In this note, we want to demonstrate that this vector space is the same as the vector space of over (with ordinary addition and scalar multiplication). We'll continue to use to represent vector addition in , and to represent scalar multiplication.

First, notice that if our field is the real numbers , our vector
space
is a one-dimensional vector space, and so *must* be
isomorphic to with ordinary addition (since they are both finite
dimensional vector spaces over the same field and have the same dimension.)

To see that, let's find a basis for
. Any positive real
number other than will do, so let's use . To show that is a
basis, we must prove that every other element of
is a linear
combination of . That means that given any
, we can find a
scalar
for which

But certainly is the desired solution. Since we found one vector which spans, is a one-dimensional vector space over the field .

If we use a subfield of for , then depending on which we choose, may or may not be an element of , and so we will need more than one element in our basis for this other vector space. For example, if , we will need an infinite basis.

The above discussion does more than show us the dimension of our vector
space. It also gives us an isomorphism between
and . To see
this, we must verify that the map is one-to-one, onto, and linear.

- If , then , and so .
- For any , we must find an so that . But works for us here, so is onto.
- At first, you might worry that the logarithm isn't a
linear map. But keep in mind that it
*is*linear with our unusual definition for addition and scalar multiplication. We have to show that for any and for any , we have

(where on the right side we are using ordinary addition and multiplication).To see that, we just expand:

as desired.

Since we have an isomorphism between and , we can think of these two vector spaces as ``the same''.