How can isomorphism be used to show equivalence between sets in linear space?

In any case, your book also said that a linear transformation, f, is "onto" if and only if the range of f is the entire codomain. That is, for every vector in the codomain, there is some vector in the domain that is mapped to it. If that is the definition of "onto" you are using, it should be easy to see that the function f:G->K^n defined byf(g(k1), g(k2), ... , g(kn))= (g(k1), g(k2), ... , g(kn)) is onto. Given any vector in K^n, say (a1, a2, ... an), let g(k1)=
  • #1
wurth_skidder_23
39
0
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)
 
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  • #2
No one out there is familiar with isomorphism?
 
  • #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.
 
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  • #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).
 
  • #5
wurth_skidder_23 said:
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).
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.

wurth_skidder_23 said:
G := Functions with values in K

f: G -> K

(g(s1),...,g(sn))
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.
we show f is one to one:

let g,h be elements of G such that f(g) = f(h)

by definition of f we get, (g(s1),...,g(sn)) = (h(s1),...,h(sn))

therefore, g(s1) = h(s1), ..., g(sn) = h(sn)
a cleaner way to assert this is that [tex]g(s_i) = h(s_i)[/tex] for all i
Therefore g = h
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.

We show Linearity:
let c1, c2 be in K and let g,h be in G

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)

Therefore G is isomorphic to K^n

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 .
 
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  • #6
wurth_skidder_23 said:
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).

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?
 

What is a linear algebra isomorphism?

A linear algebra isomorphism is a bijective linear transformation between vector spaces that preserves the algebraic structure. This means that an isomorphism maps vectors from one vector space to another in a way that maintains their linear combinations and operations.

What is the difference between an isomorphism and an automorphism?

An isomorphism is a bijective linear transformation between two vector spaces, while an automorphism is an isomorphism from a vector space to itself. In other words, an automorphism is a special case of an isomorphism where the original and target vector spaces are the same.

How do you prove that two vector spaces are isomorphic?

To prove that two vector spaces are isomorphic, you need to show that there exists a linear transformation between them that is both one-to-one and onto. This can be done by constructing an isomorphism mapping between the two vector spaces and proving its linearity and bijectivity.

What is the significance of isomorphism in linear algebra?

Isomorphism plays a crucial role in linear algebra as it allows us to study different vector spaces that have similar structures. It also helps us to simplify complex vector spaces by mapping them to simpler ones and gaining a better understanding of their properties.

What are some real-world applications of isomorphism in linear algebra?

Isomorphism has various applications in fields such as physics, engineering, and computer science. For example, in physics, isomorphisms are used to study the symmetries of physical systems. In engineering, isomorphisms are used to model and analyze structures, such as bridges and buildings. In computer science, isomorphisms are used to optimize algorithms and data structures.

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