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KonradZuse Group Title

Anyone good with inner products and their axioms?

  • one year ago
  • one year ago

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  1. tyteen4a03 Group Title
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    Just ask your question.

    • one year ago
  2. KonradZuse Group Title
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    5. Please refer to Denition 1 on page 335 of the textbook. Given u = (u1; u2; u3) and v = (v1; v2v3). Determine whether or not the following are inner products on R3. For those that are not, list the axioms that do not hold.

    • one year ago
  3. KonradZuse Group Title
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    (a) (u,v) = u[1] v[1] + u[3] v[3] (b) (u,v) = 2 2 2 2 2 2 u[1] v[1] + u[2] v[2] + u[3] v[3] (c) (u,v) = 2 u[1] v[1] + u[2] v[2] + 4 u[3] v[3] (d) (u,v) = u[1] v[1] - u[2] v[2] + u[3] v[3]

    • one year ago
  4. KonradZuse Group Title
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    I'm not 100% sure about Axiom 4, but it seems like so far 1 and 2 satify them all... thoughts? @tyteen4a03

    • one year ago
  5. KonradZuse Group Title
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    @Hero save me.

    • one year ago
  6. Hero Group Title
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    Bro, I took linear algebra already, but I'd have to re-familiarize myself with this.

    • one year ago
  7. Hero Group Title
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    Hang on for a minute

    • one year ago
  8. Hero Group Title
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    @KonradZuse, where in the book is that question?

    • one year ago
  9. KonradZuse Group Title
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    not in the book.

    • one year ago
  10. KonradZuse Group Title
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    sorry it's kind of screwed up in formatting.

    • one year ago
  11. KonradZuse Group Title
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    1 second.

    • one year ago
  12. KonradZuse Group Title
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    • one year ago
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  13. KonradZuse Group Title
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    I posted part of my answer of part b, as well as part c and d. I can post everything if oyu wihs.

    • one year ago
  14. KonradZuse Group Title
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    okay hold on.

    • one year ago
  15. KonradZuse Group Title
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    here happy?

    • one year ago
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  16. KonradZuse Group Title
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    I just want to know if I'm doing this correctly/ what exactly is axiom 4 asking?

    • one year ago
  17. KonradZuse Group Title
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    saying*

    • one year ago
  18. KonradZuse Group Title
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    @UnkleRhaukus

    • one year ago
  19. Zarkon Group Title
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    I guess I'll reply...a lot of your work is not correct..I'll let you think about that for a while

    • one year ago
  20. UnkleRhaukus Group Title
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    where are the axioms you are referencing ?

    • one year ago
  21. KonradZuse Group Title
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    one second.

    • one year ago
  22. KonradZuse Group Title
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    • one year ago
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  23. UnkleRhaukus Group Title
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    (a) \[(\textbf u,\textbf v)=u_1 v_1+u_3v_3 \]

    • one year ago
  24. KonradZuse Group Title
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    I was thinking that might fail because of the new u2v2, but not too sure?

    • one year ago
  25. UnkleRhaukus Group Title
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    \[(\textbf u,\textbf v)=u_1 v_1+u_3v_3=u_3v_3+u_1 v_1=(\textbf v,\textbf u)\] the symmetry axiom holds

    • one year ago
  26. KonradZuse Group Title
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    ok.

    • one year ago
  27. KonradZuse Group Title
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    I actually posted my work up a bit. Should I repost it?

    • one year ago
  28. KonradZuse Group Title
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    I did a and B I just was confused about axiom 4, and so far it doesn't seem like anything is going to fail. Thoughts?

    • one year ago
  29. UnkleRhaukus Group Title
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    \[(\textbf u+\textbf v,\textbf w)=(\textbf u+\textbf v)_1\textbf w_1+(\textbf u+\textbf v)_3\textbf w_3\]\[\qquad\qquad\quad=u_1w_1+v_1w_1+u_3w_3+v_3w_3\]\[\qquad\qquad\quad=u_1w_1+u_3w_3+v_1w_1+v_3w_3\]\[\qquad\qquad\quad=(\textbf u,\textbf w)+(\textbf v,\text w)\] the additivity axion holds

    • one year ago
  30. UnkleRhaukus Group Title
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    \[ (k\textbf u,\textbf v)=ku_1 v_1+ku_3v_3\]\[\qquad\quad=k(u_1 v_1+u_3v_3)\]\[\qquad\quad=k(\textbf u,\textbf v)\] the homogeneity axiom holds

    • one year ago
  31. KonradZuse Group Title
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    • one year ago
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  32. KonradZuse Group Title
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    not sure what was wrong with my work. I did exactly what the book/my other homework did for it.

