Show that the following set of elements in R^3 form subspaces.

(a) The set of all (x,y,z) such that x + y + z = 0

What is the vector i would use:

(-z-y, -z-x, -y-x)?

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- Feb 8th 2010, 12:38 PMmillerstSubspace
Show that the following set of elements in R^3 form subspaces.

(a) The set of all (x,y,z) such that x + y + z = 0

What is the vector i would use:

(-z-y, -z-x, -y-x)? - Feb 8th 2010, 12:44 PMRaoh
- Feb 8th 2010, 12:57 PMmillerst
So:

$\displaystyle

(-y-z,y,z)+(-y-z,y,z) = (-2y-2z,2y,2z)

$

So it is closed under addition.

$\displaystyle q(-y-z,y,z) = (q(-y-Z),qy,qz)$

So it is closed under multiplication.

$\displaystyle (-y-z,y,z)+(-1)(-y-z,y,z) = 0$

So zero vector is in the set?

Does this prove it? :S - Feb 8th 2010, 01:47 PMRaoh
- Feb 9th 2010, 03:08 AMHallsofIvy
This doesn't prove it is closed under addition, only that it is closed under addition of a vector to

**itself**. You need to show that it is closed under addition of any two vectors:

(-y-z, y, z)+ (-b-c, b, c)= (-(y+b)-(z+c), y+b, z+c).

Or you could do it this way: If (a, b, c) and (x, y, z) are in this set then x + y + z = 0 and a+ b+ c= 0.

Then (a, b, c)+ (x, y, z)= (a+x, b+ y, c+ z) and (a+x)+ (b+y)+ (c+ z)= (a+ b+ c)+ (x+ y+ z)= 0+ 0= 0 so this sum vector satisfies the same condition and is in the subspace.

Quote:

$\displaystyle q(-y-z,y,z) = (q(-y-Z),qy,qz)$

So it is closed under multiplication.

$\displaystyle (-y-z,y,z)+(-1)(-y-z,y,z) = 0$

So zero vector is in the set?

Does this prove it? :S