# Math Help - Subspace

1. ## Subspace

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)?

2. Originally Posted by millerst
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)?
hello
$(-y-z,y,z)=y(-1,1,0)+z(-1,0,1)$

3. So:

$

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

So it is closed under addition.

$q(-y-z,y,z) = (q(-y-Z),qy,qz)$
So it is closed under multiplication.

$(-y-z,y,z)+(-1)(-y-z,y,z) = 0$
So zero vector is in the set?

Does this prove it? :S

4. Originally Posted by millerst
So:

$

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

So it is closed under addition.

$q(-y-z,y,z) = (q(-y-Z),qy,qz)$
So it is closed under multiplication.

$(-y-z,y,z)+(-1)(-y-z,y,z) = 0$
So zero vector is in the set?

Does this prove it? :S
make it easier,
$E=\left \{ (x,y,z)\in \mathbb{R}^3\mid x+y+z=0 \right \}$ $=\left \{ y(-1,1,0)+z(-1,0,1)
\mid y,z\in \mathbb{R }\right \}=\texttt{Span}((-1,1,0),(-1,0,1))$

and that proves it.

5. Originally Posted by millerst
So:

$

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

So it is closed under addition.
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.

$q(-y-z,y,z) = (q(-y-Z),qy,qz)$
So it is closed under multiplication.

$(-y-z,y,z)+(-1)(-y-z,y,z) = 0$
So zero vector is in the set?

Does this prove it? :S