpha^n$ with coefficient$~1$ is $p_0+\cdots+p_{n-1}\alpha^{n-1}+\alpha^n=0$.
The fact that $\alpha$ is root of$~P$ means that the final relation holds. We can take $d$ minimal such that $1,\alpha,\alpha^2,\ldots,\alpha^d$ are linearly dependent over$~F$. The linear dependency involved must involve $\alpha^d$, and by scaling can be made to do so with coefficient$~1$, which gives a degree$~d$ monic polynomial$~Q$ with $\alpha$ as root. By minimality every polynomial in $F[X]$ with $\alpha$ as root, in particular$~P$, must be divisible by$~Q$ (as a nonzero quotient in the Euclidean division would have too small degree to have $\alpha$ as root). But $P$ is irreducible, so $Q=P$ and $d=n$. In particular $1,\alpha,\alpha^2,\ldots,\alpha^{n-1}$ are $F$-linearly independent, and $\alpha^n$ can be expressed uniquely as $F$-linear combination $\alpha^n=-p_0-p_1\alpha-\cdots-p_{n-1}\alpha^{n-1}$.

