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If I am correct dot product is an example of inner product on coordinate space.

I wonder if for an arbitrary inner product space with base field being $\mathbb{R}$ or $\mathbb{C}$, there always exists a coordinate system so that the inner product becomes the dot product in coordinate? What is the name of the topic regarding this question?

Thanks and regards!

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I don't fully understand - what do you mean by 'becomes the dot product?' – mixedmath Aug 2 '11 at 20:05
@mixedmath: I mean the inner product of two vectors and the dot product of their coordinates become equal. – Tim Aug 2 '11 at 20:08
Maybe in a more general sense, if your space is a manifold, and your inner-product is defined in the tangent space, you can generalize an inner-product to be a 2-form (bilinear map); then there is a way of pulling back that inner-product into $\mathbb R^n$ using the chart maps. This is part of the standard argument used to show that a $C^{\infty}$ manifold M can be made into a Riemannian manifold; specifically, the inner-product in $\mathbb R^n$ can be pulled into M; by smoothness of the chart, the positive-definiteness of the (image of the) inner-product in $\mathbb R^n$ is preserved in M. – gary Aug 2 '11 at 21:26
So, the more general topic is that of pull backs of 2-forms between manifolds. – gary Aug 2 '11 at 21:59
up vote 5 down vote accepted

For finite dimensional spaces, the answer is "yes"; this is a consequence of the Gram-Schmidt orthonormalization process: every finite dimensional inner product space over $\mathbb{R}$ or over $\mathbb{C}$ has an orthonormal basis.

Now let $\mathbf{V}$ be an inner product space, and let $\mathbf{v}_1,\ldots,\mathbf{v}_n$ be an orthonormal basis. Then $T\colon\mathbf{V}\to \mathbf{F}^n$ given by $T\mathbf{v}_i = \mathbf{e}_i$ (i.e., $T$ maps each vector in $\mathbf{V}$ to its coordinate vector relative to the orthonormal basis $\mathbf{v}_1,\ldots,\mathbf{v}_n$) is an invertible linear transformation such that for all $\mathbf{x},\mathbf{y}\in\mathbf{V}$, $\langle \mathbf{x},\mathbf{y}\rangle = T\mathbf{x}\cdot T\mathbf{y}$, where the right hand side is the standard dot product on $\mathbf{F}^n$ ($\mathbf{F}=\mathbb{R}$ or $\mathbb{C}$).

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Thanks, Arturo! How about infinite dimensional inner product spaces? – Tim Aug 2 '11 at 20:12
@Tim: For Hilbert spaces, the answer is again "yes" by a similar argument using Hilbert bases. But if the space is not complete, you may run into problems. – Arturo Magidin Aug 2 '11 at 20:19
@Tim: you can find it here (page 83) if Google lets you look. You'll find it in any introductory book on functional analysis worth its name. – t.b. Aug 2 '11 at 20:19
The axiom of choice is equivalent to Zorn's lemma. I meant: apply Zorn's lemma. I'd recall the existence proof for bases in linear algebra first, then this one will be rather easy. (order orthonormal systems by inclusion, take a maximal one by Zorn; if the maximal one were not a Hilbert basis you could find a vector orthonormal to it, contradiction). It's always the same argument. – t.b. Aug 2 '11 at 20:31
@Theo: except one has to be careful about the meaning of "basis" in the infinite dimensional space; we mean "Hilbert basis", or "linearly independent and closure of the linear span is everything." – Arturo Magidin Aug 2 '11 at 20:33

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