$V$ is a vector space over $\mathbb Q$ of dimension $3$ $V$ is a vector space over $\mathbb Q$ of dimension $3$, and $T: V \to V$ is linear with $Tx = y$, $Ty = z$, $Tz=(x+y)$ where $x$ is non-zero. Show that $x, y, z$ are linearly independent.

Let $A = \mathbb{Q}[X]$ be the polynomial ring.
Let $I = \{f(X) \in A|\ f(T)x = 0\}$.
Clearly $I$ is an ideal of $A$.
Let $g(X) = X^3 - X - 1$.
Then $g(X) \in I$.
Suppose $g(X)$ is not irreducible in $A$.
Then $g(X)$ has a linear factor of the form $X - a$, where $a = 1$ or $-1$.
But this is impposible.
Hence $g(X)$ is irreducible in $A$.
Since $x \neq 0$, $I \neq A$. Hence $I = (g(X))$.
Suppose there exist $a, b, c \in \mathbb{Q}$ such that $ax + bTx + cT^2x = 0$.
Then $a + bX + cX^2 \in I$.
Hence $a + bX + cX^2$ is divisible by $g(X)$.
Hence $a = b = c = 0$ as desired.

A more elementary version of the above proof
Suppose $x, y = Tx, z= Tx^2$ is not linearly independent over $\mathbb{Q}$.
Let $h(X)\in \mathbb{Q}[X]$ be the monic polynomial of the least degree such that $h(T)x = 0$.
Since $x \neq 0$, deg $h(X) = 1$, or $2$.
Let $g(X) = X^3 - X - 1$.
Then $g(X) = h(X)q(X) + r(X)$, where $q(X), r(X) \in \mathbb{Q}[X]$ and deg $r(X) <$ deg $h(X)$.
Then $g(T)x = q(T)h(T)x + r(T)x$.
Since $g(T)x = 0$ and $h(T)x = 0$, $r(T)x = 0$.
Hence $r(X) = 0$.
Hence $g(X)$ is divisible by $h(X)$.
But this is impossible because $g(X)$ is irreducible as shown above.
Hence $x, y, z$ must be linearly independent over $\mathbb{Q}$.
