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I have a problem. I wish to understand how following can be proven to hold; $\langle P_S(v), u_m \rangle = \sum_{j=1}^n \frac{ \langle v, u_j \rangle}{||u_j||^2} \langle u_j,u_m \rangle = \langle v , u_m \rangle $, where $P_S(v) =\sum_{j=1}^n \frac{ \langle v, u_j \rangle}{||u_j||^2} u_j $ and Suppose V is a complex inner product space and $B={u_1, u_2,..., u_n }$ is orthogonal set in $V$ with $u_j<>0$ for all j. also prove that $P_S(v)$ is a linear transformation.

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Oh boy, that was pretty unreadable! Please put the math inside dollar signs, and use \langle and \rangle instead of greater/less than signs. –  Hans Lundmark Nov 2 '10 at 12:03
    
Did you try to compute the left-hand side? –  AD. Nov 2 '10 at 13:43
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The tag is misleading, there is a <inner-product-space> tag. I suggest you should change that. –  AD. Nov 2 '10 at 13:46
    
I am not satisfy this answer please can you elaborate this question as soon as possible. –  user37273 Aug 5 '12 at 13:32

1 Answer 1

The first equality follows by definition. Now notice that $$\langle u_j, u_m \rangle = 0$$ if $i\neq j$.

Thus the sum reduces to $$\frac{\langle v, u_m \rangle}{||u_m||^2}\langle u_m, u_m \rangle $$

Now what is the definition of the norm?

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Definition of the norm is $ ||u|| = \sqrt{ \langle u,u \rangle }$ –  laovultai Nov 2 '10 at 14:04
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@alvoutila: Good, now you can finish the proof on your own. And you can also take the opportunity to learn what the phrase "rhetorical question" means. ;-) en.wikipedia.org/wiki/Rhetorical_question –  Hans Lundmark Nov 2 '10 at 14:25

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