Prove that $x,y,z$ are linearly independent. 
let T be a linear operator on the three-dimensional vector space $E$ over the field $\mathbb{Q}$. If $T(x)=y$, $T(y)=z$, and $T(z)=x+y$. Then show that $x, y, z$ are linearly independent where $x \ne 0$

My attempt: I have determined the rank of the subspace spanned by $T(x)$, $T(y)$, $T(z)$ then I have shown that the rank of the matrix formed is $3$, as the subspace formed by $x, y, z$ under L.T. is linearly independent so are the vectors $x, y, z$.
 A: Assume that $\alpha, \beta, \gamma \in \mathbb{Q}$ are scalars, not all equal to $0$, such that $\alpha x + \beta y + \gamma z = 0$. By multiplying with a common denominator we can assume that $\alpha, \beta, \gamma \in \mathbb{Z}$. Furthermore, by dividing with any common factors we can assume that $\gcd(\alpha, \beta, \gamma)=1$.
By applying $T$ and $T^2$ on $\alpha x + \beta y + \gamma z = 0$ we get the relations
$$\begin{cases} \alpha x + \beta y + \gamma z = 0\\ \gamma x + (\alpha + \gamma) y + \beta z = 0 \\ \beta x + (\beta + \gamma) y + (\alpha+\gamma)z = 0\end{cases}.$$
Define a linear map $A : E^3 \to E^3$ with
$$A\begin{bmatrix} u \\ v \\ w\end{bmatrix} := \begin{bmatrix} \alpha & \beta & \gamma \\ \gamma & \alpha+\gamma & \beta \\ \beta & \beta+\gamma & \alpha+\gamma\end{bmatrix}\begin{bmatrix} u \\ v \\ w\end{bmatrix}.$$
By definition we have
$$A\begin{bmatrix} x \\ y \\ z\end{bmatrix} = 0 \quad \text{ and } \quad\begin{bmatrix} x \\ y \\ z\end{bmatrix} \ne 0$$ since $x \ne 0$. However, the matrix $M$ in the definition of $A$ is invertible. Namely, it has determinant $$\det M = \alpha^3 - \alpha \beta^2 + \beta^3 + 2 \alpha^2 \gamma - 3 \alpha \beta \gamma + \alpha \gamma^2 - \beta \gamma^2 + \gamma^3$$
and this expression is nonzero. This can be seen by checking that $\det M \equiv 0 \pmod{2}$ if and only if $\alpha \equiv \beta \equiv \gamma \equiv 0 \pmod{2}$. The latter is impossible since we assumed $\gcd(\alpha,\beta,\gamma)=1$. There are only $8$ cases to check, or you can use Wolfram Mathematica:
Table[Mod[a^3 - a b^2 + b^3 + 2 a^2 c - 3 a b c + a c^2 - b c^2 + c^3, 2], {a, 0, 1}, {b, 0, 1}, {c, 0, 1}].
Therefore $A$ is invertible with inverse $A^{-1}\begin{bmatrix} u \\ v \\ w\end{bmatrix} := M^{-1}\begin{bmatrix} u \\ v \\ w\end{bmatrix}$ which is a contradiction since we got that $A$ is not injective.
We conclude that $x,y,z$ are linearly independent over $\mathbb{Q}$.
