Show that if $x_{t+l}=Ax_t$, then $ \rho(l)=1$ if $A>0$, and $\rho(l)=−1$ if $A<0$.

Question

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

$$\rho(l) = \sqrt{1} = \pm 1$$

Answer 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

$$\rho(l) = \sqrt{1} = \pm 1$$