if V = R^3, and S consists of all points on the xy, xz and yz planes, then S is a subspace of V. True or False

Question

loser66 · Answer

@e.mccormick

e.mccormick · Answer

Every point on xy, xz, and yz.  Hmmm...  That does sound like every possible point.

loser66 · Answer

those points are in R^2, right? it sounds like R^2 is a subspace of R^3? but it's not,

e.mccormick · Answer

Hmmm.... unless they only mean the axial planes.

(x,y,0), (x,0,z), (0,y,z)

If I add two points, (1,1,0) and (0,1,1) I get (1,2,1) which is not in (x,y,0), (x,0,z), (0,y,z).  It does not seem to be cloed in addition.

loser66 · Answer

they don't say that S is in R^3, how can you assume that xy plane's points forms (x, y , 0)?

e.mccormick · Answer

"S consists of all points on the xy, xz and yz planes"

loser66 · Answer

:) sorry, I lit you up by my silly comment. got you

e.mccormick · Answer

In 3d space, the xy plane is (x,y,0), the xz plane is (x,0,z), and the yz plane is (0,y,z) iff they mean axial plains.  And I think they do mean axial plains.

loser66 · Answer

so, the conclusion is False, right? because it doesn't satisfy the addition

e.mccormick · Answer

For it to be a subspace of anything, it would need to be closed under addition and scalar multiplication. This fails the test of addition because while the answer is in $\mathbb{R}^3$, it is not in S.  So yah, I think that makes it false by invalidating it as a subspace.

loser66 · Answer

hey, there, I study this stuff 3 times, every time , it is new to me. Wonder why I cannot get it.

loser66 · Answer

the key point is adding any point in S, the sum must be the same form of either of the original one, right? I mean the sum must be either (x, y, 0) or (0, y, z), or (x,0,z) right? right?

e.mccormick · Answer

Yah, it still has to be witin S for S to be a space of some sort.  And if it is not a space then it cant be a sub space.

loser66 · Answer

hey there, saying "witin", here don't understand, within? or else?

e.mccormick · Answer

within. Missed an h. This is a nice overview... other than the one math error: http://www.math.jhu.edu/~nitu/subspace.pdf

e.mccormick · Answer

At the very bottom of page one he has $\frac{3}{2}$ where there should be a 0.  LOL.  But the rest looks good.  On page 2 it goes over the subspace test, which I just did part of.

loser66 · Answer

Thank you very much.

e.mccormick · Answer

np.  Have fun!