    • one year ago
  33. UnkleRhaukus Group Title
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    \[(\textbf v,\textbf v)=v_1v_1+v_3v_3\]\[\qquad\quad=v_1^2+v_3^2≥0\] \[(\textbf v,\textbf v)=0\iff \textbf v=\textbf 0\] the positivity axiom holds

    • one year ago
  34. Zarkon Group Title
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    that is the problem..axiom 4

    • one year ago
  35. Zarkon Group Title
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    your work done on axiom 4 is not correct

    • one year ago
  36. Zarkon Group Title
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    remember \(u\) and \(v\) are 3-dimensional vectors

    • one year ago
  37. Zarkon Group Title
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    let \(u=v=<0,1,0>\) what is \((u,v)\)

    • one year ago
  38. Zarkon Group Title
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    or same thing ...what is \((v,v)\)

    • one year ago
  39. KonradZuse Group Title
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    I will check it out.

    • one year ago
  40. KonradZuse Group Title
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    I guess I'm still confused how to prove axiom 4....

    • one year ago
  41. Zarkon Group Title
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    you don't prove axiom 4...it is an axiom

    • one year ago
  42. KonradZuse Group Title
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    errr "satisfy axiom 4"

    • one year ago
  43. Zarkon Group Title
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    (a) is not an inner product...I gave you a counter example

    • one year ago
  44. Zarkon Group Title
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    ((0,1,0),(0,1,0))=0*0+0*0=0 but (0,1,0) is not the zero vector

    • one year ago
  45. KonradZuse Group Title
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    ic so axiom 4 isn't = 0 thus fails?

    • one year ago
  46. Zarkon Group Title
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    it fails... \((v,v)=0\Leftrightarrow v=\vec{0}\)

    • one year ago
  47. UnkleRhaukus Group Title
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    i see what you are saying now @Zarkon \[(\textbf v,\textbf v)=v_1v_1+v_3v_3\]\[\qquad\quad=v_1^2+v_3^2≥0\]\[(\textbf v,\textbf v)=0\iff \textbf v=(0,v_2,0)\not\equiv\textbf 0\]the positivity axiom does not hold in the general case

    • one year ago
  48. Zarkon Group Title
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    correct...but one only needs so show a single counter example to show that it is not an inner product (i'm lazy that way :) )

    • one year ago
  49. KonradZuse Group Title
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    I'm so confused though :)

    • one year ago
  50. KonradZuse Group Title
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    I guess I can see how (0,1,0)(0,1,0) = 1 not 0. So all we have to do is multiply and see if = 0?

    • one year ago
  51. UnkleRhaukus Group Title
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    how did you get = 1 ?

    • one year ago
  52. KonradZuse Group Title
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    I thought that's what he said above, I guess I'm just too konfused hahaha :(

    • one year ago
  53. UnkleRhaukus Group Title
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    the rule was; to add the product of the first components to the product of the third components , the rule satisfied the 4th axiom iff the rule only generates zero when the vector is the zero vector. Zarkon has shown that that the second component of the vector could be anything , therefore there are many vectors that generate zero , not just the zero vector)

    • one year ago
  54. KonradZuse Group Title
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    hmm ic.

    • one year ago
  55. KonradZuse Group Title
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    So does Axiom 4 only fail for the first one?

    • one year ago
  56. Zarkon Group Title
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    another one of your problems fails axiom 4

    • one year ago
  57. KonradZuse Group Title
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    hmm time to find it, and see if any other's fail anything....

    • one year ago
  58. KonradZuse Group Title
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    it seems like the others pass the other axioms, now to try and figure out which one fails #4 :P

    • one year ago
  59. KonradZuse Group Title
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    Does D by any chance fail Axiom 2?

    • one year ago
  60. Zarkon Group Title
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    no...but it fails another axiom

    • one year ago
  61. Zarkon Group Title
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    I'll tell you that only one of the four is an inner product

    • one year ago
  62. KonradZuse Group Title
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    geh..

    • one year ago
  63. KonradZuse Group Title
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    C also seems to fail one.

    • one year ago
  64. KonradZuse Group Title
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    my program just crashed so h/o.

    • one year ago
  65. KonradZuse Group Title
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    all of my work is gone, I'm flipping out, hold on please.

    • one year ago
  66. KonradZuse Group Title
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    SO d fails some axiom, but what about C @Zarkon ?

    • one year ago
  67. Zarkon Group Title
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    (c) is an inner product...the only one

    • one year ago
  68. KonradZuse Group Title
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    Damn now I got to look into b... Maybe that also fails axiom 4. D must fail axiom 3, maybe 4, IDk. Need to rework it, but flipping out trying to finish these other problems. Thanks for all the help I really appreciate it!

    • one year ago
  69. Zarkon Group Title
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    \[(u,v)=u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2\] \[(ku,v)=(ku_1)^2v_1^2+(ku_2)^2v_2^2+(ku_3^2)v_3^2\] \[=k^2(u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2)=k^2(u,v)\]

    • one year ago
  70. KonradZuse Group Title
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    axiom 3

    • one year ago
  71. KonradZuse Group Title
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    do you think you could help me with my other problems? I would greatly appreciate it.

    • one year ago
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