In a previous class, we saw that the positive reals $\mathbb{R}^{+}$ is a vector space over a field $\mathbb{F}$ (where $\mathbb{F}$ is $\mathbb{R}$ or a subfield of $\mathbb{R}$ such as the rationals $\mathbb{Q}$), provided that we use multiplication in $\mathbb{R}^{+}$ for vector addition, and raise our vector to the power $c$ for scalar multiplication by $c$. That is, for $\alpha,\beta \in\mathbb{R}^{+}$ and $c\in\mathbb{F}$, we set

\begin{displaymath}\alpha\boxplus\beta = \alpha\beta
c\boxdot\alpha = \alpha^c

In this note, we want to demonstrate that this vector space is the same as the vector space of $\mathbb{R}$ over $\mathbb{F}$ (with ordinary addition and scalar multiplication). We'll continue to use $\boxplus$ to represent vector addition in $\mathbb{R}^{+}$, and $\boxdot$ to represent scalar multiplication.

First, notice that if our field $\mathbb{F}$ is the real numbers $\mathbb{R}$, our vector space $\mathbb{R}^{+}$ is a one-dimensional vector space, and so must be isomorphic to $\mathbb{R}$ 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 $\mathbb{R}^{+}$. Any positive real number other than $1$ will do, so let's use $2$. To show that $\{2\}$ is a basis, we must prove that every other element of $\mathbb{R}^{+}$ is a linear combination of $2$. That means that given any $x\in\mathbb{R}^{+}$, we can find a scalar $c\in\mathbb{F}$ for which

\begin{displaymath}x = c\boxdot 2 \quad\mbox{or, equivalently,}\quad x = 2^c.\end{displaymath}

But certainly $c=\log_2 x$ is the desired solution. Since we found one vector which spans, $\mathbb{R}^{+}$ is a one-dimensional vector space over the field $\mathbb{R}$.

If we use a subfield of $\mathbb{R}$ for $\mathbb{F}$, then depending on which $x$ we choose, $\log_2 x$ may or may not be an element of $\mathbb{F}$, and so we will need more than one element in our basis for this other vector space. For example, if $\mathbb{F}= \mathbb{Q}$, 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 $\mathbb{R}^{+}$ and $\mathbb{R}$. To see this, we must verify that the map $\log_2$ is one-to-one, onto, and linear.

Since we have an isomorphism between $\mathbb{R}^{+}$ and $\mathbb{R}$, we can think of these two vector spaces as ``the same''.

Scott Sutherland March 3, 2005