What is projection? — Mathematics

Definition 20.14 University Mathematics — Year 1 · Chapter 20 — Linear Maps

Let $E = F \oplus G$ . The projection onto $F$ along $G$ maps $x = f + g$ (unique decomposition) to $p(x) = f$ ; the associated symmetry is $s(x) = f - g$ . Both are linear, and $s = 2p - \mathrm{id}$ .

The projection onto F = Vect(1,1) along G = Vect(0,1) and its symmetry, on the point M = (2,\ 0.5): sliding vertically, M hits F at p(M) = (2,2) and lands at s(M) = 2p(M) - M = (2,\ 3.5), as far above F (measured along G) as M was below. — The projection onto $F = \operatorname{Vect}(1,1)$ along $G = \operatorname{Vect}(0,1)$ and its symmetry, on the point $M = (2,\ 0.5)$ : sliding vertically, $M$ hits $F$ at $p(M) = (2,2)$ and lands at $s(M) = 2p(M) - M = (2,\ 3.5)$ , as far above $F$ (measured along $G$ ) as $M$ was below.

Examples

Example 20.16 (A projection and its symmetry, explicitly)

In $\R^2$ , project onto $F = \operatorname{Vect}(1,1)$ along $G = \operatorname{Vect}(0,1)$ . Decompose $(x, y) = a(1,1) + b(0,1)$ : the first coordinate gives $a = x$ , the second $b = y - x$ . Hence

p(x, y) = (x, x), \qquad s(x, y) = 2p(x,y) - (x,y) = (x,\ 2x - y).

Check the algebra: $p(p(x,y)) = p(x,x) = (x,x)$ , and $s(s(x,y)) = s(x, 2x - y) = (x, 2x - (2x - y)) = (x, y)$ . Geometrically, $s$ is the “oblique reflection” across the line $y = x$ in the vertical direction: it fixes $F$ pointwise and reverses $G$ . Had we projected onto the same $F$ along $G' = \operatorname{Vect}(1,-1)$ instead, the formula would change to $p'(x,y) = \bigl(\frac{x+y}2, \frac{x+y}2\bigr)$ : a projection is determined by its image and its kernel, never by the image alone.

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