In addition to the examples of vector spaces discussed in the text and elsewhere, we also discussed one example in which the operations of vector addition and scalar multiplication wouldn't normally be called addition'' or multiplication''.

• 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:

If and are positive real numbers, so is their product .
Multiplication of real numbers is associative, so no problem.
Multiplication of real numbers is commutative.
The positive real number 1 acts as the identity element for vector addition, since .
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?

Scott Sutherland February 10, 2005