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I need help determining if the following subset is a subspace of $\mathbb{F}^{3}$:

$\{(x_{1},x_{2},x_{3}) \in \mathbb{F}^3:x_{1}x_{2}x_{3} = 0$}.

Based on what I've read in my textbook, we must test that the subset is closed under vector addition and scalar multiplication (with the zero vector included).

Let $v,w \in \mathbb{F}^3$.

If $v = (x_{1},x_{2},x_{3})$ and $w = (y_{1},y_{2},y_{3})$ where $x_{1}x_{2}x_{3} = 0$ and $y_{1}y_{2}y_{3} = 0$

then $v + w = (x_{1}+y_{1}, x_{2}+y_{2}, x_{3}+y_{3}$)

The product of all ordered triples of the sum.

$(x_{1}+y_{1})(x_{2}+y_{2})(x_{3}+y_{3}) = 0$

Multiplied out:

$x_{1}x_{2}x_{3}+x_{1}x_{3}y_{2}+x_{2}x_{3}y_{1}+y_{1}y_{2}x_{3}+x_{1}x_{2}y_{3}+x_{1}y_{2}y_{3}+x_{2}y_{1}y_{3}+y_{1}y_{2}y_{3} = 0$

simplifying,

$x_{1}x_{3}y_{2}+x_{2}x_{3}y_{1}+y_{1}y_{2}x_{3}+x_{1}x_{2}y_{3}+x_{1}y_{2}y_{3}+x_{2}y_{1}y_{3}=0$

This is about where I got stuck. My guess is that the subset is not closed under addition because the given equality is only $0$ if each individual term sums to $0$.

I already know how to check scalar multiplication for this particular problem.

Thanks for the help.

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  • $\begingroup$ A problem with what you're doing is setting the first expression equal to $0$ which is something you don't know. Remember, you're checking for this. $\endgroup$ – user70962 Aug 31 '13 at 23:49
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Hint: Consider the elements $(1, 1, 0)$ and $(0, 0, 1)$, which lie in the given subset.

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  • $\begingroup$ Cool, you read the solutions manual. I get that $(1,1,1)$ will not lie in the subset. Is there a way to show this more generally based on what I was starting to do above? $\endgroup$ – St Vincent Aug 31 '13 at 23:44
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    $\begingroup$ @StVincent I have no idea what book you're using, and certainly don't have a solutions manual. My point was that you don't need to go to so much work - if it's not closed under $+$, then you only need one counterexample. The two vectors I mentioned give such a counterexample. $\endgroup$ – user61527 Aug 31 '13 at 23:47
  • $\begingroup$ I agree that it is much easier using an easy counterexample. However, is it possible to deduce that the set is not closed under addition from the work that I posted above? $\endgroup$ – St Vincent Aug 31 '13 at 23:58
  • $\begingroup$ What about without plugging in explicit vectors--would it be to complicated to try to prove that this is not a subspace symbolically? $\endgroup$ – St Vincent Sep 1 '13 at 0:08

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