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Raghav's question at Yahoo! Answers (axioms of vector space)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Here is the question:

Let V be the set of all positive real numbers: defined by
u $\$$ v = uv ($ is ordinary multiplication) and define #
by e#v = v^e. Prove that V is a vector space.

How do I go about proving this ? I know how to prove if V is a real vector space, but how do I prove if it is a vector space ?
Here is a link to the question:

Proving a set V is a vector Space? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
Last edited:

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Raghav,

Clearly, $(V,\$)$ is a commutative group (here, the zero vector is $1$) and $(\mathbb{R},+,\cdot)$ is a field. We need to prove the four properties of the scalar multiplication $\#$. Then, for all $\lambda,\mu$ real scalars and for all $u,v\in V$ vectors:

$(i)\;\lambda\#(u\;\$\;v)=\lambda\#(uv)=(uv)^ {\lambda}=u^{\lambda}v^{\lambda}=u^{\lambda}\;\$\; v^{\lambda}=(\lambda\# u)\;\$\;(\lambda\#v)$

$(ii)\;(\lambda+\mu)\#u=u^{\lambda+\mu}=u^{\lambda}u^{\mu}=(\lambda\#u)\;\$\;(\mu\#u)$

$(iii)\;(\lambda\mu)\# u=u^{\lambda\mu}=(u^{\mu})^{\lambda}=\lambda\#(\mu\#u)$

$(iv)\;1\#u=u^{1}=u$