I need to prove that this is a linear subspace:
$x-y+2z=0$ $x+y+4z=0$
How am I going to do this?
I need to prove that this is a linear subspace:
$x-y+2z=0$ $x+y+4z=0$
How am I going to do this?
What you’ve written isn’t a linear subspace of anything: it’s a pair of equations. The first step is to state the problem correctly:
Let $V$ be the set of vectors $\langle x,y,z\rangle\in\Bbb R^3$ such that $x-y+2z=0$ and $x+y+4z=0$; then $V$ is a subspace of $\Bbb R^3$.
In order to show this, you must show three things:
(It’s possible to combine the last two into a single statement, but at this point it’s probably simplest for you to keep the ideas separate.)
Is $x=0,y=0,z=0$ a solution to both of the equations $x-y+2z=0$ and $x+y+4z=0$? Clearly it is, so $\langle 0,0,0\rangle\in V$, and $V$ is not empty.
To check that $V$ is closed under vector addition, suppose that $\vec u=\langle x_1,x_2,x_3\rangle$ and $\vec v=\langle y_1,y_2,y_3\rangle$ are in $V$; this means that $x_1-x_2+2x_3=0$ and $x_1+x_2+4x_3=0$ (since $\vec u$ is in $V$), and that $y_1-y_2+2y_3=0$ and $y_1+y_2+4y_3=0$ (since $\vec v$ is in $V$. Now
$\vec u+\vec v=\langle x_1+y_1,x_2+y_2,x_3+y_3\rangle\;,$
and you need to show that this vector is in $V$. To do so, you must show that
$(x_1+y_1)-(x_2+y_2)+2(x_3+y_3)=0$ and $(x_1+y_1)+(x_2+y_2)+4(x_3+y_3)=0\;;$
I’ll leave that to you.
To check that $V$ is closed under scalar multiplication, proceed similarly: let $\vec v=\langle x_1,x_2,x_3\rangle$ be any vector in $V$ and $\alpha$ any scalar. You know that $x_1-x_2+2x_3=0$ and $x_1+x_2+4x_3=0$ (why?), and you know that $\alpha\vec v=\langle \alpha x_1,\alpha x_2,\alpha x_3\rangle$; what must you prove about $\alpha x_1,\alpha x_2$, and $\alpha x_3$ in order to show that $\alpha\vec v\in V$?