# Is this subset a subspace of $\mathbb{R}^3$?

$$\{(x, y, z)\} \space$$ 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!

• Welcome to Mathematics Stack Exchange. If $x+y+z=0$, is $cx+cy+cz=0$? – J. W. Tanner Jun 18 '20 at 20:01
• 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? – ClassicBeavs Jun 18 '20 at 20:04
• 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. – paul blart math cop Jun 18 '20 at 20:06
• 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 :) – ClassicBeavs Jun 18 '20 at 20:15
• 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 – J. W. Tanner Jun 18 '20 at 20:20

## 3 Answers

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.

• 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. – Con Jun 18 '20 at 21:53
• 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? – ClassicBeavs Jun 18 '20 at 21:59
• 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) – Con Jun 18 '20 at 22:01
• Ahh that makes a lot of sense, thank you! – ClassicBeavs Jun 18 '20 at 22:02
• Glad to be of help! – Con Jun 18 '20 at 22:03

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 $$(x_1+x_2)+(y_1+y_2)+(z_1+z_2) = 0$$.

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