# Determining vector spaces

1. Mar 13, 2013

### vampyric

1. The problem statement, all variables and given/known data

Hi there, I'm very new to vector spaces and just can't seem to figure this one problem out.

The question ask's to determine if (V,+,*) is a vector space.

I am given

V=R^2

(x,y)+(x',y')=(x+x'+1,y+y'+1)

and

λ*(x,y)=(λx+λ-1,λy+y-1) (λ∈ℝ)

for scalar multiplication

3. The attempt at a solution

I think I can work through the axioms simple enough but I cannot figure out where the +1's and -1's have come from.

and it seems to explain everything I need to know except where these mysterious $\pm$1's have come from.

Last edited: Mar 13, 2013
2. Mar 13, 2013

### Bacle2

Sorry, I don't know what your question is. What are you asking?

3. Mar 13, 2013

### Fredrik

Staff Emeritus
They didn't come from anywhere. Those equalities are (the key parts of) definitions of a non-standard addition operation on ℝ2 and a non-standard scalar multiplication operation on ℝ2. I assume that the problem is asking you to find out if ℝ2 with the operations defined by these equalities is a vector space.

A tip for next time: When you start a thread about a problem, you should at least mention what the problem is asking for.

4. Mar 13, 2013

### vampyric

Sorry about that, I was in a rush to get to class.

So I'm not sure if I get this
from definition you just simply add +1 to each coefficient...is this what you mean?

I haven't noticed it being done for the case (x,y,z), is this because it is in R^3?

Thanks! Apologies once again for my vagueness.

5. Mar 13, 2013

### vampyric

So taking this with respect to axiom 1: v+w=w+v for all v,w ∈ V

(x',y')+(x,y)=(x'+x+1,y'+y+1)

Is this correct?

6. Mar 13, 2013

### Staff: Mentor

A vector space is defined by a set of things and two operations - one that defines addition between two members of the set, and another that defines multiplication of things in the set by a scalar.

In your problem, the set is R2, and the operations are as you showed them in post 1. Since the set is R2, which consists of ordered pairs, elements of R3 aren't relevant to the problem, so there is no case for (x, y, z).

7. Mar 13, 2013

### Staff: Mentor

This is correct for (x',y')+(x,y). Now you need to show that (x, y) + (x', y') is equal to the same value. Some texts use special characters to indicate the defined operations, with $\bigoplus$ for the addition and $\bigodot$ for scalar multiplication.

For this problem you would have these definitions:
(x, y) $\bigoplus$ (x', y') = (x + x' + 1, y + y' + 1)
λ $\bigodot$ (x, y) = (λx + λ - 1, λy + y - 1)

In LaTeX these are \bigoplus and \bigodot, respectively. (I had to look them up.)

The ordinary + operator behaves as you would expect, and expressions such as λx indicate ordinary multiplication.

8. Mar 14, 2013

### vampyric

Ohhhhh right. This is actually making sense now. Thank you so much!

As for

Nevermind, I was interpreting it to apply to all ordinary + aswell and was confused as to why I hadn't seen this in R^3.

I will never underestimate the significance of notation again :)
Thanks!

9. Mar 14, 2013

### Fredrik

Staff Emeritus
You should probably use \oplus and \odot in situations like this, and \bigoplus as a replacement for the summation sigma: $$\bigoplus_{i=1}^2 x_i =x_1\oplus x_2.$$

Last edited: Mar 14, 2013
10. Mar 14, 2013

### Fredrik

Staff Emeritus
The left-hand side is defined to be equal to the right-hand side, but this is just the definition of an addition operation on ℝ2. If you want to show that v+w=w+v for all v,w ∈ V, you need to do more than just write down the definition of +. Since V=ℝ2, what you have to do is to use that definition to show that for all x,x',y,y' ∈ ℝ, we have (x',y')+(x,y)=(x,y)+(x',y').

11. Mar 14, 2013

### Staff: Mentor

My biggest concern was to help the OP distinguish between the usual definitions for addition and multiplication, and the ones in this problem.

12. Mar 15, 2013

### HallsofIvy

Staff Emeritus
Yes, and that was what Fredrik was doing. Part of the OP's difficulty was that sign for "normal" addition was being used for this new definition. It can clarify things to use a different symbol for this "new" addition.

13. Mar 15, 2013

### Fredrik

Staff Emeritus
Yes, I think we all agree about everything. I was just trying to offer a little LaTeX tip there, not suggest that anyone was wrong.