Let H be the set of points inside and the unit circle in the xy-plane. THat is, let H={[x y] |x^2+y^2

Question

Let H be the set of points inside and the unit circle in the xy-plane. THat is, let H={[x y] |x^2+y^2<=1}. Find a specific example--two vectors or a vector and a scalar-to show that H is not a subspace of R^2.

raffle_snaffle · Answer

@Loser66

raffle_snaffle · Answer

I let v=[1/root(2) ,1/root(2)]

raffle_snaffle · Answer

then v+v=[1/root(2) ,1/root(2)]+[1/root(2) ,1/root(2)]

raffle_snaffle · Answer

solution will be v=[root(2) root(2)]

raffle_snaffle · Answer

that is suppose to be sqrt

loser66 · Answer

let u = (x, y) $\in H$ so, x^2+y^2 =1 
let v=(z, t) $\in H $ so that z^2 + t^2 =1
so, u + v is supposed to be in H, right?
now see, u+v = x^2 +y^2 +z^2+t^2 =4 ( not $\leq 1) $, therefore, H is not a subspace of R^2

loser66 · Answer

sorry, =2, hehehe

raffle_snaffle · Answer

how is u+v in H?

loser66 · Answer

by definition of the subspace , V is a vector space, (this case is R^2)
H is a set in V , H is a subspace when :
1/ u in H, v in H , then u + v in H
2/ there is a scalar a such that u in H , a*u in H 
If it satisfies 2 conditions above, then H is a subspace.
Your problem is prove it is NOT a subspace, so, just pick the easiest way to show that it is NOT satisfy either of the conditions. I pick =1 to easy compute. 
if you pick < 1, it's hard to prove that the sum is >1 , fortunately, your problem  give us the case =1. so?? take advantage on it. hehehe

loser66 · Answer

woaah, Zarkon is here. let him give us his opinion. I dare not to say anything in the front of him. In other word, I am intimidated by hiiiiiim, hhehehehe.

raffle_snaffle · Answer

That helped out a lot.

zarkon · Answer

you could also use...

$$(1/2,0)\in H$$

but

$$4(1/2,0)=(2,0)\notin H$$

raffle_snaffle · Answer

How does  u+v = x^2 +y^2 +z^2+t^2 =4 ( not ≤1)

loser66 · Answer

I correct it, it is =2 , not 4 ( I am sorry for that)
hey, !!! how 2 $\leq $1?

raffle_snaffle · Answer

why did you say x^2 +y^2 +z^2+t^2 why did you include z and t that would make R^4

loser66 · Answer

2 is NOT $\leq $1 , right? so, u + v is not in H, right?

raffle_snaffle · Answer

Yes I agree. Lol

loser66 · Answer

x,y is 2 elements of vector u in H so that x^2 + y^2 =1 ,
Ok, take another circle to make it different v = x^2 + y^2 = 0.9

loser66 · Answer

|dw:1392590401697:dw|

raffle_snaffle · Answer

^Yes yes I agree.

loser66 · Answer

|dw:1392590468202:dw|