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Right now, we are learning decomposing vectors, but something I don't understand is the names given to this stuff

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For instance, in the text, the parallel component of y is said to be the orthogonal projection of y onto u. This makes no sense to me. Why is the word "orthogonal" even in there in the first place? I think I understand why they use "orthogonal" for z, but it makes no sense to me when they could just call it "orthogonal"

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Consider $$\begin{pmatrix} 1&1\\ 0&0\end{pmatrix}$$ it projects all vectors to the $x$-axis. But it doesn't project vectors onto the $x$-axis orthogonally. The vector $(0,1)^T$ is not sent to its orthogonal projection $(0,0)^T$ but is sent to $(1,0)^T$. Vectors are "diagonally" projected onto the $x$-axis.

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As Bill's answer explains, we can decompose every vector in the original space by using the projection map. This lends to an intuitive geometric interpretation of orthogonal projections.

If $p:V\to W$ is an $\color{Green}{orthogonal}$ projection down to a subspace, the fibers (pre-images) of every point $w\in W$ is perpendicular to the base (the subspace $W$). With $\dim V=2$ and $\dim W=1$:

$\hskip 0.4in$ orthogonal vs not orthogonal

The base (black line) and fibers (gray lines) can in general be viewed as higher-dimensional planes.

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There are many ways to "project" onto a subspace. This particular projection yields an orthogonal decomposition.

Given vectors $v$ and $w$ ($w \not=0$), $a = \mathrm{proj}_w(v) = \frac{v \;\cdot\; w}{|w|^2}w$ and $b = v - a = v - \mathrm{proj}_w(v)$. We have that

  • $v=a+b$
  • $a$ is parallel with $w$
  • $a$ and $b$ are orthogonal (just check that $a \;\cdot\; b=0$).

There are other "projections" for which $v=a+b$ and $a$ is parallel with $w$. For example: Let $a=w$ and $b=v-w$. Then $v=w+(v-w)=a+b$ and $a=w$ is parallel with itself (i.e $w$). But in general $v-w$ and $w$ are not orthogonal.

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I found your last paragraph very confusing! Perhaps you should say, "For example: Let $a = w$ and $b = v-w$," instead of stating a tautological equation. – Rahul Feb 22 '12 at 23:17
@Rahul: I found it a rather conventional abridgement of "Let $a,b=\dots$" – anon Feb 22 '12 at 23:25
For Rahul, some extra words. :) – Bill Cook Feb 22 '12 at 23:52

We can actually project with respect to any vector in a basis, as follows: If $v_1,\ldots, v_n$ is a basis for your vector space, then the projection of vector $a_1 v_1 + \ldots + a_n v_n$ onto $a_n$ with respect to this basis is $a_n v_n$. Note that the projection depends on the choice of basis.

If you were to take an orthogonal basis containing $u$, then the definition of orthogonal projection would coincide with the above, more general, definition of projection.

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Because line along which your orginal point travel to projective plane is orthogonal to projective plane. In this context synonim for orthogonal is perpendicular which I think is more common and well undertstood word.

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I took linear algebra with this same textbook and went through the same dilemma. I decided that the word "orthogonal" in orthogonal projection is referring to the way some vector v is being projected onto a subspace W. If you imagine the projection occurring as a drawn out process, with the point represented by v moving directly toward W on the shortest path, then the motion of the point is orthogonal to W. This would be in contrast with a "non-orthogonal," or "diagonal" projection, in which the projection of the point is not orthogonal to W. Hope this helps—it worked for me!

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