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I am new here so please don't close the question (instead please tell me how to improve it).

So I know that for a subset to be a subspace it has to satisfy the following properties:

  1. Closed under scalar multiplication.

  2. Closed under addition.

I however do not know how to go about determining whether:

$W = \{(x,y) \text{ for all in }\mathbb R^2\; |\;\frac{x}{y} = 1\}$ is a subspace of $\mathbb R^2.$

Does zero vector have to be a part of $W$ for it to be a subspace or does it only have to be a non empty set?

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  • $\begingroup$ Please use MathJax. Here is a tutorial. For tuples, use $(x,y)$ and not the Lie bracket $[x,y]$. $\endgroup$
    – Dietrich Burde
    yesterday
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    $\begingroup$ I think the downvotes and votes to close are not called for. The question is clear and shows some understanding. It's not just "do my work for me". Yes, the OP should use mathjax jojozhuang.github.io/tutorial/… $\endgroup$
    – Ethan Bolker
    yesterday
  • $\begingroup$ Hi, welcome to MSE. Generally to write a good question you should explain why you are asking this question and why it is an interesting question. $\endgroup$
    – Numeral
    yesterday
  • $\begingroup$ In $\mathbb{R}^n$, the subspaces always look like $\mathbb{R}^k$ through the origin. $\endgroup$
    – Teepeemm
    yesterday
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    $\begingroup$ You are missing one condition for something to be a subspace: it has to be non-empty. $\endgroup$
    – Arturo Magidin
    yesterday

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Once the set is nonempty and closed under those operations then $$ 0x = 0. $$ Can you finish now?

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You’re missing one key ingredient for a subset to be a subspace, it needs to contain the zero vector. Note that $(x,y) \in W$ if and only if $y$ is equal to the inverse of $x$. So it can’t contain the zero vector, because if it did $0\cdot 0 = 1$ which is impossible since $0$ doesn’t have a multiplicative inverse.

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    $\begingroup$ This does follow from "closed under scalar multiplication" when the scalar is 0. $\endgroup$
    – Teepeemm
    yesterday
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    $\begingroup$ Yes, and we could even strech it out even more by saying that a subset is a subspace if it is closed under linear combinations, but OP seemed fixated on those two axioms that he was missing the fact that if the zero fails to be in that subset, then it’s bot a subspace. $\endgroup$
    – MrGran
    yesterday
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Let $W=\{(x,y):x,y\in \mathbb{R}\land x/y=1\}\subset \mathbb{R}^2$. In order for $W$ to be a subspace of $\mathbb{R}^2$, $W$ must be a subset of $\mathbb{R}^2$ and a vector space itself. Recall the vector space axioms.

If $x/y=1$ then $(-x)/(-y)=1$. Then for all $(x,y)\in W$ there exists $(-x,-y)\in W$ such that $(x,y)+(-x,-y)=(0,0)$.

To get to the same place, let $(x,y)\in W$. Then $0(x,y)=(0,0)$.

If $(0,0)\notin W$ then

  1. There does not exist an additive identity in $W$
  2. $W$ is not closed under vector addition or scalar multiplication.

Consequently, $W$ would not be a subspace of $\mathbb{R}^2$. I hope this is instructive.

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