Need some guidance at this task:
I have $k_1a + k_2b + k_3c = 0$ for arbitrary but fixed numbers $k_1, k_2, k_3 \in \mathbb{R}$ and need to show if (let's call it $V$) is a subspace of $\mathbb{R^3}$.
$i)\ V \neq \emptyset \\ ii)\ \forall a,b,c \in V : a+b \in V \\ iii)\ \forall a \in V : \forall k \in K : k \cdot a \in V $
Okay, usually I know how to show the subspace properly, but I am confused a little bit by the $k_1, k_2, k_3$ variables.
Could anyone help me out how to show it properly?
Thanks in advance!
Let $k_1,k_2,k_3\in\mathbb{R}$ be fixed, and let
$$V=\{(x,y,z)\in\mathbb{R}^3 : k_1x+k_2y+k_3z=0\}.$$
We show that $V$ is a subspace of $\mathbb{R}^3$. Now as
$$k_1\cdot 0+k_2\cdot 0+k_3\cdot 0=0$$
we have that $(0,0,0)\in V$. Now let $(x_1,y_1,z_1),(x_2,y_2,z_2)\in\mathbb{R}^3$. Then, as
$$k_1(x_1+x_2)+k_2(y_1+y_2)+k_3(z_1+z_2)=(k_1x_1+k_2y_1+k_3z_3)+(k_1x_2+k_2y_2+k_3z_3)=0+0=0,$$
it follows that $(x_1,y_1,z_1)+(x_2,y_2,z_2)\in V$. Finally, let $\lambda\in\mathbb{R}$ and $(x,y,z)\in V$. Then, as
$$k_1(\lambda x)+k_2(\lambda y)+k_3(\lambda z)=\lambda(k_1x+k_2y+k_3z)=\lambda\cdot 0=0,$$
it follows that $\lambda (x,y,z)\in V$. Thus $V$ is a subspace of $\mathbb{R}^3$.