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 ?

  • $\begingroup$ You can take the standard basis vectors for your second question $\endgroup$
    – user210387
    Commented Jun 20, 2015 at 15:55
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    $\begingroup$ Remember you have defined $+$ as $x+y=xy$ and so $\alpha (x+y) = (xy)^{\alpha}=(x)^{\alpha}(y)^{\alpha}=\alpha x + \alpha y$ $\endgroup$ Commented Jun 20, 2015 at 15:58
  • $\begingroup$ @Rememberme The standard basis vector is $1$ which doesn't work here. $\endgroup$
    – user223391
    Commented Jun 20, 2015 at 16:02
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    $\begingroup$ 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 $\hat z=\alpha\cdot\hat x$. $\endgroup$
    – user856
    Commented Jun 20, 2015 at 17:09
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    $\begingroup$ @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...). $\endgroup$ Commented Jun 20, 2015 at 20:13

2 Answers 2


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.

  • $\begingroup$ Oh yes , my error was quite great. I should have used different notation to avoid omitting that $x+y = xy$. Thanks a lot :) $\endgroup$ Commented Jun 20, 2015 at 15:59

Remember what addition means! Note:

$$\alpha(x+y)=x^{\alpha}+y^{\alpha}=x^{\alpha}y^{\alpha}=\alpha x+\alpha y$$

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 $\pm \log_2(x)$.

  • $\begingroup$ Thank you very much! But i still don't understand how linear combinations are made. Do i use the vector space scalar multiplication or do i take standard multiplicaton? $\endgroup$ Commented Jun 20, 2015 at 16:09
  • $\begingroup$ @Transcendental I have edited my answer. $\endgroup$ Commented Jun 20, 2015 at 16:20
  • $\begingroup$ Yes yes, thank you. Base here is obviously 1. :D Thank you $\endgroup$ Commented Jun 20, 2015 at 16:25
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    $\begingroup$ @Transcendental Also, while I believe my edit inadvertedly explains what you asked here, note that linear combination is something that happens in the vector space, so you have to use vector space operations. It is a strange thing about this example that 1) the vector space is a subset of its field, 2) the vector space operations do not correspond to the field operations (but 2) only makes sense because of 1) ), but in a general vector space, you absolutely have no "access" to the field operations. $\endgroup$ Commented Jun 20, 2015 at 16:26

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