- Our vectors, which we denote as usual by greek letters , are the positive real numbers, .
- Our field of scalars (denoted by roman letters
)
can be either the reals or the rationals . We'll write but
we get a vector space with as well.
- To ``add'' two vectors, we multiply the corresponding real numbers. That is, the vector sum of and is the product .
- The scalar product of a scalar and the vector , we compute the power .

Let's check that this satisfies the necessary properties:

**additive closure**- If and are positive real numbers, so is their product .
**associativity of addition**- Multiplication of real numbers is associative, so no problem.
**commutivity of addition**- Multiplication of real numbers is commutative.
**additive identity**- The positive real number 1 acts as the identity element for vector addition, since .
**additive inverses**- For any vector
, there is another
vector
so when the two vectors are ``added'', the result
is the identity element above. In this case, the inverse of is
, since

**closure of scalar multiplication**- For any scalar and any vector , the scalar multiple is still in .
**neutrality of 1**- When we compute the scalar multiple of the
multiplicative identity in our field with any vector
,
we should get the original vector. That works fine:

**vector distributive law**- Multiplying a scalar times the sum of two
vectors and :

**scalar distributive law**- The sum of two scalars times a vector
:

So we see that is a vector space over , with an appropriate interpretation of vector addition and scalar multiplication.

Note also that in this case, a ``linear combination'' works out to be very
like factoring. For example, we can express the vector as a linear
combination of the vectors and by

Note that there may be more than one way to express the same vector as a
linear combination of two others. For example, if our underlying field is
, then there are scalars in equal to and , and
so

is also a scalar. But then if we have

we also have

since

More concretely, since

we also have

So we can express as a linear combination of and in many different ways. What is the dimension of as a vector space over ? Can you give a proof?

The situation is more complicated if we consider as a vector space
over , since
will be rational for some
and , and not for others. What do you think the dimension
of this vector space is?