# Proving if a subset is a subspace

1. Oct 2, 2011

### nayfie

1. The problem statement, all variables and given/known data

Is the following a subspace of $R^{n}$ for some n?

$W = {(x, y, z) \in R^{3} | 2x - y = 3z + x = 0}$

2. Relevant equations

A subspace of $R^{n}$ is a subset $W$ of $R^{n}$ such that;

1. $0 \in W$
2. $\forall u, v \in W; u + v \in W$
3. $\forall c \in R$ and $u \in W$; $cu \in W$

3. The attempt at a solution

I have checked that the zero vector is contained in the subset, by first letting x = 0.

2x - y = 0, therefore if x = 0, y is also equal to 0.

3z + x = 0, so if x = 0, z is also equal to 0.

The problem here is that now I have no idea how to prove that W is closed under addition and scalar multiplication.

Any help would be greatly appreciated! :)

2. Oct 2, 2011

### HallsofIvy

Remember that definitions in mathematics are "working definitions". That is, you use the precise words of definitions in proofs. Use the definitions of "addition", "scalar product", and "closed under addition and scalar product".

The scalar product of (x, y, z) with scalar k is (kx, ky, kz). Use the fact that 2x- y= 0 and 3z+ x= 0 to show that the same is true of (kx, ky, kz). For example, 2(kx)- ky= k(2x- y)= k(0)= 0. You do the other one.

Let (x, y, z) and (a, b, c) be elements of this space.
Then we know that 2x- y= 0, 3z+ x= 0, 2a- b= 0, and 3c- a= 0.
The sum of two such elements would be (x+a, y+ b, c+ z). Show that this triple satifies the same equations. 2(x+ a)- (y+ b)= (2x- y)+ (2a- b)=

In future problems, you may be asked to do more than just show a subset is a subspace.

Note that from 2x- y= 0 we get y= 2x and from 3z+ x= 0, we get x= -3z so that y= 3x= 2(-3z)= -6z. That is, we can write any member of this subset as (x, y, z)= (-3z, -6z, z)= z(-3, -6, 1) showing that it is, in fact, a one-dimensional subspace of this three-dimensional space and that {(-3, -6, 1)} is a basis for that subspace.

Last edited by a moderator: Oct 2, 2011
3. Oct 2, 2011

### I like Serena

Hi nayfie, long time no see!

I've got 2 approaches for you: one geometrical and one algebraical.

Do you know what W looks like?
That is, can you deduce its shape in terms of planes and lines and such?
(You could even draw it! )

If you pick a specific element of W, say one with x=a, can you say what y and z will be?

And say you pick another one, say with x=b.
Now add them. Does the result satisfy the equations of W?

4. Oct 2, 2011

### nayfie

I have followed both of your replies and have proven that W is indeed a subspace (as it satisfies the conditions). I find it easier to understand what's going on if I have a geometric interpretation.

Thank you both very much for the replies, you've ended a lot of frustration :)

5. Oct 2, 2011

### I like Serena

Good to know!