# Need help with proof of Vector Space (Ten Axioms)

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1. Oct 31, 2014

### Aristotle

Let S={x ∈ R; -π/2 < x < π/2 } and let V be the subset of R2 given by V=S^2={(x,y); -π/2 < x < π/2}, with vector addition ( (+) ).

For each (for every) u V, For each (for every) v ∈ V with u=(x1 , y1) and v=(x2,y2)

u+v = (arctan (tan(x1)+tan(x2)), arctan (tan(y1)+tan(y2)) )

Note: The vectors are ordered pairs of real numbers between -π/2 and π/2, and we are using non-standard vector addition.

Show that V with the designated operations forms a vector space.
Make sure that you show verification for EACH of the ten Vector Space Axioms.

My question is how would you apply the additive identity to prove that u+0=u to show that V forms a vector space?

2. Nov 1, 2014

### Fredrik

Staff Emeritus
Note: I moved this thread here from the linear algebra forum.

You simply find a $z\in V$ such that $u+z=z+u=u$ for all $u\in V$. Once you have found such a $z$, you move on to verify the next axiom.

This is a textbook-style problem, so I have moved it to homework forum. If you need further assistance, you will have to show us your own attempt to solve the problem up to the point where you're stuck.

3. Nov 1, 2014

### RUber

arctan (tan x + tan 0) = ?

4. Nov 1, 2014

### Ray Vickson

How are you defining "multiplication by a scalar": $c \cdot \vec{u}, \: c \in \mathbb{R}, \vec{u} \in V$?

5. Nov 1, 2014

### mathwonk

this seems "trivial", if looked at correctly. i.e. R^2 is a vector space under the usual operations, so if V is any set at all and f:V-->R^2 and g:R^2-->V are mutually inverse bijections, then defining a+b as g(f(a)+f(b)), for all a,b, in V, and ca = g(cf(a)), for all c in R and a in V, makes V a carbon copy of the vector space R^2. That's what your exercise is doing, with f = (tan,tan), and g = (arctan,arctan). I.e. the operations are the ones in R^2, which we know satisfy the axioms, but we are renaming the vectors (x,y) in R^2 as the pair (arctan(x), arctan(y)) in (-π/2,π/2) x (-π/2,π/2).

Checking all this stuff directly is silly in my opinion as it teaches you nothing except review of the axioms. Thus I advise you try to understand what I said here.

6. Nov 1, 2014

### HallsofIvy

Staff Emeritus
That's true if you know that f and g are "one to one, onto" functions.

7. Nov 1, 2014

### Aristotle

I figured that one out...but for the Closure Under Addition, how do I go about actually "testing" to see if u+v is in V? Kinda stuck there..
Do i choose numbers between -pi/2 to pi/2? Somebody please show me one example so I can understand. :(

8. Nov 1, 2014

### RUber

Yes, you have to choose them from the space itself which is defined on (-pi/2 , pi/2).
The areas where this could break down would be for 2 negative values close to -pi/2 or 2 positive values near pi/2.

If you can show that the sum of the tangents is still in the domain of the arctan function, and that the arctan function will still return something in (-pi\2, pi/2), you should be done.

9. Nov 1, 2014

### Aristotle

Okay so here's my approach on it. I chose my u=(x1, y1) and my v=(x2, y2) as to be u=(pi/4,/pi/6) and v=(pi/3,-pi/4) and plugged in my numbers
u+v = (arctan(tan(pi/4)+tan(pi/3)), arctan(tan(pi/6)+tan(-pi/4)) = (arctan(1+srt(3)), arctan( srt(3)/2 - 1))

Is that on the correct path? Also thank you for helping me.

10. Nov 1, 2014

### RUber

I think the broader question is what are the domain and range of the arctan function?
If you can show that the arctan( tan x + tan y ) for x,y in (-pi/2, pi/2) is in that same interval, you don't need to show it with test values.
The test values should help you to get a sense of what the function is doing...and of course if you see something that doesn't work, you have disproof by counterexample.

11. Nov 1, 2014

### Fredrik

Staff Emeritus
That's not the correct path. You don't get to choose the choose the numbers. You have to leave them arbitrary. The proof should start like this:

Let $\mathbf u,\mathbf v\in V$ be arbitrary. Let $a,b,c,d$ be elements of S such that $\mathbf u=(a,b)$ and $\mathbf v=(c,d)$. We have
$$\mathbf u+\mathbf v=(a,b)+(c,d)=\cdots$$

12. Nov 1, 2014

### Aristotle

Ah I see. So to make sure I fully understand the proof you stated...you're saying that as we let u and v be a member of V, a, b, c, d (the domain of x and y for arctan correct? e.g: S={x ∈ R; -pi/2<x<pi/2} , we can have two vectors that will satisfy to result in u+v?

13. Nov 1, 2014

### Aristotle

Well the domain for arctan is all real number..
The range, or output, of Tan–1 x is angles between –90 and 90 degrees or, in radians, between

14. Nov 1, 2014

### RUber

So then, using the method that Fredrik suggested, will the resulting sum be in your space?

15. Nov 1, 2014

### Fredrik

Staff Emeritus
I don't understand your question. Note however that it's not enough to have two vectors $\mathbf u,\mathbf v$ such that $\mathbf u+\mathbf v\in V$. We need $\mathbf u+\mathbf v\in V$ to be true for all $\mathbf u,\mathbf v$ in V.

16. Nov 1, 2014

### Aristotle

Thank you! Also last question..I'm down to my last axiom #9 of scalar multiplication where c1(c2u)=(cd)u

Heres my approach to it...by the way am I suppose to do c1(c2u) or c1(c2v) ? Or does it even matter? Im not sure if it matters whether i use vector u or v to prove this? Should I do v because I'm given c x v = (arctan (c * tan (x2) ), arctan (c *tan(y2) ) ) to prove the last 5 axioms of scalar multiplication?

c1(c2u) = c1(arctan(c2(tan(x1))), arctan(c2(tan(y1))))

= (arctan(c1(tan[arctan(c2(tan(x1)))])), arctan(c1(tan[arctan(c2(tan(y1)))])))

= (arctan(c1(c2(tan(x1)))), arctan(c1(c2(tan(y1)))))

= (arctan(c1c2(tan(x1))), arctan(c1c2(tan(y1))))

= (c1c2)(x1,y1) = (c1c2)u

17. Nov 1, 2014

### RUber

That looks right based on the definition you gave for scalar multiplication.