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Pick any three vectors of a vector space randomly (but linearly independent). Then we assign them coordinates:

$$e_1=[1 0 0]$$ $$e_2=[0 1 0]$$ $$e_3=[0 0 1]$$

Therefore now they are orthonormal because $e_i\cdot e_j = \delta_{ij}$. Then this seems to show that all sets of three linearly independent vectors are an orthonormal base, which is false. However I don't see the mistake. I know it must be something silly.

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What do you mean you assign them coordinates? – Git Gud Sep 23 '13 at 20:16
It's a way to say, "I choose them as a base of the space". – Ambesh Sep 23 '13 at 20:18
up vote 2 down vote accepted

If you have a metric on your space, then you can tell whether a chosen basis is orthonormal or not.

If you don't have a metric already you can define your basis to be orthonormal. You don't have any existing structure to prove it can't be.

So there are two different situations.

Note that the same vector space can have different metrics, all compatible with the vector operations and scalar multiplication.

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So basically. Can I choose always a metric such that makes my three vectors orthogonal?..ok I see. – Ambesh Sep 23 '13 at 20:34

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