Norm induced by inner product?

[owlpride]

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.

 

[morphism]

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.)

 

[wisvuze]

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 >

 

[morphism]

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 >

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 $$\frac14 (\|v+w\|^2 -\|v-w\|^2 + i\|v+iw\|^2 - i\|v-iw\|^2) = \langle v, w\rangle.$$

 

[blue_raver22]

The relationship between norms and inner products is a fundamental concept in linear algebra and functional analysis. In a finite-dimensional vector space over the real or complex numbers, every norm can be induced by an inner product. This means that for any norm on a finite-dimensional vector space, there exists an inner product that generates the same norm.

However, as mentioned in the content, this is not always the case for infinite-dimensional vector spaces. In these cases, there may exist norms that cannot be induced by any inner product. This is a major distinction between finite and infinite-dimensional spaces.

The reason why this distinction is important is because inner products provide a natural way to define orthogonality and angles between vectors. In finite-dimensional spaces, this allows us to define geometric concepts such as projections, orthogonality, and angles. Without an inner product, these concepts may not be well-defined.

Regarding the statement about the weak closure of the unit sphere in a finite-dimensional normed vector space, the use of an inner product is not necessary to prove this. The proof can be done using the properties of norms and the fact that any norm on a finite-dimensional vector space is induced by an inner product. However, in infinite-dimensional spaces, the use of an inner product may be necessary to prove this statement.

In summary, while every norm on a finite-dimensional vector space can be induced by an inner product, this is not always the case for infinite-dimensional spaces. The use of inner products allows us to define important geometric concepts in finite-dimensional spaces, but may not be necessary for certain proofs in these spaces.