Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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 $\mathbb{C}$).

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.