## TedG 2 years ago 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)

1. amistre64

there are certain criteria for "subspace"

2. amistre64

do you recall what they are?

3. TedG

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

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

5. TedG

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

6. TedG

sorry wasnt sure on how to put a space after text

7. amistre64

that is alot easier to read :)

8. amistre64

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

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

cu1 cu2 = c^2 u1 u2 right?

11. TedG

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

12. amistre64

im getting the indexes a little mixed up i see

13. TedG

sorry its just my latex typing isnt that quick

14. amistre64

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

15. amistre64

now im getting lost lol ... oy vey

16. TedG

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

17. TedG

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

i think that makes sense.

19. amistre64

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

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

21. TedG

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

22. amistre64

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

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

24. TedG

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

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

26. amistre64

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

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

28. amistre64

<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

in that case it definately doesnt hold right

30. amistre64

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

<-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

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

33. amistre64

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

34. TedG

well thanks for the help anyway, much appreciated