On a basis of three dimensional vector space over the field of rational numbers Let $V$ be a vector space over $\mathbb Q$ of dimension $3$. Let $u,v,w\in V$ be vectors such that $u\ne 0$ and there exists a linear map $T: V\to V$ such that $T(u)=v, T(v)=w$ and $T(w)=u+v$. 
Then how to show that $\{u,v,w\} $ is a basis of $V$ ? 
My try: It is enough to show that $\{u,v,w\} $ is linearly independent. So let $au+bv+cw=0$ for some $a,b,c\in \mathbb Q$. Then applying $T$ we get $cu+ (a+c)v+ bw=0$ and applying $T$ one more time, one gets $bu+(b+c)v+(a+c)w=0$. Writing out these equations in matrix form the corresponding determinant of the coefficient $a,b,c$ matrix is very cumbersome
https://www.wolframalpha.com/input/?i=determinant+of+%7B%7Ba%2Cb%2Cc%7D%2C%7Bc%2C+a%2Bc%2C+b%7D%2C%7Bb%2C+b%2Bc%2C+a%2Bc%7D%7D
So I'm not sure what to do next. I think I have to apply that the base field is $\mathbb Q$ somehow, but I don't exactly know how. 
Please help. 
 A: You are correct about your guess that you need to use that $\mathbb Q$ is the underlying field. My solution pretty much relies on the Rational Root Theorem (RRT).

Claim 1. $\{u, v\}$ is a linearly independent subset of $V$.
Proof. Suppose not. As $u \neq 0$, we can conclude that $v = \lambda u$ for some $\lambda \in \mathbb{Q}$. (How?)
Now, as $T$ is linear, we know that $T(v) = T(\lambda u) = \lambda T(u) = \lambda v = \lambda^2 u.$
On the other hand, we know that $T(v) = w$.
Thus, we get that $w = \lambda^2 u$.
Now substitute this in $T(w) = v + u.$
$\implies T(\lambda^2 u) = \lambda u + u$
$\implies \lambda^3 u = \lambda u + u$
$\implies (\lambda^3 - \lambda - 1) u = 0$
$\implies \lambda^3 - \lambda - 1 = 0 \qquad (\because u \neq 0)$
This is a contradiction as $\lambda = \pm 1$ is not a root and by RRT, there are no other (rational) roots.

Claim 2. $\{u, v, w\}$ is a linearly independent subset of $V$.
Proof. Suppose not. As we have already shown that $\{u, v\}$ is linearly independent, our assumption gives us that $w = a_1u + a_2v$ for some $a_1, a_2 \in \mathbb{Q}$. (How?)  
We are given that $T(w) = u + v$. Using linearity, we perform the following calculations:
$T(a_1u + a_2v) = u + v$
$\implies a_1T(u) + a_2T(v) = u + v$
$\implies a_1v + a_2(a_1u + a_2v) = u + v$
$\implies (a_1 + a_2^2 - 1)v + (a_1a_2 - 1)u = 0$
By Claim 1, it follows that $a_1 + a_2^2 = 1$ and $a_1a_2 = 1$. Using the second equation, we plug $a_1 = a_2^{-1}$ in the first equation to get:
$$1 - a_2 + a_2^3 = 0.$$
(Note: It follows from $a_1a_2 = 1$ that $a_2 \neq 0$ and thus, the substitutions are legal.)  
However, once again, by RRT, it follows that the above equation has no (rational) root and hence, $\{u, v, w\}$ is a linearly independent subset of $V$ which tells us that it is a basis.

Remark. To show that it is indeed necessary to use the properties of $\mathbb{Q}$, consider any field $\mathbb{F}$ such that $\lambda^3-\lambda-1=0$ does have a root in $\mathbb{F}$. (For example, $\mathbb{F} = \mathbb{R}$.)
Consider $V = \mathbb{F}^3.$
Let $u = [1\;0\;0]^T,\; v = \lambda_0u,\; w = \lambda_0^2u$, where $\lambda_0$ is any root of $\lambda^3-\lambda-1=0$.  
Define $T:V \to V$ as $T(x) = \lambda_0x$.
Verify that these objects do indeed satisfy all your conditions.

Both the "(How?)"s have a similar argument. Use the fact that if $S$ is a linearly independent subset of $V$ and $v \in V$, then $S \cup \{v\}$ is linearly dependent iff $v \in \operatorname{span} S$.
