Vector addition - Vector Spaces

depre87
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Homework Statement


Show if V is a vector space ([a,b,c]|ab>=0). I'm trying to test whether it is closed under
vector addition.




Homework Equations


v=[a1,b1,c1] w=[a2,b2,c2], v and w satisfy ab>=0
a1b1>=0, a2b2>=0
show (a1+a2)(b1+b2)>=0



The Attempt at a Solution


Got to a1b1 + a2b2 +a1b2 + a2b1 after expanding the above equation. Know a1b1 and a2b2 both >=0 however what do I do with the a1b2 and a2b1? I've attempted trying to factor them into a1b1 and have gotten a1b1[(b2/b1)+(a2/a1)] and know a1b1 >=0 however not sure what to do with the terms in the bracket to further prove whether or not it is closed under vector addition.
 
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Hi depre87! :smile:

Have you tried searching a counterexample?
 
I just had a look and I don't understand it enough to apply it to my question, could you enlighten me? thanks for the suggestion though.
 
Just try some numerical values of v and w and see if you can come up with a counterexample...
 
one way you could do it is to picture a & b in a 2D plane, the allowable areas are a&b both +ve, or both -ve. Now try and think of a vector addition that will take you outside of teh allowable areas (the a&b axes may be a good start)
 
Look at scalar multiplication rather than addition. If v is such a vector, what is (-1)v?

Or, similarly, every vector must have an additive inverse.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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