# $V$ 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.

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Hint: You need to show that $x$, $y=Tx$ and $z=T^2x$ are linearly independent, which means that there does not exist any nonzero polynomial $P$ in $T$ of degree at most $2$ that when applied to $x$ gives $0$. But on the other hand the fact $T^3x=x+y$ gives you a polynomial $Q$ in $T$ of degree $3$ that gives $0$ when applied to $x$. Can you see why $P$ would have to divide $Q$? Now show that this could only happen if $P$ were constant, but since $x\neq0$ you know that $P$ cannot be a constant polynomial.

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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.
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 $\mathbb{Q}$.