a belongs to R

show that the map

L: R^n------R^n>0

(R^n>0 denote the n-fold cartesian product of R>0 with itself)

(a1)

(....) ----------

(an)

(e^a1)

(.....)

(e^an)

is a isomorphism between the vector space R^n and the vector space R^n>0

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- Apr 22nd 2009, 12:55 PMak123456isomorphism question
a belongs to R

show that the map

L: R^n------R^n>0

(R^n>0 denote the n-fold cartesian product of R>0 with itself)

(a1)

(....) ----------

(an)

(e^a1)

(.....)

(e^an)

is a isomorphism between the vector space R^n and the vector space R^n>0 - Apr 22nd 2009, 02:19 PMGammaNot sure I understand
you have $\displaystyle \phi : \mathbb{R}^n \rightarrow \mathbb{R}_+^{n}$ by $\displaystyle \phi(a_1,...,a_n)=(e^{a_1},...,e^{a_n})$?

This is bijective because $\displaystyle \phi ^{-1} ((a_1, ..., a_n))= (ln(a_1),..., ln(a_n))$

I think you can check it is a morphism.