## Rowa 2 years ago how can i know this subset is a subspace: {(x,y,z) E(element) R^3(real number) : z=x+y}

1. KingGeorge

Let $$S=\{(x,y,z)\in\mathbb{R}^3\;|\;z=x+y\}$$. To prove something is a subspace, you need to satisfy three different conditions. 1. $$\vec0\in S$$. 2. If $$\vec{v},\vec{u}\in S$$, then $$\vec{v}+\vec{u}\in S$$. 3. If $$\vec{u}\in S$$, and $$c\in\mathbb{R}$$, then $$c\vec{u}\in S$$.

2. KingGeorge

The first condition is straightforward. $$\vec0=(0,0,0)$$, and $$0=0+0$$, so this is good. Can you prove the second condition yourself?

3. Rowa

so..., x+y=z, and z is an element of R^3..

4. Rowa

somehow i know thats wrong..

5. KingGeorge

Not quite. $$(x,y,z)$$ is an element of $$\mathbb{R}^3$$ such that $$x+y=z$$, and $$z$$ is an element of $$\mathbb{R}$$.

6. Rowa

oooh, ok

7. KingGeorge

And if you have two vectors $$(x_0,y_0,z_0)$$ and $$(x_1,y_1,z_1)$$ in $$S$$, then $(x_0,y_0,z_0)+(x_1,y_1,z_1)=(x_0+x_1,y_0+y_1,z_0+z_1)$Using this can you try to do the second condition?

8. Rowa

k

9. Rowa

i don't see any problem with the second condition..

10. Rowa

If there's only 1 vector (x,y,z), how can i add to see if it follows the second condition *confused*

11. Rowa

ok, i'm slowly getting it

12. KingGeorge

You need to take two vectors. So let's use the vectors I wrote out above. Let $$\vec{v}=(x_0,y_0,z_0)$$ and let $$\vec{u}=(x_1,y_1,z_1)$$. Then we need to check if $(x_0,y_0,z_0)+(x_1,y_1,z_1)=(x_0+x_1,y_0+y_1,z_0+z_1)$is still in $$S$$. To check this, we just need to check if $x_0+x_1+y_0+y_1=z_0+z_1.$However, since $$x_0+y_0=z_0$$ and $$x_1+y_1=z_1$$, (this is true since both $$\vec{v}$$ and $$\vec{u}$$ are in $$S$$, so they have that property), so it's true, and $$\vec{v}+\vec{u}\in S$$.

13. KingGeorge

That make sense?

14. Rowa

yea, i get it

15. KingGeorge

Then, can you attempt to the third condition?

16. Rowa

so..: c(x,y,z)= cx+cy+cz, cz=cy+cx ??? something like this?

17. KingGeorge

That looks pretty good. You just need to make sure that you've proven that cz=cy+cx.

18. Rowa

z=y+x => cz=cy+cx ?

19. KingGeorge

Basically. I would add the step that $$z=x+y\implies cz=c(x+y)\implies cz=cx+cy$$. But otherwise, that's the correct reasoning.

20. Rowa

nice...Thank you so much

21. KingGeorge

You're welcome.