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I think R2 is a subspace of R3 in the form(a,b,0)'.

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- Thread starter Bob
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- #1

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I think R2 is a subspace of R3 in the form(a,b,0)'.

- #2

shmoe

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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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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

[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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To begin with, for W to be a subspace of V, it must be a

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;

That said, originally, I was a little surprised by the question. It is common to

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- #7

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Thanks to both of you.

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