# explain that $\langle P_S(v), u_m \rangle = \sum_{j=1}^n \frac{ \langle v, u_j \rangle}{||u_j||^2} \langle u_j,u_m \rangle = \langle v, u_m \rangle$

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
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

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 $$\frac{\langle v, u_m \rangle}{||u_m||^2}\langle u_m, u_m \rangle$$
Definition of the norm is $||u|| = \sqrt{ \langle u,u \rangle }$ –  alvoutila Nov 2 '10 at 14:04