What is the dimension of R^{+} as a vector space over R? over Q?
Thank you in advance
Jojo,
The dimension of over is . Indeed, note that is necessarily greater than else it would be isomorphic to , which is countable. Moreover, it's obvious that . So, we have the fact that . If you're into that continuum hypothesis kind of thing you're done. If you want to see an actual proof opposed to a plausibility argument, see my blog post here.
let's exhibit a linear isomorphism between the vector space U = R+, with the vector sum:
x(+)y = xy
and the scalar multiplication:
c*x = x^{c}
and the vector space V = R, where the vector sum is ordinary real addition, and the scalar multiplication is ordinary multiplication.
we define T:U-->V by T(u) = ln(u).
to prove T is linear, we must show:
T((a*x)(+)(b*y)) = aT(x) + bT(y).
so: T((a*x)(+)(b*y)) = T((x^{a})(+)(y^{b})) = T(x^{a}y^{b}) = ln(x^{a}y^{b}) = ln(x^{a}) + ln(y^{b}) = aln(x) + bln(y)
= aT(x) + bT(y), as desired.
to prove that T is injective, we must show ker(T) = {1}, (since 1 is the 0-vector of R+).
so suppose u is in ker(T). then ln(u) = 0, hence u = e^{ln(u)} = e^{0} = 1.
to prove T is surjective, we need to find a u in R+, such that for any real number x, T(u) = x.
let u = e^{x}, which is in R^{+} (since e > 0).
then T(u) = ln(e^{x}) = x, as desired.
since R (as V) has dimension 1 over R, (the number 1 is a basis), it follows that U = R+ also has dimension 1 over R (we can take the number e as a basis for U).
if our base field is Q, it gets a bit more interesting: as drexel28 pointed out, R is infinite-dimensional over Q (in fact "more" than infinite, uncountably infinite!). the same mapping T will serve as an isomorphism, showing that R+ is also infinite-dimensional over Q.