1. ## Isomorphism

Show that two finite dimensional vector spaces over F are isomorphic iff. they have equal dimensions.

$(\Rightarrow)$
Suppose X, Y are isomorphic where X, Y are finite dimensional vector spaces.
Since f is isomorphic, f is monic and an epimorphism. Therefore, f is in a one to one correspondence. Thus, dim X = dim Y.

$(\Leftarrow)$
Suppose dim X = dim Y.
Define $f:X\to Y$ by $f(x_i)=y_i$.

If $f(x_1)=f(x_2)$, then
$f(x_1)=y_1 \ \text{and} \ f(x_2)=y_2\Rightarrow y_1=y_2$
Therefore, f is monic.

There exist $y_i\in Y$ such that $y_i=f(x_i)$. Therefore, f is an epimorphism.

Hence, f is an isomorphism.

Correct?

2. ## Re: Isomorphism

you haven't defined what the $x_i$ and the $y_i$ 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/spanning 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.

3. ## Re: Isomorphism

Originally Posted by Deveno
you haven't defined what the $x_i$ and the $y_i$ 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.
So I just need to add let $x_1,\cdots , x_n$ be a basis of X and $y_1,\cdots , y_m$ be a basis for Y for part 1 and since it is an isomorphism we will get m=n.

Let $x_1,\cdots , x_n$ be a basis of X and $y_1,\cdots , y_n$ be a basis for Y for part 2

Yes?

4. ## Re: Isomorphism

um, no. you need to prove that m = n. one way to do that, is to DEFINE yi = f(xi). since f is bijective, this shows m = n, but it does not yet show that the {f(xi)} are actually a basis. perhaps they are linearly dependent...but this can't happen because...why? perhaps they don't span all of Y....but this can't happen...why?

and again, just saying let {x1,x2,...,xn} be a basis for X, and let {y1,y2,..,yn} be a basis for Y, does not fix the problem. what is the isomorphism f for any particular (arbitrary) vector x in X?

until you say what it is, how can you even begin to assert it is a linear map (vector space homomorphism)?

5. ## Re: Isomorphism

Originally Posted by Deveno
um, no. you need to prove that m = n. one way to do that, is to DEFINE yi = f(xi). since f is bijective, this shows m = n, but it does not yet show that the {f(xi)} are actually a basis. perhaps they are linearly dependent...but this can't happen because...why? perhaps they don't span all of Y....but this can't happen...why?

and again, just saying let {x1,x2,...,xn} be a basis for X, and let {y1,y2,..,yn} be a basis for Y, does not fix the problem. what is the isomorphism f for any particular (arbitrary) vector x in X?

until you say what it is, how can you even begin to assert it is a linear map (vector space homomorphism)?
I don't understand what you are trying to get at.

6. ## Re: Isomorphism

a map from basis to basis is NOT a map from all of X into Y.

what is f(x) for an arbitrary x in X?