Linear transformation and vectors on a certain type of 3 dimensional vector space over the rational number field Let $V$ be a $3$ dimensional vector space over $\mathbb Q$ and $T$ be a linear transform on $V$ such that for some $\vec x , \vec y , \vec z \in V$ with $\vec x \ne \vec 0$ , $T(\vec x)=\vec y , T(\vec y )=\vec z , T(\vec z)=\vec x + \vec y$ , then is it true that $\vec x , \vec y , \vec z$ are linearly independent  or equivalently saying that  $\vec x , \vec y , \vec z$  generates $V$ ( since $V$ is of dimension $3$  ) ?
The answer is YES. 

Let us show first that $x$ and $y$ are linearly independent. If not, $y$
must be a multiple of $x$, $y=qx$ for some $q\in{\mathbb Q}$. We deduce
$z=T(y)=T(qx)=qT(x)=q(qx)=q^2x$, and $T(z)=x+y$ becomes $q^3x=(1+q)x$
whence $q^3=(1+q)$. But the polynomial $X^3-(1+X)$ has no rational roots,
so this is impossible.

Next, we show that $x,y$ and $z$ are linearly independent. If not, $z$
must be a linear combination of $x$ and $y$, $z=ax+by$ with $a,b\in{\mathbb Q}$.
Then $T(z)=x+y$ becomes $(ab)x+(a+b^2)y=x+y$, so $ab=1$ and $a+b^2=1$. We deduce
$b(a+b^2)=b$, or $1+b^3=b$. But the polynomial $X-(1+X^3)$ has no rational roots,
so this is impossible.
The conditions on the existence of $ \vec x$, $\vec y$ and $\vec z$ tells a lot more than tells the eye since not any $T$ satisfies this identities. In fact the problem can be reduced to the exisitence of a transformation $T$ that has the property $\det(T^3-T-T^0)=0$. Indeed in this case there exists $\vec x$ such that $(T^3-T-T^0)(\vec x)=\vec 0$ and we have $T^3(\vec x)=T(\vec z)=T(\vec x)+\vec x=\vec y+\vec x$. That such transformations exist can be seen from the companionmatrix of the polynomial $x^3-x-1$:$$T=\left(\begin{array}{rrr}%
0&amp;0&amp;1\\%
1&amp;0&amp;1\\%
0&amp;1&amp;0\\%
\end{array}\right)%
\text{so that for arbitary } \vec x=\left(\begin{array}{r}%
a\\%
b\\%
c\\%
\end{array}\right) \text{ we have }% 
\vec y = \left(\begin{array}{c}%
c\\%
a+c\\%
b\\%
\end{array}\right)% 
\text{ and } \vec z = \left(\begin{array}{c}%
b\\%
b+c\\%
a+c\\%
\end{array}\right)%
\text{ with } T(\vec z)=\left(\begin{array}{c}%
a+c\\%
a+b+c\\%
b+c\\%
\end{array}\right)%
=\left(\begin{array}{r}%
a\\%
b\\%
c\\%
\end{array}\right) +  \left(\begin{array}{c}%
c\\%
a+c\\%
b\\%
\end{array}\right)% 
$$ and $\vec x, \vec y, \vec z$ linearly indepentent for $\vec x\neq \vec 0$. It is clear that any transformation similar to $T$ also meet this requirements. Remains the question if there are transformations $T$ such that $T^3-T-T^0$ has rank $&lt;0$ instead of rank $0$ as it is the case here. Whith your permission I would open this here as a new question.