I am trying to understand a piece of text from a Time Series textbook (Brickel's Time Series : Theory and Methods):
Let $H$ denotes a Hilbert space, $P$ denotes the projection operator and $\bar{sp}$ denotes closed span.
Suppose $H_1 = \bar{sp}\{X_2,...,X_n\}$ and $H_2 = \bar{sp}\{X_1 - P_{H_1}X_1\}$ are orthogonal subspaces of $H_n = \bar{sp}\{X_1,...,X_n\}$. Moreover it is easy to see that for any $Y \in L^2(\Omega, F,\mathcal{P})$, $P_{H_n} Y = P_{H_1}Y + P_{H_2}Y$. Hence
$$P_{H_1}X_{n+1} + P_{H_2}X_{n+1} = P_{H_1}X_{n+1} + a(X_1 - P_{H_1}X_1)\;\;\;\;\;(1)$$,
where $$a = \frac{<X_{n+1},X_1 - P_{H_1}X_1>}{||X_1 - P_{H_1}X_1||^2}$$
My question is how does the LHS of equation 1 go to RHS of equation 1?
Equation (1) is true because $$ P_{H_2}X_{n+1} = a(X_1 - P_{H_1}X_1) $$ with the coefficient $a$ from the question. This is a general fact about orthogonal projections. To understand it, it's very helpful to be familiar with them. :-)
In the particular situation from the question, we can argue as follows. Let's put $$ Z := X_1 - P_{H_1}X_1. $$ Then $H_2$ is spanned by $Z$ (no closure required since finite-dimensional subspaces are automatically closed - this fact is not entirely obvious, but I'll just gloss over this now :-).
If $Z = 0,$ then $H_2 = \{0\}$ and $P_{H_2} = 0,$ and from the formula for $a$ from the question, we get $a = \frac00,$ which is not really well defined. Since $Z = 0,$ we can take $a = 0,$ but this case isn't really interesting anyway. So let's assume $$ Z \neq 0. $$ Then, $$ P_{H_2}X_{n+1} = \alpha Z \in H_2 $$ where $\alpha$ is to be found. We're going to show $\alpha = a.$
Now, one way to state the orthogonality of the projection $P_{H_2}$ is to say that $X_{n+1} - P_{H_2}X_{n+1}$ is orthogonal to $H_2,$ or equivalently (since $H_2$ is spanned by $Z$) $X_{n+1} - P_{H_2}X_{n+1}$ is orthogonal to $Z.$ In symbols, $$ \langle X_{n+1} - P_{H_2}X_{n+1} , Z \rangle = 0. $$ If that's not "geometrically intuitive", consult the Wikipedia page about orthogonal projections cited above. Plugging in our observation from above, we get $$ 0 = \langle X_{n+1} - \alpha Z , Z\rangle = \langle X_{n+1},Z \rangle - \alpha\langle Z,Z\rangle. $$ Solving for $\alpha,$ we get $$ \alpha = \frac{\langle X_{n+1},Z \rangle}{\langle Z,Z\rangle} = \frac{\langle X_{n+1},Z \rangle}{\|Z\|^2} = a, $$ as desired.