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subspace

Hi guys
is W ={(x,y,z)∈R3 |x2 +y2 +z2 =0} a subspace or not? if not, how do i prove it
hope someone can help
thanks
Reply 1
Avatar for Vesniep
Vesniep
15
Original post by 123321123
Hi guys
is W ={(x,y,z)∈R3 |x2 +y2 +z2 =0} a subspace or not? if not, how do i prove it
hope someone can help
thanks


what is x2+y2+z2?
Do you mean x2+y2+z2=0 x^2+y^2+z^2=0 in which case you only have one point namely the origin.
Reply 2
Avatar for 123321123
123321123
OP
2
Original post by Vesniep
what is x2+y2+z2?
Do you mean x2+y2+z2=0 x^2+y^2+z^2=0 in which case you only have one point namely the origin.


HI thanks for your reply. yeah I mean the square of x y z. I understand it is just a point(0,0,0), so how do I prove this is not a subspace by one the three conditions?
Hope to hear from you soon
Reply 3
Avatar for Vesniep
Vesniep
15
Original post by 123321123
HI thanks for your reply. yeah I mean the square of x y z. I understand it is just a point(0,0,0), so how do I prove this is not a subspace by one the three conditions?
Hope to hear from you soon


But it is a subspace !
1) The zero vector is contained in the subspace
2) and 3) Any lineal combination λu⃗+μv⃗ \lambda \vec{u} + \mu \vec{v} where those vectors are in the subspace is equal to 0⃗∈W \vec{0} \in \mathbb{W}
Reply 4
Avatar for Vesniep
Vesniep
15
Original post by 123321123
HI thanks for your reply. yeah I mean the square of x y z. I understand it is just a point(0,0,0), so how do I prove this is not a subspace by one the three conditions?
Hope to hear from you soon


did the question mention that W is not a subspace?
Reply 5
Avatar for 123321123
123321123
OP
2
Original post by Vesniep
did the question mention that W is not a subspace?


Nope it asks weather it's a subspace or not. Can I state it is just a point of (0,0,0) before I go through all three conditions? because this would be easier to prove the set is under vector addition.
Reply 6
Avatar for Vesniep
Vesniep
15
Original post by 123321123
Nope it asks weather it's a subspace or not. Can I state it is just a point of (0,0,0) before I go through all three conditions? because this would be easier to prove the set is under vector addition.


Yes that's what you do.

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