Why is this called the orthogonal projection of $u$ on $W$ if $proj_Wu$ is not orthogonal to $u$?

Question

Why is this called the orthogonal projection of $\mathbf u$ on $W$ if $proj_W \mathbf u$ is not orthogonal to $\mathbf u$?

enter image description here

Think of $u$ falling over $W$ only by the action of gravity. So the fall is orthogonal to $W$. — Sigur, May 16 '14 at 19:18
Possible duplicate of Why is it called “Orthogonal Projection”? Why not just “Projection”? from the sidebar. The answers there would be helpful for answering this question, I think. — epimorphic, Feb 19 '15 at 04:54

score 1 · Answer 1 · answered May 16 '14 at 19:14

1

As written in the text, $u = w_1 + w_2$ where $w_1 \in W$ and $w_2$ is orthogonal to $W$.

It is called an "orthogonal" projection because the difference $w_2 = u-w_1$ between $u$ and its projection is itself orthogonal to $W$.

(That it deserves to be called a projection is because $\mathrm{proj}_W(u) \in W$, and $\mathrm{proj}_W(\mathrm{proj}_W(u)) = \mathrm{proj}_W(u)$.)

answered May 16 '14 at 19:14

Thomas Belulovich

3,891

I am not fully comprehending. It says "orthogonal projection of u on $W$, but u is not orthogonal to $W$. – user5826 May 16 '14 at 19:20
$w_1$ is called an orthogonal projection of $u$ because $w_1$ differs from $u$ by a vector $w_2 = u-w_1$ that is orthogonal to $W$. – Thomas Belulovich May 16 '14 at 19:22

score 0 · Answer 2 · edited May 16 '14 at 19:47

0

Simply because $(\operatorname{proj}_W \mathbf u - \mathbf u)$ is orthogonal to W, i.e. its projection on $W$ is a zero-length vector ("orthogonal" to it). So if you make an analogy with 3D space, to make this projection you drop a perpendicular on the "plane" $W$.

edited May 16 '14 at 19:47

Shaun

44,997

answered May 16 '14 at 19:18

Shady_arc

291

Why is this called the orthogonal projection of $u$ on $W$ if $proj_Wu$ is not orthogonal to $u$?

2 Answers2