Why W is not a subspace of $\mathbb{R}^2$?

Question

The set $W = \{(x,y)\in\mathbb{R}^2\mid y = x \vee y = -x\}$ is not a subspace of $V=\mathbb{R}^2$. I know $0\in W$, but I'm not sure if the others conditions are satisfied.

Answer 1

There are many approaches to this. I just give a couple examples.

1.

$(1,1), (1,-1) \in W$ but their sum $(2,0) \not \in W$ so it is not a vector space.

2.

The line $(t,t)$ with $t\in \mathbb R$ is a subspace of $V$ of dimension $1$. If $Y=-X$ adds new points and $W$ is a vector space, then it must have dimension $2$ (becuse we are inside $\mathbb R ^2$) and so $W = \mathbb R^2$. But you can easily find many points not in $W$, for example $(0,1)$ doesn't satisfies $Y=X$ nor $Y = -X$, so we can conclude that $W$ is not a vector space.

Answer 2

To be a subspace it must be closed under vector addition and scalar multiplication. But for instance $(1,1)+(2,-2)=(3,-1) $, showing that it's not closed under addition.

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Alternatively, since $(1,1),(1,-1)\in W$ are independent, if it were a subspace it would be two-dimensional, hence be all of $\mathbb R^2$. But it isn't, as there are plenty of elements of $\mathbb R^2$ which are not either of the form $(x,x)$ or $(x,-x)$.

Answer 3

Proposition

Let $V$ be a vector space, where $V_{1}\leq V$ and $V_{2}\leq V$.

Then $V_{1}\cup V_{2}$ is a vector space if, and only if, $V_{1}\subseteq V_{2}$ or $V_{2}\subseteq V_{1}$.

Proof

The implication $(\Leftarrow)$ is straight forward.

Having said that, let us prove the converse direction $(\Rightarrow)$ by contradiction.

Suppose that $V_{1}\cup V_{2}$ is a vector space and $V_{1}\not\subseteq V_{2}$ and $V_{2}\not\subseteq V_{1}$.

Then we can take $x\in V_{1} - V_{2}\subseteq V_{1}\cup V_{2}$ and $y\in V_{2} - V_{1}\subseteq V_{1}\cup V_{2}$.

Therefore we can deduce that $w = x + y \in V_{1}\cup V_{2}$, whence we conclude there are three possibilities:

$$(w\in V_{1} - V_{2})\vee(w\in V_{1}\cap V_{2})\vee(w\in V_{2} - V_{1})$$

In the first case, we would have $y = w - x\in V_{1}$.

In the second case, we would have $x = w - y\in V_{2}$ and $y = w - x\in V_{1}$,

Finally, in the third case we would have $x = w - y \in V_{2}$.

In all of them, we obtain a contradiction.

Hence we conclude the original claim is true, and we are done.

Solution

At your case, $V = \mathbb{R}^{2}$, $V_{1} = \{(x,y)\in\mathbb{R}^{2}\mid y = x\}$ and $V_{2} = \{(x,y)\in\mathbb{R}^{2}\mid y = -x\}$.

Since $V_{1}\not\subseteq V_{2}$ and $V_{2}\not\subseteq V_{1}$, the union $V_{1}\cup V_{2}$ is not a subspace of $V$.

Hopefully this helps!