## TedG Group Title W := {(x_1; x_2; x_3) in V l (x_2)=(x_1)(x_3)} Is W a subspace of R^3 (subspace test) one year ago one year ago

1. amistre64 Group Title

there are certain criteria for "subspace"

2. amistre64 Group Title

do you recall what they are?

3. TedG Group Title

I know show 0 vector is in set, closure under addition and closure under multiplication, i have shown the first two but for the last i am unsure on whether i have done it right so just wanted to check

4. amistre64 Group Title

V l (x_2)=(x_1)(x_3)} that is hard to parse thru, can you clear it up any?

5. TedG Group Title

$W:= \{(x_{1}, x_{2}, x_{3}) \in V \text{such that} x_{2}=x_{1}x_{3}\} )$

6. TedG Group Title

sorry wasnt sure on how to put a space after text

7. amistre64 Group Title

that is alot easier to read :)

8. amistre64 Group Title

if you can prove that: $c_1\vec X+c_2\vec Y\in W$that should satisfy the addition and multiplcation requirements in one fell swoop

9. TedG Group Title

oh ok so i have done it seperately, but under multiplication can it be said that for cu=c(u1,u2,u3) then cu=(cu1,cu2,cu3) and if we let say v= cu that conditions are met? as cu2=cu1cu3 implies v2=v1v3 if that makes sense.

10. amistre64 Group Title

cu1 cu2 = c^2 u1 u2 right?

11. TedG Group Title

yh see thats where i was not sure as then surely that wouldnt work

12. amistre64 Group Title

im getting the indexes a little mixed up i see

13. TedG Group Title

sorry its just my latex typing isnt that quick

14. amistre64 Group Title

$\vec u=c_1\vec x_1+c_2\vec x_2+c_3\vec x_3$ $\vec u=c_1\vec x_1+c_2\vec x_1\vec x_3+c_3\vec x_3$ $k\vec u=k(c_1\vec x_1)+k(c_2\vec x_1\vec x_3)+k(c_3\vec x_3)$

15. amistre64 Group Title

now im getting lost lol ... oy vey

16. TedG Group Title

ok i think i got it it let me type it in latex so it is clear.

17. TedG Group Title

Suppose $$c \in R$$ and $$\mathbf{v} \in V$$ $c\mathbf{v} = c(v_{1}, v_{2}, v_{3}) =(cv_{1}, cv_{2}, cv_{3})$Now if we let $$\mathbf{u} \in V$$ where $$\mathbf{u} = c\mathbf{v}$$ which is valid as it is given that $$(x_{1}, x_{2}, x_{3}) \in V$$ and $$V$$ is the vector space $$R^3$$ so multiplying by a scalar will still be in the reals. Thus, we have that $cv_{2} =cv_{1}cv_{3}$ and as $$\mathbf{u} = c\mathbf{v}$$ it can be shown that $u_{2} =u_{1}u_{3}$

18. TedG Group Title

i think that makes sense.

19. amistre64 Group Title

with my limited recollection, that does make sense to me as well. but wouldnt we want: u in W? or is the line definition of v2 = v1v3 sufficient for that ....

20. amistre64 Group Title

im not even sure what that definition would actually entail ... but for the sake of definitions :)

21. TedG Group Title

yh you are right. i want u in W not V

22. amistre64 Group Title

by defining u in W, u = cv, and W is of the defintion x2= x1x3; cv2 = c^2 v1v2 for all c seems to defy closure to me

23. amistre64 Group Title

..but knowing me, id want a 2nd opinion :) @.Sam. you any decent at these things?

24. TedG Group Title

yh thats the reason i came on here to ask in the first place the c^2 might mean it is not a subspace.

25. amistre64 Group Title

if we could visualize a x1x3 product ... we would have something to compare with

26. amistre64 Group Title

if its a "dot product" 3<1,1,1> 3<4,1,0> ---------- 3(4+1+0) doesnt seem to make sense to me

27. amistre64 Group Title

cross product the only other thing i can thing of for a test

28. amistre64 Group Title

<1,1,1>x<4,1,0> = <-1,4,-3> <3,3,3>x<12,3,0> = <-9,36,-27> = 3<-3,12,-9>

29. TedG Group Title

in that case it definately doesnt hold right

30. amistre64 Group Title

thats what im wondering about we can still get <-1,4,-3> back again, by factoring out another 3, and since 3^2 is a real number ... im just not sure if thats acceptable or not :/

31. amistre64 Group Title

<-1,4,3> , if we are defining the product of x1x3 as a cross, IS in W, and any real scalar of it is as well

32. TedG Group Title

hmm i guess i will just have to email my lecturer see what he says

33. amistre64 Group Title

well, good luck with it :) i think ive stared myself into the 50s bracket of an IQ test with this one ....

34. TedG Group Title

well thanks for the help anyway, much appreciated