- #1

- 32

- 6

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.

- I
- Thread starter Santiago24
- Start date

- #1

- 32

- 6

- #2

Staff Emeritus

Science Advisor

Gold Member

- 5,544

- 1,479

Sure. For example, there is a set theoretic bijection between ##\mathbb{Q}## and ##\mathbb{Q}^2##, since they are both countable. So you could take elements of ##\mathbb{Q}##,l and define addition/multiplication by mapping them to ##\mathbb{Q}^2##, do you normal addition/multiplication there, then mapping the result back to ##\mathbb{Q}##. The result is turning ##\mathbb{Q}## into a two dimensional vector space, but the definition of addition and multiplication are going to be some gibberish function you've never seen before (they probably won't even be continuous in the normal topology)

Last edited:

- #3

- 5,695

- 2,475

I think ##R^2## under this new operation of addition and the usual multiplication by scalar, remains a vector space with neutral element (0,0) but the opposite of an element ##(x,y)## is not ##(-x,-y)## it is instead ##(-y,-x)## and everything else remains the same.

- #4

Homework Helper

2022 Award

- 2,667

- 1,272

I think there are probably infinite ways on how exactly we can define the two operations in such a way to satisfy the properties of a vector space. For example let's take the vector space ##R^2##. I think we can "redefine" the operation of addition as $$(x_1,y_1)\mathbf{+}(x_2,y_2)=(x_1+y_2,y_1+x_2)$$ (instead of ##(x_1+x_2,y_1+y_2)##.

Vector addition must be commutative. This operation is not: [tex]

(x_2, y_2) + (x_1, y_1) = (x_2 + y_1, y_2 + x_1) \neq (x_1, y_1) + (x_2, y_2).[/tex] Let's consider what conditions we need in order for [tex]\mathbf{x}_1 \oplus \mathbf{x_2} \equiv A_1\mathbf{x}_1 + A_2\mathbf{x}_2[/tex] to be an abelian group on [itex]\mathbb{R}^2[/itex], where [itex]A_1[/itex] and [itex]A_2[/itex] are square matrices and operations on the right hand side have their usual meaning. (Your example has [itex]A_1 = I[/itex] and [itex]A_2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}[/itex].)

Commutativity appears to require [tex]

(A_1 - A_2)\mathbf{x} = 0[/tex] for all [itex]\mathbf{x} \in \mathbb{R}^2[/itex], so [itex]A_1 = A_2 = A[/itex].

For associativity we need [tex]

A(A \mathbf{x}_1 + A\mathbf{x}_2) + A\mathbf{x}_3 = A\mathbf{x}_1 + A(A \mathbf{x}_2 + A \mathbf{x}_3)[/tex] for every [itex]\mathbf{x}_1[/itex], [itex]\mathbf{x}_2[/itex] and [itex]\mathbf{x}_3[/itex]. From this it follows that [tex]A^2 = A,[/tex] so [itex]A[/itex] is either the identity matrix or zero, and taking zero violates the inverse axiom.

- #5

- 5,695

- 2,475

what about if we redefine addition as multiplication, I mean $$(x_1,y_1)\oplus (x_2,y_2)=(x_1x_2,y_1y_2)$$ then this I think satisfies commutativity and associativity , the neutral element becomes (1,1) (instead of 0,0) and the opposite of (x,y) is (1/x,1/y). Is then ##R^2-(0,0)## a vector space under this redefined addition (and the usually defined multiplication by scalar)? What do you think @pasmith ?

- #6

- 5,695

- 2,475

Well anyway I still believe there infinite ways to define the two operations of the vector space in such a way that the properties of a vector space hold.

- #7

Homework Helper

2022 Award

- 2,667

- 1,272

what about if we redefine addition as multiplication, I mean $$(x_1,y_1)\oplus (x_2,y_2)=(x_1x_2,y_1y_2)$$ then this I think satisfies commutativity and associativity , the neutral element becomes (1,1) (instead of 0,0) and the opposite of (x,y) is (1/x,1/y). Is then ##R^2-(0,0)## a vector space under this redefined addition (and the usually defined multiplication by scalar)? What do you think @pasmith ?

That fails: (x,0) has no inverse. You can avoid that by taking your set of vectors to be [itex](\mathbb{R} \setminus \{0\})^2 \subsetneq \mathbb{R}^2 \setminus \{(0,0)\}[/itex], but you still have a problem with distributivity of scalar multiplication: [tex]a(\mathbf{x}_1 \oplus \mathbf{x}_2) = (ax_1x_2, ay_1y_2) \neq (a^2x_1x_2,a^2y_1y_2) = (a\mathbf{x}_1) \oplus (a\mathbf{x}_2)[/tex] (and you would need to assign a meaning to [itex]0\mathbf{x}[/itex] since (0,0) is not one of your vectors).

Edit: [itex]a(x,y) = (x^a, y^a)[/itex] might work if you restrict your vectors to [itex](0,\infty)^2[/itex].

- #8

- 5,695

- 2,475

Yes this redefinition of scalar multiplication would fix the distributivity problem and also assign (1,1) the neutral element to ##0\mathbf{x}##. And I don't see any other problems.Edit: a(x,y)=(xa,ya) might work if you restrict your vectors to (0,∞)2.

Share:

- Replies
- 4

- Views
- 827

- Replies
- 4

- Views
- 365

- Replies
- 4

- Views
- 899

- Replies
- 5

- Views
- 866

- Replies
- 1

- Views
- 152

- Replies
- 2

- Views
- 936

- Replies
- 5

- Views
- 914

- Replies
- 24

- Views
- 1K

- Replies
- 48

- Views
- 6K

- Replies
- 38

- Views
- 4K