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Let $V$ be a finite dimensional vector space over $\Bbb R$, with positive definite scalar product. Let $\{v_1, v_2, ... v_m\}$ be the set of elements of $v$ of norm $1$. and mutually perpendicular. (i.e $<v_i, v_j>=0, i \neq j$) Assume that for every $v \in V$ we have $$||v||^2 = \sum_{i=1}^m <v, v_i>^2$$ Show that $\{v_1, v_2, ... v_m\}$ is basis of $V$.

Let $c_1, c_2, ... c_m $ be the component of $v$ along $v_1, v_2, ... v_n$, when we get $||v||^2 = c_1^2 + c_2^2 + ... +c_m^2$. But I don't see how I can show it? Any hints?

Thanks in advance!!

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up vote 2 down vote accepted

Hint: Suppose that $\displaystyle \sum_{i=1}^m c_i v_i = 0$. You need to show that $c_1=\cdots=c_m=0$, because then you'll have a set of $m$ linearly independent vectors (which must therefore be a basis). Use the fact that the $v_i$s are mutually perpendicular.

For a further hint, hover your mouse over the greyed out box below

What is $\left\langle \displaystyle\sum_{i=1}^m c_i v_i, v_j \right\rangle$ for each $j$?

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Can I dot this $||v||^2 = c_1^2 + c_2^2 + ... +c_m^2 = \left\langle (c_1, c_2, ... c_m), (c_1, c_2, ... c_m) \right \rangle = <v,v>$ –  hasExams Jan 31 '13 at 18:14
    
Is it intentional that math formulas are seen without hovering? :) –  Ilya Jan 31 '13 at 18:32
    
@testuser: Yes; but here we've set $v=0$, so what can you say about the $c_j$s? –  Clive Newstead Jan 31 '13 at 18:48
    
@Ilya: I think that may depend on your browser or ad-blocking or something... but in the end it doesn't matter that much. –  Clive Newstead Jan 31 '13 at 18:48
    
No, it doesn't - I rather tried to get how this thing works. –  Ilya Jan 31 '13 at 18:51

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