# Determine whether the subset $\{(x_1,x_2,x_3) \in \mathbb{F}^3:x_1x_2x_3 = 0\}$ is a subspace of $\mathbb{F}^{3}$

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.

• 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. – user70962 Aug 31 '13 at 23:49

Hint: Consider the elements $(1, 1, 0)$ and $(0, 0, 1)$, which lie in the given subset.
• 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? – St Vincent Aug 31 '13 at 23:44
• @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. – user61527 Aug 31 '13 at 23:47