I am reading in chapter one of Yousef Saad's "Iterative Methods for Sparse Linear Systems" about projectors (Projection Operators) and direct sums. (section 1.12 here: https://www-users.cse.umn.edu/~saad/IterMethBook_2ndEd.pdf)
There is a transition (a "Therefore," in fact) in Saad's development that is stumping me. First I will summarize the points that I do understand.
A projector is a linear, idempotent map, $\mathbb{C^n}$ into $\mathbb{C^n}$, some $n \geqslant 1.$
$I-P$ is a projector when $P$ is.
$\ker P = \text{ran } (I-P),$ and $\ker (I-P) = \text{ran } P.$
$\mathbb{C^n}=\ker P \oplus \text{ran } P.$
Those are properties of any given projector $P$.
Next we take a different set of givens. We take any two subspaces $M,S$ and assume $\mathbb{C^n}=M \oplus S.$ I won't give the details here, but we can construct the obvious projector $P$ onto $M$, via the unique decomposition afforded by the direct sum. And, we can show that $P$ is the only projector that satisfies $M=\text{ran } P$ and $S=\ker P.$ Now I quote Saad:
For any $x$, the vector $Px$ satisfies the conditions $$Px \in M,$$ $$x-Px \in S.$$ The linear mapping $P$ is said to project $x$ onto $M$ and along or parallel to the subspace $S$.
That passage is great, no problem. He continues:
If $P$ is of rank $m$, then the range of $I-P$ is of dimension $n-m$.
Also great; this is just rank-nullity, I believe.
Therefore, it is natural to define $S$ through its orthogonal complement $L=S^\perp$ which has dimension $m$. The above conditions that define $u=Px$ for any $x$ now become $$u \in M,$$ $$x-u \perp L.$$
I cannot understand how this last block follows from the preceding discussion. I do not even understand why Saad is talking about defining $S$, for I thought it was fixed, given above.
I know orthogonal projectors are important, but I feel I am missing something important on this page of the book. Why is it natural to introduce $L$ at this point? What is the relevance of $\text{rank } P = m = \dim L$?
What Saad is saying makes sense but it seems like an unnecessarily convoluted way to set things up in my opinion.
The "standard" view is to define a projector in terms of $M=\operatorname{ran}(P)$ and $S=\operatorname{ker}(P)$. Saad's view is defining a projector in terms of $M=\operatorname{ran}(P)$ and $L=(\operatorname{ran}(I-P))^\perp$. The "standard" view amounts to asking "what is the surface that is parallel to all paths from $x$ to $Px$?". Saad's view amounts to asking "what is the surface that is perpendicular to all paths from $x$ to $Px$?" See Figure 1.1.
Now let's massage that definition of $L$ a bit. The fundamental subspaces theorem tells you $(\operatorname{ran}(I-P))^\perp=\operatorname{ker}((I-P)^*)$. The identity is self-adjoint so this is $\operatorname{ker}(I-P^*)$. Now $P^*$ is also idempotent, so this is $\operatorname{ran}(P^*)$.
Thus we are defining a projector in terms of its range and the range of its adjoint. Saad's view says that you get an orthogonal projector when you set these equal to each other. This ultimately amounts to choosing $S=M^\perp$; in this regard $M$ is given and $S$ is not. If $M$ and $S$ are both given then you have no freedom in selecting the projector at all.