Given a single stationary series $x_t$ with zero mean and autocovariance function $\gamma(h)$ and autocorrelation function $\rho(h)$
Show that if $x_{t+l}=Ax_t$, then $ \rho(l)=1$ if $A>0$, and $\rho(l)=−1$ if $A<0$.
What I have done is
$$ \text{By definition we know that}\quad \rho(l) = \frac{\gamma(l)}{\gamma(0)}$$
So for $x_{t+l}=Ax_t$ we have
$$\gamma(l) = cov(x_t, x_{t+l}) = cov(x_t, Ax_t) = A\gamma(0)$$
and it follows that
$$\rho(l) = A\frac{\gamma(0)}{\gamma(0)} = A$$
which is not what I want. Where did I go wrong?
EDIT: Can I say that because of stationarity that
$$\frac{x_{t+l}}{x_t} = \pm 1$$
EDIT 2: How about this?
Because of stationarity
$$Var(x_t) = Var(x_{t+l}) = Var(Ax_t) = A^2Var(x_t)$$
which implies that $A^2 = 1$, and thus
$$\rho(l) = \sqrt{1} = \pm 1$$
$$ \text{By definition we know that}\quad \rho(l) = \frac{\gamma(l)}{\gamma(0)}$$
So for $x_{t+l}=Ax_t$ we have
$$\gamma(l) = cov(x_t, x_{t+l}) = cov(x_t, Ax_t) = A\gamma(0)$$
and it follows that
$$\rho(l) = A\frac{\gamma(0)}{\gamma(0)} = A$$
Because of stationarity
$$Var(x_t) = Var(x_{t+l}) = Var(Ax_t) = A^2Var(x_t)$$
which implies that $A^2 = 1$, and thus
$$\rho(l) = \sqrt{1} = \pm 1$$