Is ℝ^2 with Custom Scalar Multiplication a Vector Space?

[ilyas.h]

Homework Statement

The set ℝ^2 with vector addiction forms an abelian group.

a ∈ ℝ,

x = $$\binom{p}{q}$$

we put: a ⊗ x = [latex]\binom{ap}{0}[/latex] ∈ ℝ^2; this defines scalar multiplication

ℝ × ℝ^2 → ℝ^2
(p, x) → (p ⊗ x)

of the field ℝ on ℝ^2.

Determine which of the axioms defining a vector space hold for this abelian group ℝ^2 with this scalar multiplication.

Homework Equations

The Attempt at a Solution

a ⊗ x = $$\binom{ap}{0}$$

i have no idea where to begin. We'd have to look at the axioms for the vector space and go from there? which ones would i want to prove are false?

 

[Dick]

Homework Statement

The set ℝ^2 with vector addiction forms an abelian group.

a ∈ ℝ,

x = $$\binom{p}{q}$$

we put: a ⊗ x = $$\binom{ap}{0}$$ ∈ ℝ^2; this defines scalar multiplication

ℝ × ℝ^2 → ℝ^2
(p, x) → (p ⊗ x)

of the field ℝ on ℝ^2.

Determine which of the axioms defining a vector space hold for this abelian group ℝ^2 with this scalar multiplication.

Homework Equations

The Attempt at a Solution

a ⊗ x = $$\binom{ap}{0}$$

i have no idea where to begin. We'd have to look at the axioms for the vector space and go from there? which ones would i want to prove are false?

Use the "tex" tag instead of "latex". And yes, you'd look at all of the vector space axioms. You might want to pay particular attention to the ones involving the scalar product.

 

[ilyas.h]

Use the "tex" tag instead of "latex". And yes, you'd look at all of the vector space axioms. You might want to pay particular attention to the ones involving the scalar product.

i see that you have to 'use' the axioms, but how do i 'use' them?

 

[HallsofIvy]

You don't "use" the axioms, you determine whether or not the axioms are true. What are those axioms and what are they in terms of the definitions given here?

 

[ilyas.h]

You don't "use" the axioms, you determine whether or not the axioms are true. What are those axioms and what are they in terms of the definitions given here?

(a, x) → (a ⊗ x)

a ⊗ ($$\binom{p}{q}\binom{r}{s}$$) = a ⊗ $$\binom{pr}{qs}$$ = $$\binom{apr}{0}$$

(a ⊗ $$\binom{p}{q})\binom{r}{s}$$) = $$\binom{ap}{0}\binom{r}{s}$$ = $$\binom{apr}{0}$$is this correct? I am trying to prove multiplicative distributuvuty.

 

[HallsofIvy]

"multiplicative distributivity" means that a(u+ v)= au+ av for scalar a and vectors u and v. I don't see any addition of vectors in that. In fact, you have not stated what the vectors are and what the scalar is. Perhaps you mean "associativity"? That a(bv)= (ab)v for scalars a and b and vector v?

(Using "itex" instead of "tex" let's you keep everything on one line.)

 

[Dick]

a ⊗ ($\binom{p}{q}\binom{r}{s}$)

im trying to prove multiplicative distributuvuty.

Where did you find that axiom? In a vector space multiplication of vectors is not generally defined. I'll give you a big hint. Can you find an axiom that would tell you what 1⊗$\binom{p}{q}$ should be? State it. That's the scalar product of 1 with a vector.

 

[ilyas.h]

⊗

Where did you find that axiom? In a vector space multiplication of vectors is not generally defined. I'll give you a big hint. Can you find an axiom that would tell you what 1⊗$\binom{p}{q}$ should be? State it. That's the scalar product of 1 with a vector.

"multiplicative distributivity" means that a(u+ v)= au+ av for scalar a and vectors u and v. I don't see any addition of vectors in that. In fact, you have not stated what the vectors are and what the scalar is. Perhaps you mean "associativity"? That a(bv)= (ab)v for scalars a and b and vector v?

(Using "itex" instead of "tex" let's you keep everything on one line.)

I think i understand now. I got help irl. It's a bit funky because you have to assume their multiplication is true, when it appears not to be.

