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KonradZuse

  • 2 years ago

Anyone good with inner products and their axioms?

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  1. tyteen4a03
    • 2 years ago
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    Just ask your question.

  2. KonradZuse
    • 2 years ago
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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.

  3. KonradZuse
    • 2 years ago
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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]

  4. KonradZuse
    • 2 years ago
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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

  5. KonradZuse
    • 2 years ago
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    @Hero save me.

  6. Hero
    • 2 years ago
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    Bro, I took linear algebra already, but I'd have to re-familiarize myself with this.

  7. Hero
    • 2 years ago
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    Hang on for a minute

  8. Hero
    • 2 years ago
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    @KonradZuse, where in the book is that question?

  9. KonradZuse
    • 2 years ago
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    not in the book.

  10. KonradZuse
    • 2 years ago
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    sorry it's kind of screwed up in formatting.

  11. KonradZuse
    • 2 years ago
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    1 second.

  12. KonradZuse
    • 2 years ago
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  13. KonradZuse
    • 2 years ago
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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.

  14. KonradZuse
    • 2 years ago
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    okay hold on.

  15. KonradZuse
    • 2 years ago
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    here happy?

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

  17. KonradZuse
    • 2 years ago
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    saying*

  18. KonradZuse
    • 2 years ago
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    @UnkleRhaukus

  19. Zarkon
    • 2 years ago
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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

  20. UnkleRhaukus
    • 2 years ago
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    where are the axioms you are referencing ?

  21. KonradZuse
    • 2 years ago
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    one second.

  22. KonradZuse
    • 2 years ago
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  23. UnkleRhaukus
    • 2 years ago
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    (a) \[(\textbf u,\textbf v)=u_1 v_1+u_3v_3 \]

  24. KonradZuse
    • 2 years ago
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    I was thinking that might fail because of the new u2v2, but not too sure?

  25. UnkleRhaukus
    • 2 years ago
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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

  26. KonradZuse
    • 2 years ago
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    ok.

  27. KonradZuse
    • 2 years ago
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    I actually posted my work up a bit. Should I repost it?

  28. KonradZuse
    • 2 years ago
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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?

  29. UnkleRhaukus
    • 2 years ago
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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

  30. UnkleRhaukus
    • 2 years ago
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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

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

  33. UnkleRhaukus
    • 2 years ago
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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

  34. Zarkon
    • 2 years ago
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    that is the problem..axiom 4

  35. Zarkon
    • 2 years ago
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    your work done on axiom 4 is not correct

  36. Zarkon
    • 2 years ago
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    remember \(u\) and \(v\) are 3-dimensional vectors

  37. Zarkon
    • 2 years ago
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    let \(u=v=<0,1,0>\) what is \((u,v)\)

  38. Zarkon
    • 2 years ago
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    or same thing ...what is \((v,v)\)

  39. KonradZuse
    • 2 years ago
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    I will check it out.

  40. KonradZuse
    • 2 years ago
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    I guess I'm still confused how to prove axiom 4....

  41. Zarkon
    • 2 years ago
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    you don't prove axiom 4...it is an axiom

  42. KonradZuse
    • 2 years ago
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    errr "satisfy axiom 4"

  43. Zarkon
    • 2 years ago
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    (a) is not an inner product...I gave you a counter example

  44. Zarkon
    • 2 years ago
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    ((0,1,0),(0,1,0))=0*0+0*0=0 but (0,1,0) is not the zero vector

  45. KonradZuse
    • 2 years ago
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    ic so axiom 4 isn't = 0 thus fails?

  46. Zarkon
    • 2 years ago
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    it fails... \((v,v)=0\Leftrightarrow v=\vec{0}\)

  47. UnkleRhaukus
    • 2 years ago
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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

  48. Zarkon
    • 2 years ago
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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 :) )

  49. KonradZuse
    • 2 years ago
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    I'm so confused though :)

  50. KonradZuse
    • 2 years ago
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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?

  51. UnkleRhaukus
    • 2 years ago
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    how did you get = 1 ?

  52. KonradZuse
    • 2 years ago
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    I thought that's what he said above, I guess I'm just too konfused hahaha :(

  53. UnkleRhaukus
    • 2 years ago
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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)

  54. KonradZuse
    • 2 years ago
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    hmm ic.

  55. KonradZuse
    • 2 years ago
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    So does Axiom 4 only fail for the first one?

  56. Zarkon
    • 2 years ago
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    another one of your problems fails axiom 4

  57. KonradZuse
    • 2 years ago
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    hmm time to find it, and see if any other's fail anything....

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

  59. KonradZuse
    • 2 years ago
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    Does D by any chance fail Axiom 2?

  60. Zarkon
    • 2 years ago
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    no...but it fails another axiom

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

  62. KonradZuse
    • 2 years ago
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    geh..

  63. KonradZuse
    • 2 years ago
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    C also seems to fail one.

  64. KonradZuse
    • 2 years ago
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    my program just crashed so h/o.

  65. KonradZuse
    • 2 years ago
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    all of my work is gone, I'm flipping out, hold on please.

  66. KonradZuse
    • 2 years ago
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    SO d fails some axiom, but what about C @Zarkon ?

  67. Zarkon
    • 2 years ago
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    (c) is an inner product...the only one

  68. KonradZuse
    • 2 years ago
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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!

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

  70. KonradZuse
    • 2 years ago
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    axiom 3

  71. KonradZuse
    • 2 years ago
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    do you think you could help me with my other problems? I would greatly appreciate it.

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