Question

how can i know if {(x,y,z) E R^3 : z=xy} is a subspace? using : (for every constant) c,a E R(real number), (for every) w1, w2 E W : cw1 + aw2 E W

Answer 1

yeah

Answer 2

yeah...? what yeah?

Answer 3

yeah

Answer 4

frickin trolls

Answer 5

It is not a subspace!

Answer 6

how'd you find out using the condition i used

Answer 7

(1,1,1) is belong to the set because 1=1*1. But 2(1,1,1)=(2,2,2) is not in the set since \( 2\neq 2\cdot 2\)

Answer 8

ok... what about : z=x+y

Answer 9

it is a subspace. You need to check it though :D

Answer 10

if its (1,1,1) then it has to be 1=1+1, right? which is obviously wrong..

Answer 11

if you want to say that it is not a subspace you can work with example. But if you want to show that it is a subspace you need to work in general

Answer 12

i see...because i proved it IS a subspace, but when i think of it as (1,1,1) it doesn't make sense

Answer 13

Let (a,b,c), (d,e,f) in the set. It means c=a+b and f=d+e. Let k,l be any scalar. Is k(a,b,c)+l(d,e,f) in the set? (how to check it? check it the third component is equal or not with the sum of the first two components)

Answer 14

ka+kb=kc and ld+le=lf

Answer 15

yes, but what do you need to show is that (ka+ld,kb+le,kc+lf) is in the set.

Answer 16

ok so.. how

Answer 17

what is the criterion if a vector belong to your set?

Answer 18

uh?

Answer 19

i don't understand what you mean

Answer 20

z needs to equal x+y ..that?

Answer 21

yes, exactly!

Answer 22

but you need to read it as third component = first component + second component
is that true for the vector (ka+ld,kb+le,kc+lf) ?

Answer 23

no

Answer 24

why not ?

Answer 25

oh yes, it is

Answer 26

why it is? :D

Answer 27

first component was ka then ld, then kb and le, etc..

Answer 28

you need to show that kc+lf = (ka+ld)+(kb+le)

Answer 29

which is true..

Answer 30

lets say: w1=(x1, y1, z1) and w2=(x2,y2,z2) then
k(x1,y1,z1) + c(x2,y2,z2) = (kx1, ky1, kz1) + (cx2, cy2, cz2) = (kx1+cx2, ky1+cy2, kz1+kz2) so... (kx1+cx2) + (cy1+cy2) = kz1+kz2

Answer 31

QED?

Answer 32

almost you need to give an argument why (kx1+cx2) + (cy1+cy2) = kz1+kz2 is true

Answer 33

hoooow xD

Answer 34

oh wait

Answer 35

i kinda f'd up with the c and k

Answer 36

k(x1,y1,z1) + c(x2,y2,z2) = (kx1, ky1, kz1) + (cx2, cy2, cz2) = (kx1+cx2, ky1+cy2, kz1+cz2) so... (kx1+cx2) + (cy1+cy2) = kz1+cz2

Answer 37

thats corrected

Answer 38

still a typo :). It should be (kx1+cx2) + (ky1+cy2) = kz1+cz2

Answer 39

lol

Answer 40

anyway, so its: kx1+ky1= kz1 and cx2 + cy2= cz2

Answer 41

that it?

Answer 42

good! lets write all the argument neatly

Answer 43

Let W={(x,y,z) : z=x+y}. Let : w1=(x1, y1, z1) and w2=(x2,y2,z2) are elements of W. Then z1=x1+y1 and z2=x2+y2. Now we want to show that k(x1,y1,z1)+c(x2,y2,z2) in W. Notice that k(x1,y1,z1)+c(x2,y2,z2)=(kx1+cx2,ky1+cy2,kz1+cz2). Now kz1+cz2=k(x1+y1)+c(x2+y2)=(kx1+cx2) +(ky1+cy2). Therefore k(x1,y1,z1)+c(x2,y2,z2) in W.

Answer 44

i'm so smart..

Answer 45

seriously though, thanks!

Answer 46

the original question kinda got discarded, but its ok