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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

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  1. amistre64 Group Title
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    there are certain criteria for "subspace"

    • one year ago
  2. amistre64 Group Title
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    do you recall what they are?

    • one year ago
  3. TedG Group Title
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    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

    • one year ago
  4. amistre64 Group Title
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    V l (x_2)=(x_1)(x_3)} that is hard to parse thru, can you clear it up any?

    • one year ago
  5. TedG Group Title
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    \[ W:= \{(x_{1}, x_{2}, x_{3}) \in V \text{such that} x_{2}=x_{1}x_{3}\} ) \]

    • one year ago
  6. TedG Group Title
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    sorry wasnt sure on how to put a space after text

    • one year ago
  7. amistre64 Group Title
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    that is alot easier to read :)

    • one year ago
  8. amistre64 Group Title
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    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

    • one year ago
  9. TedG Group Title
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    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.

    • one year ago
  10. amistre64 Group Title
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    cu1 cu2 = c^2 u1 u2 right?

    • one year ago
  11. TedG Group Title
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    yh see thats where i was not sure as then surely that wouldnt work

    • one year ago
  12. amistre64 Group Title
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    im getting the indexes a little mixed up i see

    • one year ago
  13. TedG Group Title
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    sorry its just my latex typing isnt that quick

    • one year ago
  14. amistre64 Group Title
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    \[\vec u=c_1\vec x_1+c_2\vec x_2+c_3\vec x_3\] \[\vec u=c_1\vec x_1+c_2\vec x_1\vec x_3+c_3\vec x_3\] \[k\vec u=k(c_1\vec x_1)+k(c_2\vec x_1\vec x_3)+k(c_3\vec x_3)\]

    • one year ago
  15. amistre64 Group Title
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    now im getting lost lol ... oy vey

    • one year ago
  16. TedG Group Title
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    ok i think i got it it let me type it in latex so it is clear.

    • one year ago
  17. TedG Group Title
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    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 \[ u_{2} =u_{1}u_{3} \]

    • one year ago
  18. TedG Group Title
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    i think that makes sense.

    • one year ago
  19. amistre64 Group Title
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    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 ....

    • one year ago
  20. amistre64 Group Title
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    im not even sure what that definition would actually entail ... but for the sake of definitions :)

    • one year ago
  21. TedG Group Title
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    yh you are right. i want u in W not V

    • one year ago
  22. amistre64 Group Title
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    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

    • one year ago
  23. amistre64 Group Title
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    ..but knowing me, id want a 2nd opinion :) @.Sam. you any decent at these things?

    • one year ago
  24. TedG Group Title
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    yh thats the reason i came on here to ask in the first place the c^2 might mean it is not a subspace.

    • one year ago
  25. amistre64 Group Title
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    if we could visualize a x1x3 product ... we would have something to compare with

    • one year ago
  26. amistre64 Group Title
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    if its a "dot product" 3<1,1,1> 3<4,1,0> ---------- 3(4+1+0) doesnt seem to make sense to me

    • one year ago
  27. amistre64 Group Title
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    cross product the only other thing i can thing of for a test

    • one year ago
  28. amistre64 Group Title
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    <1,1,1>x<4,1,0> = <-1,4,-3> <3,3,3>x<12,3,0> = <-9,36,-27> = 3<-3,12,-9>

    • one year ago
  29. TedG Group Title
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    in that case it definately doesnt hold right

    • one year ago
  30. amistre64 Group Title
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    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 :/

    • one year ago
  31. amistre64 Group Title
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    <-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

    • one year ago
  32. TedG Group Title
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    hmm i guess i will just have to email my lecturer see what he says

    • one year ago
  33. amistre64 Group Title
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    well, good luck with it :) i think ive stared myself into the 50s bracket of an IQ test with this one ....

    • one year ago
  34. TedG Group Title
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    well thanks for the help anyway, much appreciated

    • one year ago
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