How do I determine if this is a subset of a subspace?

So I know that for a subset to be a subspace it has to satisfy the following properties:

1. Contain the zero vector
2. Closed under scalar multiplication

I however do not know how to about determining whether:

W = {[x,y,z] for all in R^3 |x-2y+3z = 0}

is a subspace of R^3

• Did you mean "is a subspace" on the last line? Isn't it obvious that $W$ is a subset? Aug 11 '14 at 13:25
• This is a collection of ordered triples of reals, so it is a subset of $\mathbb{R}^3$. Aug 11 '14 at 13:25

Take any $\mathbf{x} = (x,y,z)^T$ in $W$. $\mathbf{x}$ satisfies $x - 2y + 3z = 0$. Now take the vector $a \mathbf{x}$, it satisfies $(ax - 2ay - 3az) = a(x-2y-3z) = a 0 = 0$. So the second condition is satisfied. Now take two vectors $\mathbf{x}_1 =(x_1,y_1,z_1)$ and $\mathbf{x}_2 = (x_2,y_2,z_2)$ in $W$. Both satisfy the equation. As the equation is linear, $c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2$ also satisfies the equation $\forall c_1,c_2 \in \mathbb{R}$. Finally, it is trivial to show that $\mathbf{0} = (0,0,0)$ also satisfies the equation.
One way to do this is consider the function $f:\mathbb{R}^3 \to\mathbb{R}$ given by $$f(x,y,z) := x-2y+3z$$ Show that this function is linear. Now check that for any linear function $f$ as above, the set $$\{(x,y,z)\in \mathbb{R}^3 : f(x,y,z) = 0\}$$ is a subspace of $\mathbb{R}^3$ by checking all the axioms.
It is rather straight forward, all you have to do is follow the three conditions: 1. $0$ abviousliy belongs to W since : $0-2*0+3*0 = 0$ 2. if you take $(x,y,z)$ in W and multiply by a scalar a you get - $ax-a2y+a3z = a(x-2y+3z) = a*0 = 0$. 3. and for the third if you take $(a,b,c),(d,e,f)$ in W and take their sum you get $(a+d,b+e,c+f) = (a+d)-2(b+e)+3(c+f) = (a+2b-3c)+(e-2d+3f) = 0+0 = 0$ so their sum is in the subspace as well.