Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$V$ 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.

share|improve this question

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer

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 $\mathbb{Q}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.