Why R2 is not a subspace of R3?

  • Thread starter Bob
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  • #1
Bob
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I think R2 is a subspace of R3 in the form(a,b,0)'.
 

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  • #2
shmoe
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R^2 is isomorphic to the subset (a,b,0) of R^3, but it's also isomorphic to infinitely many other subspaces of R^3 (any 2 dimensional one). As such, there's no canonical embedding, and you don't usually think of R^2 as being contained in R^3.

A more obvious explanation is the vector (a,b) is not the same as the vector (a,b,0). 2 components vs 3 components, so they are different objects.
 
  • #3
HallsofIvy
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Shmoe is correct. However, it is common to speak of isomorphic things as if they were the same thing. Most mathematicians would say (with "abuse of terminology") that R2 is a subspace of R3, understanding that what they really mean is that it is isomorphic to one.
 
  • #4
MaxManus
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I know that it is an old thread, but I still don't get why R^2 is not a subspace of R^3. Is it only because R^3 has 3 components and R^2 only 2 components? Is it possible to use the three conditions to show that R^2 is not a subspace of R^3?
1. The zero vector, 0, is in W.
2. If u and v are elements of W, then the sum u + v is an element of W;
3. If u is an element of W and c is a scalar from K, then the scalar product cu is an element of W;
 
  • #5
lanedance
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I think the point in the threads above is that R^2 & R^3 are different objects, before you can discuss whether R^2 is a subspace of in R^3 you need to "embed" R^2 in R^3 by defining an isomorphism between a subset of R^3 & all of R^2, the obvious one being
[tex] (a,b) \in \mathbb{R}^2 \leftrightarrow (a,b, 0) \in \mathbb{R}^3 [/tex]

however as schmoe pointed out there are infinite ways to do it eg. another isomorphsim toa subpsapce of R^3 is
[tex] (a,b) \in \mathbb{R}^2 \leftrightarrow (a,0,b) \in \mathbb{R}^3 [/tex]

i think the key here is, before you can discuss whether elements of R^2 are closed under addition in R^3, you first need to know how you map an element of R^2 into R^3 (the isomorphism)

if you've done that, you should be able to show using the 3 subspace criteria, that R^2 is isomorphic to a subspace of R^3. Then as pointed out, many people would be happy to accept the abuse of terminology and say R^2 is a subspace of R^3, implying there is an isomorphism to a subspace of R^3

however, as an example what if you chose to embed R^2 in R^3 by
[tex] (a,b) \in \mathbb{R}^2 \leftrightarrow (a,b,1) \in \mathbb{R}^3 [/tex]
then clearly the zero vector is not in the embedded R^2, so it is not a subspace of R^3
 
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  • #6
HallsofIvy
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I know that it is an old thread, but I still don't get why R^2 is not a subspace of R^3. Is it only because R^3 has 3 components and R^2 only 2 components? Is it possible to use the three conditions to show that R^2 is not a subspace of R^3?
1. The zero vector, 0, is in W.
2. If u and v are elements of W, then the sum u + v is an element of W;
3. If u is an element of W and c is a scalar from K, then the scalar product cu is an element of W;
To begin with, for W to be a subspace of V, it must be a subset of V. Things in R^2 are of the form (a, b), with two components while things in R^3 are of the form (a, b, c) with three components. Members of R^2 are not members of R^3 so R^2 is not a subset of R^3.

That said, originally, I was a little surprised by the question. It is common to think of R^2 as being a subset of R^3 using the obvious isomorphism to a subspace of R^3: (a, b)-> (a, b, 0). Strictly speaking, it is not R^2 that is a subspace of R^3, it is that subspace. But one has to very strict!
 
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  • #7
MaxManus
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Thanks to both of you.
 

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