Definition of Orthogonal Projector onto a subspace $S$ unclear.

Question

I am self studying Matrix Computation and just want to clear a seemingly easy conceptual definition.

Firstly, my notes did not define what an orthogonal projector $P$ onto a subspace $S$ means (Internet mostly say what an orthogonal projection is), after reading through numerous times, my understanding is as follows

$P$ is an orthogonal projector onto $S$ if $P$ is the unique projector such that $\text{Range}(P) = S$, and that $\text{Range}(P)+\text{Null}(P) = \mathbb{R}^n$, $\text{Range}(P) \cap \text{Null}(P) = \{0\}$ and in particular, we must have $$\text{Range}(P) = \text{Null}(P)^{\perp}$$

So my question is: Is my definition of $P$ being a orthogonal projector onto a subspace $S$ correctly phrased?

Note: Sometimes the definitions in textbooks may not be clear to a novice student like me and hence I am asking something fundamental to make sure I did not interpret the definition wrong.

Answer 1

Yes. Easier, if you construct an orthonormal basis $e_1,\ldots,e_n$, where $e_1,\ldots,e_r$ is an orthonormal basis of $S$ and $e_{r+1},\ldots,e_n$ is an orthonormal basis of $S^\perp$, then $$ P\left(\sum_{j=1}^nc_je_j\right)=\sum_{j=1}^rc_je_j. $$ Or, equivalently, $$ Px=\sum_{j=1}^r\langle x,e_j\rangle\,e_j. $$

Answer 2

The fact that you used the $\perp$ in your answer suggests that you are talking about orthogonal projections. These are linear maps $P\colon\mathbb{R}^n\longrightarrow S$ such that, for each $v\in\mathbb{R}^n$, $v-P(v)$ is orthogonal to each element of $S$.

The general concept of projection needs a complementary space $T$ of $S$, that is, a subspace $T$ of $\mathbb{R}^n$ such that $S\cap T=\{0\}$ and that $S+T=\mathbb{R}^n$. Then each $v\in\mathbb{R}^n$ can be written as $s+t$, with $s\in S$ and $t\in T$, in one and only one way. Then we define $P(v)=s$. That's the projection of $\mathbb{R}^n$ onto $S$ along $T$.