Dimension of a vector space $U = \{(x,y,z)\in\mathbb{R}^{3}: x-y+2z=0\}$ Let $U = \{(x,y,z)\in\mathbb{R}^{3}: x-y+2z=0\}$
My lecturer says without explanation: 'Intuitively $U$ has dimension $2$'. Can someone explain his intuition?
 A: The map
$$f\colon \Bbb R^3\rightarrow \Bbb R,\quad (x,y,z)\mapsto x-y+2z$$
is a non zero linear form so
$$U=\ker f$$
is an hyper-plan of $\Bbb R^3$ so $\dim U=2$.
Edit: We can find the dimension of $U$ using this geometric method which's pretty intuitive:
First notice that $U$ is a subspace of $\Bbb R^3$ so its dimension is less or equal $2$ (clearly $U\ne\Bbb R^3$). Moreover:


*

*the intersection of $U$ with $x-y$ plane i.e. the plane $\{z=0\}$ is 
$$\{(x,y,z)\in\Bbb R^3\;:\; x=y\}$$
which is a line and

*the intersection of $U$ with $x-z$ plane i.e. the plane $\{y=0\}$ is 
$$\{(x,y,z)\in\Bbb R^3\;:\; x=-2z\}$$
which is also a line so $U$ contains two non identical lines hence $U$ is a plane and its dimension is $2$.

A: The intuition behind is you have three variables combinated linearly so one of them is combination of the other two therefore you have two "degrees of freedom"
$$(x,y,z)=\left(x,y,\frac{y-x}{2}\right)$$
$$(x,y,z)=x\left(1,0,\frac{-1}{2}\right)+y\left(0,1,\frac{1}{2}\right)$$
Then all the $(x,y,z) \in U$ are linear combination of $\left(1,0,\frac{-1}{2}\right)$ and $\left(0,1,\frac{1}{2}\right)$ and since they  are independent the dimension is $
2$.
