1
$\begingroup$

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

$\endgroup$
6
  • $\begingroup$ Welcome to Mathematics Stack Exchange. If $x+y+z=0$, is $cx+cy+cz=0$? $\endgroup$ – J. W. Tanner Jun 18 '20 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 '20 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 '20 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 '20 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 '20 at 20:20
2
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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 '20 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 '20 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 '20 at 22:01
  • $\begingroup$ Ahh that makes a lot of sense, thank you! $\endgroup$ – ClassicBeavs Jun 18 '20 at 22:02
  • $\begingroup$ Glad to be of help! $\endgroup$ – Con Jun 18 '20 at 22:03
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.