Dimension of a basis of a subset I am just wondering why if in $\mathbb{R}^3$ any equation of the form $ax+by+cz = 0$ will be a subspace of dimension two. Is it because it is equal to zero, therefore it is a plane through the origin hence in $\mathbb{R}^3$ it will be of dimension two? If I'm wrong can you please explain.
Thanks
 A: Your reasoning is correct. To fill in the technical details: You could see this as the kernel of the mapping $\mathbb{R}^3\rightarrow\mathbb{R}$ represented by the matrix $(a,b,c)$. If one of $a,b,c$ is nonzero, then this matrix has rank one and by the dimension formula the kernel has dimension $3-1=2$. 
(Otherwise it has of course dimension three by the same formula or because your equation reduces to $0=0$)
A: We have that
$$ax + by + cz = 0 \Leftrightarrow (a,b,c) \cdot (x,y,z) = 0 \\\Leftrightarrow (a,b,c) \perp (x,y,z) \Leftrightarrow \forall \alpha \in \mathbb{R}:\alpha(a,b,c) \perp (x,y,z)$$
Then the set of solutions of this equation is the orthogonal complement of $\operatorname{span}\{(a,b,c)\}$ which gives us that it is dimension $2$.
A: Consider $Y=\{(x,y,z) \in \mathbb{R}^3:ax+by+cz=0\}$. It's linear subspace of $\mathbb{R}$, because:
1.$(0,0,0) \in Y$.
2.If $v,w \in Y$ then $v+w \in Y$.
3.If $v \in Y$ and $\alpha \in \mathbb{R}$, then $\alpha v \in Y$.
You always should check these three conditions to tell that something is linear subspace.
For example, $Z=\{(x,y,z) \in \mathbb{R}^3:ax+by+cz=1\}$ isn't linear subspace, because if $v=(x_1,x_2,x_3) \in Z$ then $ax_1+bx_2+cx_3=1$, but for $2v=(2x_1,2x_2,2x_3)=(y_1,y_2,y_3)$ you have $ay_1+by_2+cy_3=2$.
