Understanding Vector Spaces with Mysterious Coefficients

[vampyric]

Homework Statement

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)

for addition on V

and

λ*(x,y)=(λx+λ-1,λy+y-1) (λ∈ℝ)

for scalar multiplication

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.

I have looked at https://www.physicsforums.com/showthread.php?t=388909

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

Please haallp!

 

[Bacle2]

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

 

[Fredrik]

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.

 

[vampyric]

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

So I'm not sure if I get this
For an addition operation (x,y)+(x',y')
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.

 

[vampyric]

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.

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?

 

[Mark44]

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

So I'm not sure if I get this
For an addition operation (x,y)+(x',y')
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?

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).

 

[Mark44]

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?

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.

 

[vampyric]

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.

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

As for

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).

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

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

 

[Fredrik]

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.)

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

 

[Fredrik]

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?

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').

 

[Mark44]

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

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

 

[HallsofIvy]

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.

 

[Fredrik]

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.