Roots within the unit circle and non-stationarity

I am quite new to time series analysis and I am delving for the first time into stationary processes. I don't seem to understand the concepts of non-stationarity and the presence of roots within the unit circle. Particularly, these are my questions:

1. Where does the characteristic equation come from (i.e. what is its meaning) and why is it important? (I know the way it is built, I just do not understand why it is built that way)

2. Why can we say that a process is stationary if and only if the roots of the characteristic equation are outside of the unit circle?

And, CE obtained by the backshift operator is a way of analyses. This is a matter of which transform you use. You could get the CE of $$y_t+\alpha y_{t-1}+\beta y_{t-2}=\epsilon_t$$ as $$1+\alpha B+\beta B^2$$ or if you use $$\mathcal{Z}$$ transform, that'd be $$1+\alpha z^{-1}+\beta z^{-2}=0$$, which requires your roots, i.e. $$z$$'s, should be inside the unit circle if you want to be stationary, instead of outside as in $$B$$, since roots obtained by $$B$$ and $$z$$ are reciprocals of each other.
2. If we go back to what you're accustomed to, i.e. using the back-shift operator, $$B$$, we can try to find the roots' relation to stationarity. Consider the time series $$y_t=2y_{t-1}+\epsilon_t$$, we use the previous output, double it, and add it up with $$\epsilon_t$$ to obtain current output. The process is clearly exploding. The CE of this is $$1-2B=0$$, which yields $$|B|=1/2<1$$. But, if it were $$y_t=0.5y_{t-1}+\epsilon_t$$, the root would be $$|B|=2>1$$. This is for just an intuition why would you need $$|B|>1$$ for being stationary.
Now consider the general case, $$y_t=\alpha y_{t-1}+\epsilon_t$$. We can obtain the following equation by back-substitution: i.e. $$y_t=\alpha^2y_{t-2}+\alpha\epsilon_{t-1}+\epsilon_t$$, and so... $$y_t=\sum_{i=0}^\infty{\alpha^i\epsilon_{t-i}}$$ mean and variance of $$y_t$$ would be $$E[y_t]=\sum_{i=0}^{\infty}{\alpha^iE[\epsilon_{t-i}]}=\mu_\epsilon\sum_{i=0}^\infty\alpha^i=\frac{\mu_\epsilon}{1-\alpha} \ \ \text{iff} \ \ |\alpha|<1 \rightarrow |B|>1$$ since CE is $$1-B\alpha=0\rightarrow B=1/\alpha$$. Similarly for the variance, we have $$\sigma_Y^2=var(y_t)=\sum_{i=0}^{\infty}\alpha^{2i}\sigma_{\epsilon}^2=\frac{\sigma_{\epsilon}^2}{1-\alpha^2} \ \ \text{iff} \ \ |\alpha^2|<1 \rightarrow |\alpha=1/B|<1$$
The auto-correlation is $$r_Y(k)=E[y_t y_{t-k}]:$$ $$E[y_t y_{t-1}]=E[(\alpha y_{t-1}+\epsilon_t)y_{t-1}]=\alpha\sigma_Y^2+\mu_\epsilon\mu_Y$$ $$E[y_t y_{t-2}]=E[(\alpha y_{t-1}+\epsilon_t)y_{t-2}]=\alpha r_Y(1)+\mu_\epsilon\mu_Y=\alpha^2\sigma_Y^2+\mu_\epsilon\mu_Y(1+\alpha)$$ ... $$E[y_t y_{t-k}]=\alpha^k\sigma_Y^2+\mu_\epsilon\mu_Y(1+\alpha+...+\alpha^{k-1})$$ which is irrespective of $$t$$, and conditioned on $$\sigma_Y^2$$ and the means don't depend on $$t$$.
This is for AR(1), but AR(k) can be reduced down to a series of AR(1)'s: $$(1-\alpha B)(1-\beta B)y_t=\epsilon_t \rightarrow x_t = (1-\beta B)y_t=x_t \ \ \ \& \ \ (1-\alpha B)x_t=\epsilon_t$$ and the analysis can be performed recursively. Here, we're actually referring to weak stationarity but if $$\epsilon_t$$ is Gaussian as usual (i.e. which is the noise process is assumed to be distributed with typically), weak stationarity corresponds to stationarity.