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I have been studying projection from Strang's linear algebra book. The projection part is here.

The projection $p$ onto a line is given below.

enter image description here

The line from $b$ to $p$ is perpendicular to the vector $a$. This is the dotted line marked $e=b-p$ for the error of the above figure.

I have two questions:

(1) how this error($e=b-p$) is calculated?

(2) and why this error $e$ belong to orthogonal complement of Subspace ,$S^\perp$ (when we are talking about projection onto any subspace $S$) ?

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The vector $b$'s projection on the subspace $S$ spanned by vector $a$, is given by

$ p = Proj_a(b) = P \ b $

where

$ P = \dfrac{ a a^T}{ a^T a } $

Hence, the error is

$ e = b - p = \left( I - \dfrac{{aa}^T}{a^T a} \right) b $

Now the orthogonal complement of $S$ is the set of vectors $x$ such that

$ a^T x = 0 $

To check if $e$ belongs to this orthogonal complement, substitute $e$ for $x$:

$ a^T \left( I - \dfrac{{aa}^T}{a^T a} \right ) b = \left( a^T - \dfrac{ (a^T a) a^T }{a^T a} \right) b = 0 $

So indeed $e \in S^\perp $.

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  • $\begingroup$ I don't understand why $e$ is equal to $b-p$? Is it any vector rule for $b-p$? I mean to ask why perpendicular length is $b-p$? $\endgroup$ Commented Apr 24, 2024 at 9:53
  • $\begingroup$ It's only natural to define it this way. The error is (by definition) the difference between the original value and the new value. That's just what it is. It is a definition. $\endgroup$ Commented Apr 24, 2024 at 10:50
  • $\begingroup$ I understand the error value(original value and the new value.). But still my question is why perpendicular length is $b-p$? $\endgroup$ Commented Apr 24, 2024 at 19:41

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