Projector

Let $$\mathcal X$$ and $$\mathcal Y$$ be complementary subspaces of a vector space $$V$$ so that each $$\mathbf v \in \mathcal V$$ can be uniquely resolved as $$\mathbf{v = x+y}$$, where $$\mathbf x \in \mathcal X$$ and $$\mathcal Y$$. The unique linear operator $$\mathbf P$$ is called the projector onto $$\mathcal X$$ along $$\mathcal Y$$, and $$\mathbf P$$ has the following properties (p386 in ):


 * $$\mathbf P^2 = \mathbf P$$
 * $$\mathbf{I-P}$$ is the complementary projector onto $$\mathcal Y$$ along $$\mathcal X$$.
 * $$R(\mathbf P) = \{\mathbf x : \mathbf{Px = x}\}$$   <--  this is a property used frequently in proofs
 * $$R(\mathbf P) = N (\mathbf{I-P}) = \mathcal X$$ and $$R(\mathbf{I-P})=N(\mathbf P) = \mathcal Y$$
 * If $$\mathcal V = \mathbb R^n$$ or $$\mathbb C^n$$, then $$\mathbf P$$ is given by:

$$\mathbf P = [\mathbf X | \mathbf 0] [\mathbf X | \mathbf Y]^{-1} = [\mathbf X | \mathbf Y] \begin{pmatrix} \mathbf I & \mathbf 0 \\ \mathbf 0 & \mathbf 0 \end{pmatrix} [\mathbf X | \mathbf Y]^{-1}$$
 * (The formula above does not depend on basis choice $$\mathbf X$$ or $$\mathbf Y$$).

Note that the projector definition in this page is not necessarily an Orthogonal Projector; it is more general, and sometimes the projector here is referred to an Oblique Projector.

A concise characterization
The following is a simple and powerful characterization of a projector (p387 of )

A linear operator $$\mathbf P$$ on a vector space $$\mathcal V$$ is a projector iff $$\mathbf P^2 = \mathbf P$$

Interesting facts
For any projector $$\mathbf P$$
 * $$\text{rank} (\mathbf P) = \text{trace}(\mathbf P)$$ (see Exercise 5.9.13 )
 * $$||\mathbf P||_2 \geq 1$$ for every non-zero projector, and $$||\mathbf P||_2 = 1$$ iff $$R(\mathbf P) \perp N(\mathbf P)$$ (see Exercise 5.9.8 ; see also Matrix norm)
 * $$\mathbf P$$ is similar to the matrix $$\begin{pmatrix} \mathbf I_r & \mathbf 0 \\ \mathbf 0 & \mathbf 0 \end{pmatrix}$$ where $$r$$ is the rank of $$\mathbf P$$. (See formula above with $$\mathbf X$$ and $$\mathbf Y$$ to see why this holds).
 * The weighted average of a set of projectors is a projector (given that the sum of weights is 1; see Exercise 5.9.12)