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I have a previous post here. There is a part b to that question and it asks:
Let $x=(1,1,1)^T$. Write x as a linear combination of $u_1, u_2, u_3$ using Parseval's formula to compute $||x||$.

I know how to compute $||x||$, it's simply the magnitude. However I am totally unsure how do to x as a linear combo, I've read through my book and tried looking online with no luck. Any ideas?

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Hint: Since you have orthonormal basis $\{u_1,u_2,u_3\}$, you can express any vector $x$ as a linear combination of the basis vectors as $$x=\alpha_1 u_1+\alpha_2 u_2+\alpha_3 u_3$$ where you have to determine the coefficients (scalars) $\alpha_i$'s.

Since, $u_i$'s are orthonormal, $=u_i^t.u_j=\delta_{ij}$ (=1, only if $i=j$, else $0$). So,

$$x^t.u_i==(\sum\alpha_j u_j^t).u_i=\alpha_i$$.

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    you completely lost me after the "Since, $u_i$'s are". Can you please rephrase it a little differently, i'm not getting it2012-10-12
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    Sorry... I assumed you know about inner product, otherwise how could the Perseval's identity come? The $$ symbol stands for "inner product of x and y" which, in $\mathbb{R}^3$ can be taken as $x^T.y$. Is it clear now?2012-10-12
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    inner product was mentioned in class but I not too familiar with it. But from what I'm reading from the book and online, isn't inner product just a dot product, just a more generalized version. And the book says perseval's formula but i have not idea what that is2012-10-12
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    Inner product and dot product are different terminology of the same thing $$. Perseval's formula says $\|x\|^2=\sum\alpha_i^2$2012-10-12
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    okay so the first step I is determine the coefficients $\alpha_i$'s which can be found by doing $$. Am I right so far?2012-10-12
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    Yes....correct!2012-10-12
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    okay great I solved it, all thanks to you. $\alpha_1=-\frac {\sqrt{2}}{3}$ and $\alpha_2=\frac {5}{3}$ and $\alpha_3=0$2012-10-12
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    Excellent! You are welcome.2012-10-12