- #1

Bob

- 29

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Bob
- Start date

- #1

Bob

- 29

- 0

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

- #2

shmoe

Science Advisor

Homework Helper

- 1,994

- 1

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

Science Advisor

Homework Helper

- 43,021

- 970

- #4

MaxManus

- 277

- 1

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

Homework Helper

- 3,304

- 2

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

Last edited:

- #6

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

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

Last edited by a moderator:

- #7

MaxManus

- 277

- 1

Thanks to both of you.

Share:

- Last Post

- Replies
- 4

- Views
- 276

- Last Post

- Replies
- 6

- Views
- 369

- Replies
- 61

- Views
- 2K

- Last Post

- Replies
- 7

- Views
- 840

- Last Post

- Replies
- 6

- Views
- 270

- Replies
- 4

- Views
- 450

- Replies
- 12

- Views
- 472

- Last Post

- Replies
- 3

- Views
- 454

- Replies
- 15

- Views
- 790

- Replies
- 26

- Views
- 1K