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

$$\alpha\boxplus\beta = \alpha\beta \qquad\mbox{and}\qquad c\boxdot\alpha = \alpha^c$$

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

$$x = c\boxdot 2 \quad\mbox{or, equivalently,}\quad x = 2^c.$$

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.

• If $\log_2 x = \log_2 y$, then $2^{\log_2 x} = 2^{\log_2 y}$, and so $x=y$.
• For any $y\in\mathbb{R}$, we must find an $x\in\mathbb{R}^{+}$ so that $\log_2{x} = y$. But $x=2^y$ works for us here, so $\log_2$ 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 $\alpha,\beta \in\mathbb{R}^{+}$ and for any $c\in\mathbb{F}$, we have
  
  $$\log_2\left( (c\boxdot\alpha) \,\boxplus\,\beta\right) = c\log_2 \alpha + \log_2 \beta,$$
  
  
  (where on the right side we are using ordinary addition and multiplication).

  To see that, we just expand:
  

  $$\begin{array}{rcl} \log_2\left( c\boxdot\alpha \,\boxplus\,\... ...\beta) \\ &=& c \log_2 \alpha + \log_2(\beta) \\ \end{array}$$
  
  
  as desired.

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