Question

Show that W is not a subspace of \(V = R^3\)
a) W = {(a,b,c): a≥0}
b) W = {(x,y,z): x^2 +y^2 +z^2 ≤1}

Answer 1

sorry - I have no knowledge of this particular topic in mathematics :(

Answer 2

maybe @mukushla can help?

Answer 3

Did subspace mean : subset ?@higgs

Answer 4

since mukushla is offline right now :

@waterineyes will help u

Answer 5

@mathslover this question is under linear algebra and a subspace is not a subset

Answer 6

for a vector space is \[\vec{w}\in W\]

then \[c\vec{w}\in W\]

show this doesn't always work for your \(W\)

Answer 7

*for a vector space if

Answer 8

here \(c\in\mathbb{R}\)

Answer 9

I think I understand this now after Zarkon's hint. Zarkon I can see to apply your method to (b) but to (a)?

Answer 10

to both

Answer 11

yes I agree it should be applied to both but I just can't se /how/ to apply it to (a)

Answer 12

let c=-1

Answer 13

hang on

Answer 14

yes - I /just/ got it - thx! :)

Answer 15

I thought that for it to be a subspace, W has to be closed under addition AND scalar multiplication? (As well as passing through the origin)
Is Zarkon just doing scalar multiplication?

Answer 16

yes ! scalar multiplication for the first one !
(1,0,0) in W but -1 (1;0;0) =(-1;0;0) doesn't belong to W

Answer 17

for the second you can take the same vector ! (1,0,0) is in W but 2(1,0,0) isn't !

Answer 18

@cherylim23

Why would I need to show both?

Answer 19

There are two conditions that need to be satisfied: closed under addition AND closed under scalar multiplication. What Zarkon did was only the latter

Answer 20

I usually do it simultaneously, such that cW + mP is a subspace of W

Answer 21

when showing something is not a vector space one only need to to show that at least one of the two fails.