Vector space without a scalar product

Question

In linear algebra the terms vector space and scalar product always (at least for me) appear together. Can you give me an example of a vector space without a scalar product? Does the sentence Let $V$ be a vector space without a scalar product make any sense? Have it any application in mathematics?

Answer 1

I think this is a terminology problem.

By definition, a vector space always has a scalar multiplication, where you multiply a vector with a scalar and get a vector.

However, vector spaces can optionally come with a inner product (or dot product) where you multiply two vectors and get a scalar.

The latter operation is not required for vector spaces, and it is easy to make a vector space without one -- just don't define it. For example "the vector space of infinite sequences of real numbers, with coordinatewise addition and (scalar) multiplication".

"Scalar product" could in principle mean either concept, but would most often be an inner product.

Whenever we have a vector space, we can use the Axiom of Choice to prove that it is possible to choose some inner product that satisfies the required axioms. But, as the words are usually used, the vector space doesn't "have" an inner product until we decide to make such a choice.

Answer 2

You firstly want your underlying field to be characteristic $0$ (else the condition $\langle x,x\rangle = 0$ may have nontrivial solutions). The simplest fields of characteristic $0$ are $\Bbb R$ and $\Bbb C$. If $V$ is a finite dimensional vector space over $\Bbb R$ or $\Bbb C$, you can define an inner product on it in a way similar to the Euclidean inner product. Even if you do not explicitly state that there is an inner product on the set, it still exists. So in this setting, it is not possible.

We must then venture into the realm of infinite dimensional vector spaces, and here it is easily possible. Let's take our vector space $V$ to be the set of sequences of real numbers such that the sequence is bounded, i.e. if $(x_n)_{n=1}^{\infty}$ is a sequence, then $\sup_n |x_n| < \infty$. There is not an inner product on this vector space that is compatible with the norm structure.

That said, one can define an inner product on this vector space but it will not be compatible with the norm structure. Suppose $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ are in $V$, then define $\langle (x_n),(y_n)\rangle$ by

$$\langle (x_n),(y_n)\rangle = \sum_{n=0}^{\infty}\frac{1}{2^n}x_ny_n.$$

This is an inner product on $V$. The existence of the norm is not an accident and this leads to H. Malcolm's answer above.