Originally Posted by

**Deveno** you haven't defined what the $\displaystyle x_i$ and the $\displaystyle y_i$ ARE.

what is the dimension of a vector space? it is the cardinality of any basis. therefore, bases should figure prominently in your proof. i don't see the word "basis" once.

for example, to prove that two isomorphic vector spaces X,Y have the same dimension, one way is to pick a basis {x1,x2,...xn} for X.

since we have an isomorphism f, show that {f(x1),f(x2),...,f(xn)} is a basis for Y. we know that it has the same cardinality, because f is bijective,

so linear independence is the only thing to prove.

conversely, if X,Y have the same dimension, you need to show that a map which maps a basis {x1,x2,...,xn} for X bijectively to a basis {y1,y2,...,yn} for Y. can be extended to

an isomorphism of X with Y. this map must be surjective (a consequence of the {yi} being a spanning set), injective (a consequence of the {xi} being linearly independent),

and linear (which suggests that we extend our basis map linearly). even if you merely failed to specify that the {xi} and {yi} are bases, showing that f is bijective just on those sets,

does not make f into a linear map on all of X. you need to say HOW you are defining f on an arbitrary element of X.