## norm induced by inner product?

On a finite-dimensional vector space over R or C, is every norm induced by an inner product?

I know that this can fail for infinite-dimensional vector spaces. It just struck me that we never made a distinction between normed vector spaces and inner product spaces in my linear algebra course on finite-dimensional vector spaces.

Why I actually care about it: I wonder why the unit sphere in a finite-dimensional normed vector space is weakly closed. Obviously the statement should be that a sequence in a finite-dimensional space converges weakly if and only if it converges strongly, but I'm not sure how to go about this without using an inner product.

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire

Recognitions:
Homework Help
 Quote by owlpride On a finite-dimensional vector space over R or C, is every norm induced by an inner product?
No. The parallelogram law can and does fail for a variety of norms (e.g. all the p-norms except when p=2).

 Why I actually care about it: I wonder why the unit sphere in a finite-dimensional normed vector space is weakly closed. Obviously the statement should be that a sequence in a finite-dimensional space converges weakly if and only if it converges strongly, but I'm not sure how to go about this without using an inner product.
On a finite-dimensional vector space, all norm topologies are equivalent. So you can always assume your norm topology comes from a norm that is induced by an inner product. (In fact it's also true that in this case the weak topology and any norm topology are the same.)

 Blog Entries: 1 It should be noted that if a norm satisfies the parallelogram law, then you can always uncover an inner product from the norm via the "polarization identity" 1/4 ( | v + w |^2 - | v - w |^2 ) = < v , w >

Recognitions:
Homework Help
Just for the sake of completeness: that form of the polarization identity is only valid if we're working over R; over C, the identity becomes $$\frac14 (\|v+w\|^2 -\|v-w\|^2 + i\|v+iw\|^2 - i\|v-iw\|^2) = \langle v, w\rangle.$$