# Dot product in an orthonormal basis

I am told that for two vectors $v,w$ in a vector space $V$ $$v \cdot w = \underline{v}^T \underline{w}$$ only in an orthonormal basis. This is clear if the basis is the standard $(1,0,\dots,0), (0,1,\dots,0),\dots$ etc but surely when this basis changes (and is still orthonormal) the two vectors $v,w$ will still be the same and so $v \cdot w$ won't change but the coordinates $\underline{v}, \underline{w}$ with respect to that basis will change, so $\underline{v}^T \underline{w}$ will also change?

Also, for a Euclidean space, where you have a vector space and a positive definite symmetric bilinear form, what is the 'dot product'? Is it even definable if you don't know what the basis is?

I know these two questions might be slightly contrasting but I can't get my head around the relation between dot product and bilinear forms!

Thanks!

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## 1 Answer

Actually, a basis change doesn't matter: consider an arbitrary orthonormal basis $\{e_i\}_{i = 1}^{n}$ for $\mathbb{R}^n$ (though you could easily generalize to any finite dimensional inner product space). If $v = \sum a_i e_i$ and $w = \sum b_i e_i$, then $$v\cdot w = \left( \sum_i a_i e_i \right) \cdot \left( \sum_j b_j e_j \right) = \sum_i \sum_j a_i b_j (e_i \cdot e_j) = \sum_i a_i b_i = \underline{v}^T \underline{w}$$

As for your second question, the proof above also works for any inner product on a finite dimensional real vector space, and thus shows that there is only one inner product on such a space, namely the dot product. For complex vector spaces, the definition of an inner product changes slightly (it becomes conjugate-linear in one factor), but the result is the same: there is only one (up to isometry) Hilbert space of a given dimension (which is the cardinality of any given orthonormal basis).

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So what exactly is the difference between the 'dot product' and a bilinear form on a non-Euclidean space with a non-orthonormal basis? –  user26069 Apr 8 '12 at 14:40
@user26069: not sure I understand the question. As I've proved above, on any finite dimensional real vector space, there is no difference between the dot product and a positive definite symmetric bilinear form (i.e. an inner product). General bilinear forms is a completely different story (there are lots of them, for example on a finite dimensional vector space). –  Martin Wanvik Apr 8 '12 at 15:20
Oh ok I understand, so when you're not on a Euclidean space (so don't have a positive definite symmetric bilinear form) the bilinear form and dot product are simply defined differently? –  user26069 Apr 8 '12 at 15:26
@user26069: Every inner product (note that the dot product on $\mathbb{R}^n$ is an inner product) is a bilinear form, but not every bilinear form on a vector space is an inner product (see en.wikipedia.org/wiki/Inner_product_space). –  Martin Wanvik Apr 8 '12 at 16:29