1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Linear Algebra Isomorphism

  1. Aug 31, 2006 #1
    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


    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. jcsd
  3. Aug 31, 2006 #2
    No one out there is familiar with isomorphism?
  4. Aug 31, 2006 #3
    :confused: 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
  5. Sep 1, 2006 #4
    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).
  6. Sep 1, 2006 #5
    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 [itex] s_i[/itex] for i =1,2,...,n.
    Let G be the vector space of all functions mapping S into K.

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

    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 [itex]s_i \mapsto a_i[/itex], 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
  7. Sep 1, 2006 #6


    User Avatar
    Science Advisor

    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[itex]\ne[/itex] 1?
    What about the function g(k2)= 1, g(kn)= 0 is n[itex]\ne[/itex] 2?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook