Question

Prove that R^n and R^m are isomorphic if and only if m=n

Answer 1

Given: R^n and R^m are isomorphic
n =m

Since we're given a biconditional statement, we must prove two claims:
If R^n and R^m are isomorphic, then m=n
If m = n, then R^n and R^m are isomorphic

Answer 2

R^n and R^m are isomorphic if and only if their dimensions are equal.

Answer 3

Two mathematical structures are said to be isomorphic if there is an isomorphism between them.

Answer 4

Yeah... isn't that enough, @UsukiDoll ?

Answer 5

Oh wait, that's what you're trying to prove, lol
NVM

Answer 6

I just want to make sure that it's correct. that's all/

Answer 7

R^n and R^m are isomorphic if and only if their dimensions are equal.

Answer 8

I know how to solve it

Answer 9

OH GO AWAY SOTY!

Answer 10

so anyway....that one sentence is the proof...umm is that all I need to write?

Answer 11

you can probably do a proof by contradiction

Answer 12

:( really?

Answer 13

case 1 . n > m, case 2 . n < m , both lead to contradiction

Answer 14

That one sentence looks like that which you are trying to prove, @UsukiDoll :(

Answer 15

well if they don't have equal dimensions, there is no way that R^n and R^m can be isomorphic

Answer 16

Yeah... isn't that what you are trying to prove in the first place?

Answer 17

oh got one! like perl said earlier.... and mine as well. If the dimensions aren't equal, then they can't be isomorphic. the only way for R^m and R^n to be isomorphic is when the dimensions are equal m=n

Answer 18

If N > M or N < M, then the dimensions aren't equal.

Answer 19

would that be right @perl

Answer 20

we have to define dimension.

Answer 21

define dimension?

Answer 22

I have an idea... use this fact
Let T be a linear transformation...
\[\huge T: U \rightarrow V\]
\[\huge U = \mathbb{R}^n\]\[\huge V=\mathbb{R}^m\]

And use the fact that that
\[\huge \dim[T(U)]\le \dim(V)\]

Answer 23

Or not... never mind...

Answer 24

no my class isn't using linear transformations for it

Answer 25

oh...

Answer 26

ack how is dimension defined?

Answer 27

Dimension is the cardinality (number of elements) of the basis (which is the maximal linearly independent set) or the linearly independent spanning set of a Vector Space.

Answer 28

o-o

Answer 29

Simply put...
\[\huge \dim(\mathbb{R}^n) =n\]

Answer 30

Ok. HOw's this?
R^n and R^m are isomorphic if and only if their dimensions are equal. However if n > m or n < m, then the dimensions aren't equal and R^n and R^m can't be isomorphic

Answer 31

As long as you're allowed to use the fact that two spaces are isomorphic only if their dimensions are equal, then you're good to go.

Answer 32

yeah I am.

Answer 33

because it's not fancy stuff. I used inner product on a problem that was in chapter one and the professor was fuming.

Answer 34

he was like we didn't learn this. it's a bad idea to put advanced things.

Answer 35

Okay. then it's all good :)

Answer 36

basically what you are proving is the following.
R^m and R^n are isomorphic if and only their dimensions are equal

Answer 37

this sounds like an advanced proof

Answer 38

but we are learning about isomorphics. that's section 4.8 in the book. we just finished chapter 4

Answer 39

This is the question:
Prove that R^n and R^m are isomorphic if and only if m=n

Answer 40

@perl real question ^

Answer 41

My answer:
R^n and R^m are isomorphic if and only if their dimensions are equal. However if n > m or n < m, then the dimensions aren't equal and R^n and R^m can't be isomorphic

Answer 42

I'm wondering if its correct. that is all

Answer 43

I'm wondering the same thing :D

Answer 44

That's not a proof.

Answer 45

so where do I begin to prove this biconditional statement?

Answer 46

I suppose we can stick to tradition. Let m = n

Answer 47

If R^n and R^m are isomorphic, then m=n If m = n, then R^n and R^m are isomorphic

Answer 48

Then show that R^n is isomorphic to R^m

Answer 49

That's the easy bit.

