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I did some linear algebra exercise and did the following: Give an example of a nonempty subset $U$ of the xy-plane with the property that $U$ is closed with respect to scalar multiplication but $U$ is not a subspace.

I believe the set $U=\{(x,y): xy=0\}$ should provide such an example. Are there any better examples? More visual ones?

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    $\begingroup$ Your example is perfect. Visually, it represents the union of the $x$ and $y$ axes (which is a union of two lines through the origin that are not collinear). It fails to be a subspace because it isn't closed with respect to vector addition. $\endgroup$
    – Adriano
    Aug 7, 2013 at 10:29
  • $\begingroup$ @Adriano Sorry, I don't see how it is the union of the x and the y axis? Could you say a bit more? $\endgroup$
    – newb
    Aug 7, 2013 at 10:37
  • $\begingroup$ @newb: The points for which $xy=0$ are exactly the ones that are either on the $x$ axis or on the $y$ axis (or both). $\endgroup$ Aug 7, 2013 at 10:38
  • $\begingroup$ @HenningMakholm Oh yes, right. Thank yoU! $\endgroup$
    – newb
    Aug 7, 2013 at 10:40

2 Answers 2

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Your example is perfect. Visually, it represents the union of the $x$ and $y$ axes (which is a union of two lines through the origin that are not collinear). It fails to be a subspace because it isn't closed with respect to vector addition.

To see your example is the union of the $x$ and $y$ axes, observe that: $$ \begin{align*} U &= \{(x,y): xy=0\} \\ &= \{(x,y): x=0 \text{ or } y=0\} \\ &= \{(x,y): x=0\} \cup \{(x,y):y=0\} \\ &= (y\text{-axis}) \cup (x\text{-axis}) \end{align*}$$

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Any finite (or countable!) set of lines through the origin, at least two of which are not collinear.

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