Find if W1 is a linear subspace, being
W1={(x;y;z) e R^2/ Sqrt(x+y+z)=0 } @Mathematics

Question

Find if W1 is a linear subspace, being
W1={(x;y;z) e R^2/  Sqrt(x+y+z)=0 }   @Mathematics

turingtest · Answer

Is this linear algebra?

anonymous · Answer

Yes, its linear algebra

anonymous · Answer

Please, nobody in linear algebra helps me!

zarkon · Answer

did you mean $$W_1=\{(x;y;z) \in R^3|\sqrt{x+y+z}=0 \}$$

zarkon · Answer

note that $$\sqrt{x+y+z}=0 $$
is the same as $$x+y+z=0 $$

zarkon · Answer

you got it?

turingtest · Answer

Zarkon can you look at this problem when you get the chance? http://openstudy.com/#/updates/4ec2cec0e4b032a1a5ab20d0

anonymous · Answer

Yes, if the vector Wn=(0,0,0), then $$\sqrt{Wn}$$ = $$\sqrt{0+0+0}=0$$. 2) If u\[inW1, and V \in W1, then \sqrt{ux+uy+uz}=0    , and  \sqrt{vx+vy+vz}=0, you get (u+v)=(\sqrt{(ux+vx)+(uy+vy)+(uz+vz)}=0, => (u+v)inW1. Finally I prove: \alpha \in \mathbb{R}, au=(aux; auy; auz), \sqrt{aux+auy+auz} =0, \sqrt{a}*\sqrt{ux+uy+uz} =0, \sqrt{a}*0=0. I'm right?]