I want to find the integral curves of $V=(y-z)\partial_x+(z-x)\partial_y+(x-y)\partial_z$.
I have solved the system $$\dot{\gamma}=\begin{pmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{pmatrix}\gamma$$ The eigenvalues of the matrix are $0,\pm i\sqrt{3}$, so I obtained $$\gamma(t)=a+b\cos(\sqrt{3}t)+c\sin(\sqrt{3}t)$$ where $a,b,c\in\mathbb{C}^3$.
But we require the image of $\gamma$ to be in $\mathbb{R}^3$. So can I simply restrict $a,b,c$ to $\mathbb{R}^3$? Or is a different approach needed?
On
Given
$$ M =\left( \begin{array}{ccc} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \\ \end{array} \right) $$
the system
$$ \dot\gamma = M\cdot\gamma $$
is equivalent to
$$ \dot\gamma = \vec\omega\times\gamma $$
where $\vec\omega = -(1,1,1)$
then
$$ \vec\omega\cdot\dot\gamma =\vec \omega\cdot(\vec\omega\times\gamma) = 0 $$
and after integration follows
$$ \vec\omega\cdot\gamma = C_0\Rightarrow \vec\omega\cdot(\gamma-\gamma_0) = 0 $$
which means that $\gamma$ evolves onto a fixed plane with normal $\vec\omega$.
By orthogonality of sine and cosine, there is no linear addition of 1, sine and cosine that would cancel out the imaginary part. Therefore, since gamma is a pure real, then a, b and c are pure reals.