Let V be 3-dimensional vector space over $\mathbb{Q}$. Let $f\in End(V)$. Let $x,y,z\in V$ have the following attributes: $x\not= 0, f(x)=y, f(y)=z$ and $f(z)=x+y$. Prove that the vector system x,y,z is linearly independent.
My idea was to first show that if x is not 0, then y, z cannot be zero either (correct me if this is not true). Then I supposed that x,y are linearly dependent: $x=ky$. Then $f(x)=f(ky)\Rightarrow y=k\cdot f(y)\Rightarrow y=kz.$ Then apply $f$ one more time and get: $f(y)=f(kz)\Rightarrow z=k\cdot f(z)\Rightarrow z=k(x+y)\Rightarrow kx+ky-z=0.$ Multiply the last equation by $z$: $z(\frac{kx+ky}{z}-1)=0$. Since $z\not=0$ then $\frac{kx+ky}{z}-1=0$. Can I show that this does not have any possible solutions over $\mathbb{Q}$? And so $x,y$ must be linearly independent?
Then I supposed that $z=kx+my$.From this I get $f(z)=f(kx+my)\Rightarrow x+y = ky+mz\Rightarrow x+y-ky-mz=0$. Can (and if yes how) I show that this also doesn't have any possible solutions in $\mathbb{Q}$.
Am I on the right track here or should I prove this some other way. Any help is appreciated.
Suppose that $ y = m x$ for $m \ne 0$. Since $f(x) = y$ we then have that $f(x) = m x$. Since $f(y) = z$ this gives us $m f(x) = z$. Apply f to both sides and recall that $f(z) = x+ y$ to get $$ m f(f(x)) = x+y$$ Thus $$ m f(y) = x+m x$$ So we get $$m f(m x) = x + m x$$ This gives us $$m^2 y = x + m x$$ which is $$m^3 x = (1+m) x$$ The only way this can happen is if $m$ is not a rational number. Thus $x$ and $y$ are linearly independent.
Now suppose that $z = a x + b y$ with $a$ and $b$ rational. Then $$x+y = f(z) = a f(x) + bf(y)$$ $$x + y = a y + b z$$ $$x + y = a y + b a x + b^2 y$$ $$(1 - b a) x + (1 - a - b^2) y =0$$ Since $x$ and $y$ are linearly independent then $1-b a = 0$ and $1 - a - b^2 = 0$. There are no rational solutions this system of equations. Hence z is not in the span of x and y.