R3 Subspace: Proving S={(x,y,z): √3x = √2y} is a Subspace of R3

[Stefenng]

R3 Subspace - Urgent

Homework Statement

Prove that S={(x,y,z):\sqrt{}3 x=\sqrt{}2 y is a subspace of R³

I'm really confuse with this and I still don't know how to proved it.
Can anyone help me with this?
I really a newbie in this. ><

Homework Equations

The Attempt at a Solution

Should I change the \sqrt{}3x=\sqrt{}2y into x =\sqrt{}2/\sqrt{}3y first?

What should I do to prove that it is null vector, closed under addition and closed under scalar multiplication?

 

[LCKurtz]

You have to just show two things:

If you have two points from R³ in S their sum is in S

If you have a point from R³ in S then a constant times it is in S.

 

[Stefenng]

err...I don't get it still... @.@

let
(x,y,z) = ( <br /> \sqrt{}2/\sqrt{}3 <br /> y,y,z)

= y ( <br /> \sqrt{}2/\sqrt{}3 <br />,1,0) + z( 0,0,1) ?

 

[Mark44]

You need two elements in your set S, not just one.

For example,
u = (x₁, (sqrt(6)/2)x₁, z₁)
and
v = (x₂, (sqrt(6)/2)x₂, z₂)


*Note that instead of solving for x in terms of y, as you did, I solved for y in terms of x, and simplified sqrt(3)/sqrt(2) to sqrt(6)/2.

u and v are elements of S. Can you show that u + v is also an element of S?
Can you show that k*u is an element of S?

 

[Stefenng]

ok...I start to understand a little bit d...

if I substitute inside just like wad u say mark, I find that it is weird,

the equation I try :

sqrt (3) x = sqrt (2 ) y
sqrt (3)/ sqrt (2 )x = y

u = ( 1, sqrt(3)/sqrt(2) 1 , c )
v = ( 2, sqrt(3)/sqrt(2) 2 , d )

u + v = ( 1, sqrt(3)/sqrt(2) , c ) + ( 2, 2 ( sqrt(3)/sqrt(2) ) , d )
= ( 1 +2 , sqrt(3)/sqrt(2) + 2(sqrt(3)/sqrt(2)) , c+d)
= ( 3 , 3 ( sqrt(3)/sqrt(2) ) , c+d )

3 = 3 ( sqrt(3)/sqrt(2) )
hence, x = y ( sqrt(3)/sqrt(2) ?
and about the multiple part, what is K?

 

[Mark44]

ok...I start to understand a little bit d...

if I substitute inside just like wad u say mark, I find that it is weird,

the equation I try :

sqrt (3) x = sqrt (2 ) y
sqrt (3)/ sqrt (2 )x = y

u = ( 1, sqrt(3)/sqrt(2) 1 , c )
v = ( 2, sqrt(3)/sqrt(2) 2 , d )

You need to show that for any points/vectors u and v in S, then u + v is also in S, so you can't pick values for x.

u + v = ( 1, sqrt(3)/sqrt(2) , c ) + ( 2, 2 ( sqrt(3)/sqrt(2) ) , d )
= ( 1 +2 , sqrt(3)/sqrt(2) + 2(sqrt(3)/sqrt(2)) , c+d)
= ( 3 , 3 ( sqrt(3)/sqrt(2) ) , c+d )

3 = 3 ( sqrt(3)/sqrt(2) )
hence, x = y ( sqrt(3)/sqrt(2) ?
and about the multiple part, what is K?

k is any arbitrary constant - you don't get to pick it. You need to show that if u is in S, then ku is also in S, for any k.

 

[Stefenng]

u = (a, (sqrt(6)/2)a, c)
v = (f, (sqrt(6)/2)f, e)

u + v = ( a + f, sqrt (6 ) ( a+f) , c+ e )
a+f = sqrt(6) ( a+f ) ??

as for multiple part,

k(a, (sqrt(6)/2)a, c) =(ka, ka (sqrt(6)/2), kc)

ka = ka (sqrt(6)/2), hence subspace S is the subset or R3?

 

[Mark44]

u = (a, (sqrt(6)/2)a, c)
v = (f, (sqrt(6)/2)f, e)

u + v = ( a + f, sqrt (6 ) ( a+f) , c+ e )
a+f = sqrt(6) ( a+f ) ??

Almost. u + v = (a + f, sqrt(6)/2(a + f), c + e)
Since the y coordinate equals sqrt(6)/2 times the x coordinate, u + v is in S whenever u and v are in S.

The key thing about this subset S is that for any vector u = (a, b, c), u is a member of set S provided that b = sqrt(6)/2 * a.

as for multiple part,

k(a, (sqrt(6)/2)a, c) =(ka, ka (sqrt(6)/2), kc)

ka = ka (sqrt(6)/2), hence subspace S is the subset or R3?

Again, since the y coordinate is sqrt(6)/2 times the x coordinate, ku is in S whenever u is in S, for any arbitrary scalar k.

One thing not mentioned is that every subspace must include the 0 vector. That's easy to show, since 0 * (a, sqrt(6)/2 * a, c) = (0, 0, 0). With (0, 0, 0), 0 = sqrt(6)/2 * 0.

Therefore, S is a subspace of R³.

 

[Stefenng]

O.O"

Solved?