# Proof that $\mathbb{R}^+$ is a vector space

I was doing some beginner linear algebra tasks and stumbled upon this one:

Proove that $\mathbb{R}^+$ is a vector space over field $\mathbb{R}$ with binary operations defined as $a+b = ab$ (where $ab$ is multiplication in $\mathbb{R}$ and $\alpha *b =b^\alpha$, where $b \in \mathbb{R}$ and $\alpha \in \mathbb{R}$.

It's easy to prove that $(\mathbb{R}^+,+)$ is an Abelian group and i will leave that part of proof out. However, when proving the following property of vector spaces, there seems to be a problem:

$\alpha (x+y) = \alpha x + \alpha y$ ( where $x,y \in \mathbb{R}^+$ and $\alpha \in \mathbb{R}$)

By definition: $$\alpha (x+y) =(x+y)^\alpha$$

and $$\alpha x + \alpha y = x^\alpha + y ^\alpha$$

In general case: $(x+y)^\alpha \ne x^\alpha + y^\alpha$ so this appears not to be a vector space, but even the solution in textbook states it is ( this property proof is completely omitted). Could this be author's error or did I make a mistake?

If the mistake is mine, I would like to ask and additional question ,which should probably be posted in a separate thread: How would i find one base of this vector space. By defintion, I need to find a positive real number who's linear combination would generate all positive real numbers. This is quite simple but should i use this vector space operations to form linear combinations or general multiplication and addition i.e. would a linear combination of $a \in \mathbb{R}^+$ be $b=5a$ or would that be $b=a^5$. If it's the latter, is it safe to assume that any positive number other than 1 is a base vector ?

• You can take the standard basis vectors for your second question – user210387 Jun 20 '15 at 15:55
• Remember you have defined $+$ as $x+y=xy$ and so $\alpha (x+y) = (xy)^{\alpha}=(x)^{\alpha}(y)^{\alpha}=\alpha x + \alpha y$ – Joaquin Liniado Jun 20 '15 at 15:58
• @Rememberme The standard basis vector is $1$ which doesn't work here. – user223391 Jun 20 '15 at 16:02
• Fun fact: This unusual vector space over $\mathbb R^+$ is just the image of $\mathbb R$ (with the usual vector space structure) under $\exp$. That is, each $x\in\mathbb R^+$ has a corresponding $\hat x\in\mathbb R$ via $x=\exp\hat x$, and the vector space operations are defined by copying them over: $z=x\oplus y$ iff $\hat z=\hat x+\hat y$, and $z=\alpha\odot x$ iff $\hat z=\alpha\cdot\hat x$. – user856 Jun 20 '15 at 17:09
• @Rahul It is also funny, because when viewing $\mathbb{R}$ and $\mathbb{R}^+$ as different vector spaces as we do here, the $\exp$ function is linear. When I started learning abstract mathematics, I wondered what is the reason for the list notation, eg. saying that $(V,+,\mathbb{F}\cdot)$ is a vector space rather than $V$. This is a pretty nontrivial example, since whether we consider $\exp$ a linear function or not depends on the choice of different linear structures on the same set (well $\mathbb{R}^+$ is only the half of it, but whatever...). – Bence Racskó Jun 20 '15 at 20:13

By definition $\alpha\odot(x\oplus y)=(xy)^\alpha$ and $\alpha\odot x\oplus\alpha \odot y=x^\alpha\oplus y^\alpha=x^\alpha y^\alpha$, and those two are the same. I denoted vector space addition and scalar multiplication by $\oplus$ and $\odot$ for distinguishability.

Edit: Though avid19 already answered this, any nonzero vector will provide a one-element basis for $\mathbb{R}^+$, however, in this case the zero vector is $1$, since $1\oplus x=1x=x$.

We can check this by the following. Let $g$ be a nonzero (eg. $\neq 1$) element of $\mathbb{R}^+$, and let $x$ be an arbitrary element of $\mathbb{R}^+$. And also let $\alpha\in\mathbb{R}$ be a scalar. In this case the equation $$\alpha\odot g=x$$ is written as $$g^\alpha=x.$$ Taking the $g$-base logarithm of both sides (remember $g$ and $x$ are larger than zero): $$\alpha=\log_gx,$$ which by the properties of logarithm functions, always exists. Thus given a non $1$ element of $\mathbb{R}^+$, we can always find a scalar, which when multiplied together by vector space scalar multiplication , results in any desired vector, so $\{g\}$ is a generating set.

It is also linearly independent, since there is only one element in it, and it isn't the zero vector, so $\{g\}$ is a basis.

• Oh yes , my error was quite great. I should have used different notation to avoid omitting that $x+y = xy$. Thanks a lot :) – Transcendental Jun 20 '15 at 15:59

$$\alpha(x+y)=x^{\alpha}+y^{\alpha}=x^{\alpha}y^{\alpha}=\alpha x+\alpha y$$
As for a basis, you can take any (non-zero) vector (also, what is $\vec{0}$ in this space? Hint: it's not $0$). For example, $2$ can be a basis vector. Every nonzero number $x$ can be reached by $\pm \log_2(x)$.