# Linear Algebra Isomorphism

1. Aug 31, 2006

### wurth_skidder_23

Sets in Linear Space

I am trying to show the set of all row vectors in some set K with dimension n is the same as the set of all functions with values in K, defined on an arbitrary set S with dimension n. I am using isomorphism to show this, but I can't determine how to show that the isomorphism is onto.

My work so far:

G := Functions with values in K

f: G -> K

(g(s1),...,g(sn))

one to one

f(G) = f(H)

(g(s1),...,g(sn)) = (h(s1),...,h(sn))

therefore, g(s1) = h(s1), ..., g(sn) = h(sn)

preserves structure

f(c1*G + c2*H) = (c1*g(s1) + c2*h(s1),...,c1*g(sn) + c2*h(sn))
= c1*(g(s1),...,g(sn)) + c2*(h(s1),...,h(sn))
= c1*f(G) + c2*f(H)

Last edited: Aug 31, 2006
2. Aug 31, 2006

### wurth_skidder_23

No one out there is familiar with isomorphism?

3. Aug 31, 2006

### nocturnal

Huh? This is not very clear. Is this the statement you are trying to prove?

Proposition: Let K,S be n-dimensional vector spaces over a Field F. Let G be the set of all functions mapping S to K. Then G is isomorphic to K.

If so, then it's false. For one the set of all linear transformations mapping S to K is a subspace of G of dimension n^2, so G can't possibly have the same dimension as K which is a necessary and sufficient condition for finite vector spaces to be isomorphic.

Last edited: Aug 31, 2006
4. Sep 1, 2006

### wurth_skidder_23

Sorry, my book isn't very clear, either. That was part of the problem. Here is the full problem statement.

Examples of Linear Spaces.
(i) Set of all row vectors: (a1,...,an), aj in K; addition, multiplication defined componentwise. This space is denoted as K^n.
(iii) Set of all functions with values in K, defined on an arbitrary set S.
Show that if S has n elements, (i) is the same as (iii).

5. Sep 1, 2006

### nocturnal

The books version makes a little more sense :tongue:

You have the right idea: Demonstrate an Isomorphism between these two vector spaces. I have added some notes.

This is pretty poorly defined. I can guess what you mean but I shouldn't have to. Here's one way to phrase this:

Suppose S is a set of n elements $s_i$ for i =1,2,...,n.
Let G be the vector space of all functions mapping S into K.

Define $$f:G \rightarrow K^n$$ such that $$\forall g \in G, g \mapsto (g(s_1),\ldots,g(s_n))$$

We show f is an isomorphism.
---------
My modifications are in bold.
To show f is onto, you let y = (a1, a2, ... , an) be an arbitrary element of K^n and show that there is some element of G that gets mapped to y by f. This is trivial, but if you want you could write something like this: let g be the function $s_i \mapsto a_i$, then g is a function mapping S into K and thus is an element of G. By the definition of f we have f(g) = y.

Another way to prove G is isomorphic to K^n is to show that
dim(G) = dim(K^n) = n. You can do this by showing that a basis for G has n elements. As an exercise (if you feel like it), give a basis for G, ie: provide a subset of G which is linearly independent and spans G .

Last edited: Sep 1, 2006
6. Sep 1, 2006

### HallsofIvy

Staff Emeritus
As nocturnal said, this is quite different, and much clearer than what you said. I imagine your book also said, earlier, that K is a field. If S is a set with n elements, say k1, k2, ... kn, then the "functions" are functions that assign f(k[subthe]i[/sub]= ap, etc., members of K.
Do you understand that the set of vectors {(1, 0, 0, ..., 0), (0, 1, 0, ..., 0), ... , (0, 0, ..., 1)} form a basis for K^n? What about the function
f(k1= 1, f(kn)= 0 for n n$\ne$ 1?
What about the function g(k2)= 1, g(kn)= 0 is n$\ne$ 2?