Answer 50

If R^n and R^m are isomorphic, then m=n
If m = n, then R^n and R^m are isomorphic

Answer 51

Obviously the hard part is showing that no isomorphism exists for n not equal to m

Answer 52

that's the contradiction. If m doesn't equal to n, then R^n and R^m are NOT isomorphic

Answer 53

Okay, @UsukiDoll
Clean slate...
Now...
suppose m = n
It is easy to show that R^n is isomorphic to R^m
Right?

Answer 54

But that's what you're trying to prove is it not. You are just restating the problem.

Answer 55

I wish I could just plug in numbers... if m = 4 and n = 4, but that defeats the purpose of a proof

Answer 56

Relax, @UsukiDoll
I'm all about proofs (preferably Abstract Algebra)
But maybe we can do this together :D

Answer 57

So, suppose m = n

Answer 58

sure.

Answer 59

at least I attempted to do a proof, not beg for answers only.

Answer 60

Look that direction is trivial, if n = m then clearly there is an isomorphism just use the identity.

Answer 61

the identity of R^n and R^m?

Answer 62

Yeah, just putting it up for the sake of completion.
Now the other way around is not nearly as easy,
assuming that an isomorphism from R^n to R^m exists...

Answer 63

UsukiDoll if n=m one isomoprhism between the two spaces is just the identity matrix

Answer 64

So... one direction is clear to you?
That
\[\huge m = n \rightarrow \mathbb{R}^n \cong \mathbb{R}^m\]

Answer 65

@UsukiDoll

Answer 66

huh yeah I'm here

Answer 67

That much is clear, though?
So we can proceed to the second part?

Answer 68

yup

Answer 69

Okay, now suppose
\[\huge \mathbb{R}^n \cong \mathbb{R}^m\]

Answer 70

I'm pretty sure the other direction is only possible with a fair bit of abstract algebra.

Answer 71

What the! But I'm in Intro to Linear Algebra O_O

Answer 72

The two fields are related... to an extent... a Vector Space is a group, after all.

Answer 73

@___@

Answer 74

Relax, @UsukiDoll
Now... suppose \[\huge \mathbb{R}^n \cong \mathbb{R}^m\]
okay...?

Answer 75

This type of problem is not easy. You need to somehow show that no isomoprhism can exist if \(n \neq m\)

Answer 76

Possibly a lot easier than you might think @Alchemista ;)

Answer 77

well somehow this question was inside my elementary linear algebra book

Answer 78

@UsukiDoll
I proceed only at your signal :)

Answer 79

there are two claims here
claim 1 : two real finite-dimensional vector spaces that are isomorphic must have the same dimension.
claim 2 . if the dimensions of two real finite dimensional vector spaces are the same then they are isomorphic

Answer 80

there are also 'infinite dimensional' vector spaces, R^infinity

Answer 81

If R^n and R^m are isomorphic, then m=n
If m = n, then R^n and R^m are isomorphic
yes that is true we are given two claims

Answer 82

Yes perl, we already dealt with one direction (claim 2).

Answer 83

ok :)

Answer 84

And I think we are assuming finite dimensionality here.

Answer 85

right, try to prove claim 1 using contradiction.
so you have two isomorphic vector spaces, but they have different dimension. can this lead to something absurd

Answer 86

you mean should two isomorphic vector spaces have different dimensions, the results would be catastrophic.

Answer 87

yes, find a catastrophe

Answer 88

if m isn't equal to n --->

Answer 89

then R^m and R^n aren't isomorphic?

Answer 90

so you are given R^n, and R^m which are isomorphic. let n > m

Answer 91

or let n < m

Answer 92

either way works , yes

Answer 93

then isomorphic of R^n and R^m doesn't exist

Answer 94

you have to prove that

Answer 95

how?

Answer 96

maybe the values of n and m are different

Answer 97

just a question @UsukiDoll : how do u define isomorphic without using linear transformations?