# Isomorphic Finite Dimensional Vector Spaces

1. Nov 9, 2013

### TheOldHag

I'm going through the text "Linear Algebra Done Right" 2nd edition by Axler. Made it to chapter 4 with one problem I'm unable to understand fully. The theory that two vector spaces are isomorphic if and only if they have the same dimension. I can see this easily in one direction, that is, isomorphic vector spaces will have the same dimension, but it seems I can imagine vector spaces having the same dimension but for which they are not isomorphic. Perhaps one vector space would be 2-tuples over the integers but the with modular operators limiting the number of elements in the space and another vector space of integer 2-tuples without modular operators. The latter space would have more element in it so how would any map from the former to the latter be surjective? And yet do they not have the same dimension?

I'm guessing he is making assumptions about the operations on the vector spaces and they are over R or C. However, this assumption is made on many of the theories this particular theory doesn't state V explicitly and so I'm not sure what part of this theory the assumptions fall under.

Any clarification on this would be appreciated.

2. Nov 9, 2013

### jgens

Let V and W be n-dimensional vector spaces over the same base field F. Fix bases {e1,...,en} and {f1,...,fn} for V and W respectively. Define a linear transformation φ:V→W by setting φek = fk and extending by linearity. This is an isomorphism.

3. Nov 9, 2013

### Office_Shredder

Staff Emeritus
It's a vector space, so it's a vector space over a field.. R, C, Q or any other random field will work. Both vector spaces have to be over the same field but that's the only requirement - note your attempt at a counterexample is not a vector space because there is no field of scalars ( well, when you mod out if it's by a prime number there is)

4. Nov 9, 2013

### TheOldHag

That answers it for me. Basically, I haven't gone over abstract algebra yet so I think in this case I just have to move forward and assume the notion of field will clear up that detail later. Thanks.

5. Nov 9, 2013

### jgens

Since Axler apparently does not introduce abstract fields, just read my post again with the first line changed to: "Let V and W be n-dimensional real or complex vector spaces." The argument is the same from there.

6. Nov 9, 2013

### TheOldHag

Your post has gotten me thinking about the very definition of liner map. I do better with more rigorous text only because they don't omit tedious and trivial details. For instance, he defines a linear map as having homogeneity and additivity and leaves it to the reader to deduce that V and W have to be vector spaces over the same field by the fact that the constant 'a' appears in both side of the equation of homogeneity. Yes, it should be obvious but I do better when it is spelled out. Perhaps it would have been more clear if he qualified that a linear map was a function between vector spaces over the same field.

Nevertheless at the beginning of each chapter he states the assumption that F is R or C and that V is a vector space over F. However, the theorem I was confused about didn't mention V or F at all.

From what I can tell though, this is a great book for self study.