Hello I am trying to understand the proof in my notes and I don't get it. I am looking for someone to show me the proof or if possible tell me the name of this theorem so I can look it up. This is not homework as I believe the proof is supposed to be commonly known , but I can't seem to find the correct name.

I have it written as the "Orthogonal Lemma"

and it states;

Let $\{w_1,\ldots,w_r\}$ be an orthogonal set in an vector space $V$. Given any $v \in V$ , write $$w_{r+1}=\frac{\langle v,w_1\rangle}{\langle w_1,w_1\rangle}w_1-…-\frac{\langle v,w_r\rangle}{\langle w_r,w_r\rangle}w_r$$

Then, $w_{r+1}$ is orthogonal to $\{w_1,\ldots,w_r\}$, and secondly, if $v \notin\operatorname{span}\{w_1,\ldots,w_r \}$ then $w_{r+1} \neq 0$ , $\{w_1,\ldots,w_r,w_{r+1}\}$ is orthogonal and moreover $\operatorname{span}\{w_1,\ldots,w_r,v\}=\operatorname{span}\{w_1,\ldots,w_r,w_{r+1}\}$ .

But yea Im just not understanding my proof. I understand the importance of it because I use it for GS process and projection theorem etc, but can anyone explain/provide the proof or provide an alternate name to this? Thank you all

  • $\begingroup$ < and > mean "less than" and "greater than", and produce spacing correct for that meaning only; to make angle brackets, use \langle and \rangle. $\endgroup$ – Zev Chonoles Apr 19 '15 at 18:17
  • $\begingroup$ That $w_{r+1}$ is orthogonal to $w_1, \dots, w_r$ is just a direct computation using the assumptions (i.e. that $w_1, \dots, w_r$ are orthogonal). What exactly do you not understand? Furthermore, I think you mean to write $w_{r+1} = v - \dots$, where $\dots$ is what you are writing. With your definition of $w_{r+1}$, you have $w_{r+1} \in {\rm{span}}\{ w_1, \dots, w_r\}$, which is not good. $\endgroup$ – PhoemueX Apr 19 '15 at 18:19
  • $\begingroup$ Isn't it supposed to be $w_{r+1} = v - \frac{\langle v,w_1\rangle}{\langle w_1,w_1\rangle}w_1-\dots$? Otherwise your $w_{r+1}$ is defined as a linear combination of $w_i$'s since each $\langle \bullet,\bullet\rangle$ is a scalar. $\endgroup$ – JMoravitz Apr 19 '15 at 18:22
  • $\begingroup$ @JMoravitz Thanks!! I think this is why I was confused for so long! $\endgroup$ – Quality Apr 19 '15 at 18:27

As for a proof...let $\{w_1,w_2,\dots,w_r\}$ be an orthogonal set and consider $\langle w_{r+1}, w_i\rangle$. You get then that:

$\langle w_{r+1},w_i\rangle \\= \langle v - \frac{\langle v,w_1\rangle}{\langle w_1,w_1\rangle} w_1 - \dots,w_i\rangle ~~~~\text{by definition of}~w_{r+1}\\= \langle v, w_i\rangle - \langle\frac{\langle v,w_1\rangle}{\langle w_1,w_1\rangle}w_1,w_i\rangle - \dots - \langle\frac{\langle v,w_i\rangle}{\langle w_i,w_i\rangle}w_i,w_i\rangle-\dots~~~\text{by linearity of inner product}\\ = \langle v, w_i\rangle - \frac{\langle v,w_1\rangle}{\langle w_1,w_1\rangle}\langle w_1,w_i\rangle - \dots - \frac{\langle v,w_i\rangle}{\langle w_i,w_i\rangle}\langle w_i,w_i\rangle-\dots~~~\text{by linearity of inner product}\\ =\langle v,w_i\rangle - 0 - 0 -\cdots - \frac{\langle v, w_i\rangle}{\langle w_i, w_i\rangle} \langle w_i,w_i\rangle - 0 - \cdots~~~\text{by orthogonality of}~\{w_1,\cdots,w_r\}\\ =\langle v,w_i\rangle - \langle v,w_i\rangle = 0$

Hence, $w_{r+1}$ is orthogonal to $\{w_1,\cdots, w_r\}$

Furthermore, if $v\notin \text{span}\{w_1,\cdots, w_r\}$, then there does not exist any linear combination of $w_1,\dots,w_r$ that results in $\alpha_1 w_1 + \alpha_2 w_2 + \dots \alpha_r w_r = v$. In particular when $\alpha_i = \frac{\langle v,w_i\rangle}{\langle w_i,w_i\rangle}$. We have then that $w_{r+1} = v - \frac{\langle v,w_1\rangle}{\langle w_1,w_1\rangle}w_1 - \dots \neq 0$

The rest of the lemma follows quickly from definition of span and noting that $\{w_1,w_2,\dots,w_{r+1},v\}$ is a linearly dependent set (since you can write $w_{r+1}$ as in its definition as a linear combination of the others), and the dependency couldn't be a result of the first $r$ entries.


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