There is no vector multiplication axiom in a vector space (you can't multiply vecturs, duh). The first distributivity axiom is as follows:

a, b ∈ Field
u, v ∈ vector space

1st law: v(a + b) = va + vb

proof as in the question...

prove: $\binom{p}{q}$⊗(a + b) = $\binom{p(a+b)}{0}$

proof:

compare with $\binom{p}{q}$⊗(a+b) = a⊗$\binom{p}{q}$ + b⊗ $\binom{p)}{q}$

= $\binom{ap}{0}$ + $\binom{bp}{0}$

= $\binom{ap + bp)}{0}$

= $\binom{p(a+b)}{0}$

therefore the first law of distributivity holds for this multiplication.

Am I going along the right lines? And the ⊗ represents the "interaction" (or binary operation) between the field and the vector space.

 

[Dick]

⊗I think i understand now. I got help irl. It's a bit funky because you have to assume their multiplication is true, when it appears not to be.

There is no vector multiplication axiom in a vector space (you can't multiply vecturs, duh). The first distributivity axiom is as follows:

a, b ∈ Field
u, v ∈ vector space

1st law: v(a + b) = va + vb

proof as in the question...

prove: $\binom{p}{q}$⊗(a + b) = $\binom{p(a+b)}{0}$

proof:

compare with $\binom{p}{q}$⊗(a+b) = a⊗$\binom{p}{q}$ + b⊗ $\binom{p)}{q}$

= $\binom{ap}{0}$ + $\binom{bp}{0}$

= $\binom{ap + bp)}{0}$

= $\binom{p(a+b)}{0}$

therefore the first law of distributivity holds for this multiplication.

Am I going along the right lines? And the ⊗ represents the "interaction" (or binary operation) between the field and the vector space.

Yes, you are getting the hang of it. And, yes, that proves that axiom. Now look through the list of axioms and see if you can find one that's not going to be true. There is one.

 

[ilyas.h]

Yes, you are getting the hang of it. And, yes, that proves that axiom. Now look through the list of axioms and see if you can find one that's not going to be true. There is one.

I think I found the one that doesn't work, it's the additive inverse.

0 in Field.
v in Vector space.

0 + v = v0 ⊗ $\binom{p}{q}$ = $\binom{0p}{0}$ = $\binom{0}{0}$.

Am I correct? thanks. From what I can see, I think this is the only false axiom.

EDIT: I might be wrong, since I am adding an element from the field (zero) into the vector space: 0 + v = v, or perhaps the zero is meant to be the zero vector (?)./

 

[ilyas.h]

⊗

Yes, you are getting the hang of it. And, yes, that proves that axiom. Now look through the list of axioms and see if you can find one that's not going to be true. There is one.

Actually, I think I know now. It's the v + (-v) = 0.

$\binom{p}{q}$ + (-1)⊗$\binom{p}{q}$

= $\binom{p}{q}$ + $\binom{-1p}{0}$

= $\binom{p - p}{q}$

$\binom{0}{q}$ =/= $\binom{0}{0}$

so not true.

 

[Dick]

⊗

Actually, I think I know now. It's the v + (-v) = 0.

$\binom{p}{q}$ + (-1)⊗$\binom{p}{q}$

= $\binom{p}{q}$ + $\binom{-1p}{0}$

= $\binom{p - p}{q}$

$\binom{0}{q}$ =/= $\binom{0}{0}$

so not true.

You're getting closer and that is a valid problem but how do you know (-1)⊗v should be (-v)? I actually don't know exactly what your axiom list looks like. There are variations. Do you have an axiom that says (1)⊗v=v?

 

[ilyas.h]

You're getting closer and that is a valid problem but how do you know (-1)⊗v should be (-v)? I actually don't know exactly what your axiom list looks like. There are variations. Do you have an axiom that says (1)⊗v=v?

here are the axioms that i used:

http://mathworld.wolfram.com/VectorSpace.html

and yes, there is a 1v = v, which i proved rather simply (within the question) on my own. Therefore, -1v must equal -v, so my proof from above should be correct (?)/

v + (-v) =
v + (-1)*v==> v + (-1)⊗v

 

[Dick]

here are the axioms that i used:

http://mathworld.wolfram.com/VectorSpace.html

and yes, there is a 1v = v, which i proved rather simply (within the question) on my own. Therefore, -1v must equal -v, so my proof from above should be correct (?)/

v + (-v) =
v + (-1)*v==> v + (-1)⊗v

Yes, it's correct. It's just a little roundabout. Why don't you just show 1⊗v=v isn't generally true? That's enough. And that's what's REALLY going wrong. Not the additive inverse. v+(-v)=0 is correct. It's just that (-1)⊗v isn't equal to -v.