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$\{(x, y, z)\} \space$ with $\space x + y + z = 0$

Working through some problems in a textbook and I'm not very confident about checking if subsets are subspaces. I know that for a subset to be a subspace of $\space \mathbb{R}^3 \space$ it must be closed under addition and scalar multiplication but I'm not sure how to check this with examples. Any help would be appreciated!

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  • $\begingroup$ Welcome to Mathematics Stack Exchange. If $x+y+z=0$, is $cx+cy+cz=0$? $\endgroup$
    – J. W. Tanner
    Jun 18, 2020 at 20:01
  • $\begingroup$ Thank you for the welcome! :D If x + y + z = 0, then I believe that any scalar multiple must also be equal to zero. So it is closed under scalar multiplication? $\endgroup$
    – ClassicBeavs
    Jun 18, 2020 at 20:04
  • $\begingroup$ Correct, how about the addition? That is, if $x + y + z = 0$ and $a + b + c = 0$, is $(x + a) + (y + b) + (z + c) = 0?$ By the way, if you manage to figure this out yourself you can and should answer your own question. $\endgroup$
    – paul blart math cop
    Jun 18, 2020 at 20:06
  • $\begingroup$ Your comments have helped a lot, thanks guys! If x + y + z = 0, and a + b + c = 0, then (x+a) + (y+b) + (z+c) = 0. As a general strategy to checking if a subset is a subspace, should I try to find the general case as you have done? Also, this is an aside, how do you format your comments to change the font of your expressions? Thanks again :) $\endgroup$
    – ClassicBeavs
    Jun 18, 2020 at 20:15
  • $\begingroup$ Here is a MathJax tutorial; and to show that a subset is a subspace, you must show that it is closed under addition and scalar multiplication for any vectors in general, not just for particular examples $\endgroup$
    – J. W. Tanner
    Jun 18, 2020 at 20:20

3 Answers 3

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With the help of these comments, I now have the answer! The subset IS a subspace of R3. To check if it is closed under scalar multiplication: If $x + y + z = 0$, then the following is true for any scalar multiple: $ax + ay + az = 0$ To check for addition: If $x + y + z = 0$ and $a + b + c= 0$ , then $(x+a)+(y+b)+(z+c)=0$ Therefore the subset is closed under scalar multiplication and addition, and is therefore a subspace of R3.

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  • $\begingroup$ Little nittpick: This is actually not enough to show that this is a subspace. You also need to show that it is non-empty (which is clear here, but still). The empty set is closed und both addition and scalar mutliplication, but is not a subspace since ot does not have a neutral element gor the addition. That is the reason why this extra condition is needed. $\endgroup$
    – Con
    Jun 18, 2020 at 21:53
  • $\begingroup$ Interesting, thank you! To show that it is non-empty, would it be sufficient to say that, for example: $(1,1,-2)$ is a vector in the subset, therefore the subset is non-empty? Is giving an example of a non-zero vector in the subset enough to show that it is non-empty? $\endgroup$
    – ClassicBeavs
    Jun 18, 2020 at 21:59
  • $\begingroup$ Yes, that would be fine. Since either way the zero vector has to be included, you can also always check whether the zero vector is in the subset. So, not only a non-zero vector, but rather any vector suffices. Usually one defines subspace by the three conditions: 1) zero vector included 2) closed under addition 3) closed under scalar multiplication - it just turns out that these three axioms are equivalent to the the axioms where we replace zero vector by any vector in 1) $\endgroup$
    – Con
    Jun 18, 2020 at 22:01
  • $\begingroup$ Ahh that makes a lot of sense, thank you! $\endgroup$
    – ClassicBeavs
    Jun 18, 2020 at 22:02
  • $\begingroup$ Glad to be of help! $\endgroup$
    – Con
    Jun 18, 2020 at 22:03
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Well, if $x+y+z = 0$ then $t x+t y +t z = 0$.

If $x_k+y_k+z_k = 0$ for $k=1,2$ then $(x_1+x_2)+(y_1+y_2)+(z_1+z_2) = 0$.

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It is because the subset is the inverse image of the subspace $\{0\}$ under the linear map $(x,y,z)\mapsto x+y+z$.

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