Is $2x-y = 0$ a plane or line in 3d?

Question

I am going through Strang's linear algebra video 1 (https://www.youtube.com/watch?time_continue=1131&v=ZK3O402wf1c), and he seemed to suggest that the equation:

$2x - y = 0$

is a plane in 3 dimensions. However, isn't $z$ zero in this equation? Therefore, this would describe a line in 3d not a plane. I think his logic was that the omitted term is $0z$, so $z$ can be anything.

That makes sense, but I also don't understand how the same equation can be a line in 2d but a plane in 3d.

Answer 1

$z$ can be anything at all in this case, yes. The reason this is a line in 2d and a plane in 3d is because in 3d there's "room" for it to grow into a plane. Similarly, this equation considered in 4d space would give a 3-dimensional hyperplane, and in $n$d space it gives an $n-1$ dimensional hyperplane, because there is "room to grow" into in all these situations.

Answer 2

$z$ is not always zero because $z$ can take any value. The equation that you have defines the plane by two directions. One of them is given by the line in the $xy$ plane, the other is along the $z$ axis.

Answer 3

There are a few ways to see this.

$1.$ $2x-y=0$ defines a line in the $(x,y)-$plane which is then shifted in $z$ because if $(x_0,y_0)$ satisfies $2x-y$, then we are free to choose $z_0$ so that $(x_0,y_0,z_0)$ satisfies the equation. So, we get a line in the $(x,y)-$plane shifted in the $z$ direction forming a plane.

$2.$ This is the kernel of the linear map $T:\mathbb{R}^3\to \mathbb{R}$ given by $T(x,y,z)=2x-y$. Now, $\dim\operatorname{range}(T)=1$, so that by the rank nullity theorem, $\dim\ker(T)=2$. This tells us that $$ \dim\ker(T)=\dim\{(x,y,z)\in \mathbb{R}^3: 2x-y=0\}=2.$$ So, the dimension of the solution set to that equation is $2-$dimensional, i.e. a plane.

$3.$ Alternatively, we can view the set of solutions of the equation $2x-y=0$ as the set of elements $(x,y,z)$ in $\mathbb{R}^3$ such that $$ (2,-1,0)\cdot(x,y,z)=0.$$ That is, the solutions to the equation are the set of vectors in $\mathbb{R}^3$ orthogonal to $(2,-1,0)$. Such vectors form a plane.

Answer 4

First, let's look at why $2x-y=0$ represents a line in 2D. All the points $(a, 2a)$ satisfies the equation, giving you:

If we allow for a 3rd coordinate, we get all the points $(a, 2a, c)$ where $c$ can be any thing and the equation will still hold true, which gives you the following:

In fact, if you were to restrict $z$ to be $0$, then you do in fact get a line in 3 space.

Answer 5

The confusion appears to arise from Strang conflating equations and sets. I.e. "$2x-y=0$" isn't really a plane or a line; it's just an equation. Instead it's the set $\{(x,y,z) \in \mathbb{R}^3 | 2x-y=0\}$ that is a plane. (If you're unfamiliar with this set notation, read it as, "The set of triples $(x,y,z)$ such that $2x-y=0$.")

Note that this set is different to both $\{(x,y,z) \in \mathbb{R}^3 | 2x-y=0, z=0\}$ (the line in 3-D space given by the same equation from before and also the extra condition that $z=0$) and the set $\{(x,y) \in \mathbb{R}^2 | 2x-y=0\}$ (the line given by the same equation in 2-D space).

Once you're more experienced, you'll be able to work out the sets other mathematicians are talking about from context without them having to spell out the whole thing (e.g. Strang talking about $2x-y=0$ in 3 dimensions is clearly referring to $\{(x,y,z) \in \mathbb{R}^3 | 2x-y=0\}$ rather than any other set). But until then, the extra formalism of the set notation can help you understand the difference between each of the things you had confused.