Is $\{(x,y) \in \Bbb{R}^2 : x^2 -y^2 =0\}$ a subspace of $\mathbb{R^2}$?

Question

Prove or Disprove : $\{(x,y) \in \Bbb{R}^2: x^2 -y^2 =0\}$ is a subspace of $\mathbb{R^2}$

My attempts : $ x^2-y^2 = (x-y)(x+y)=0$

$(x-y)=0$ or $(x+y) =0$

I thinks it is a subspaces because $(x- y) = 0 $

Any hints ...

Answer 1

Since $x^2-y^2=(x-y)(x+y)=0$, then that subspace contains $(1,-1)$ and $(1,1)$, then it should contains any linear combinations of them, that it should coincides with the entire space $\Bbb R^2$. But $(1,0)$, for instance, does not belong to that subspace since $1^2-0^2\neq 0$. Then it is not a subspace of $\Bbb R^2$.

Answer 2

The definition of "subspace"doesn't in any way reference the equations used to define the space, so "because $(x-y) = 0$" doesn't really mean anything.

There is a short list of requirements that you need to check:

• Is the zero vector there?
• If you have a vector in your subset, is any scaling of that vector in that set?
• If you have two vectors in your subset, is their sum also in that set?

Those are the things you need to check. And to make checking that easier, you should look at what elements are actually in your set: it's all vectors in $\Bbb R^2$ which are either of the form $(a, a)$ or of the form $(a, -a)$.

• Is $(0, 0)$ on one of the two forms mentioned above?
• If we have a vector $v$ on one of the two forms mentioned above, is $rv$ of one of the two forms, no matter which real number $r$ is?
• If $u$ and $v$ are both vectors on one of the two forms mentioned above (no matter which form either of them has), is $u+v$ also of such a form?