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KonradZuse Group Title

6.3 #2 Which of the sets in Exercise 1 are orthonormal with respect to the Euclidean inner product on R^2 a = (0, 1), 2, 0; b = (-1/sqrt(2), 1/sqrt(2)), 1/sqrt(2), 1/sqrt(2); c = (-1/sqrt(2), -1/sqrt(2)), 1/sqrt(2), 1/sqrt(2); d = (0, 0), 0, 1;

  • 2 years ago
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  1. KonradZuse Group Title
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    Now it says in the book that A set of two or more vectors in a real inner product space is said to be orthogonal if all pairs of distinct vectors in the set are orthogonal. An orthogonal set in which each vector has norm 1 is said to be orthogonal.

    • 2 years ago
  2. KonradZuse Group Title
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    Now in the problem before this I found that a,b, and d were orthogonal vectors. So would I use those 3 only, or all 4? I'm not sure what the latter part means "An orthogonal set in which each vector has norm 1 is said to be orthogonal." I would think we wouldn't know it was orthogonal, and isn't saying an orthogonal set is orthogonal kind of redundant??? If it meant that we know the vectors are already orthogonal then I would assume the 3 a b, and d.

    • 2 years ago
  3. KonradZuse Group Title
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    @mahmit2012 @kropot72 @asnaseer

    • 2 years ago
  4. KonradZuse Group Title
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    • 2 years ago
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  5. mahmit2012 Group Title
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    b & c is a orthonormal set. a with (1,0) also is a orthonormal, but d is a (0,0) and it can not be in a orthonormal set.

    • 2 years ago
  6. KonradZuse Group Title
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    so I basically compare 1 at a time to each other...?

    • 2 years ago
  7. KonradZuse Group Title
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    in the book's example they compare each of them to each other it seems.(u1,u3) = (u2,u1) = (u2,u3) = 0

    • 2 years ago
  8. KonradZuse Group Title
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    I guess they do 2 at a time though.....

    • 2 years ago
  9. KonradZuse Group Title
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    @mahmit2012 should I try a and C and b and D? I'm also not 100% exact what you were saying about (0,0) for d?

    • 2 years ago
  10. KonradZuse Group Title
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    Also wouldn't a b and D be an orthogonal set since each individual is orthogonal?

    • 2 years ago
  11. KonradZuse Group Title
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    @UnkleRhaukus

    • 2 years ago
  12. KonradZuse Group Title
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    I think that according to the definition I would have to just use the ones that make sense... C shouldn't be anything since it's not orthogonal, and according to definition 1 that states: A set of two or more vectors in a real inner product space is said to be orthogonal if all pairs of distinct vectors in the set are orthogonal.

    • 2 years ago
  13. KonradZuse Group Title
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    @Outkast3r09 save me :)

    • 2 years ago
  14. KonradZuse Group Title
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    Definition 1 stating "A set of two or more vectors in a real inner product space is said to be orthogonal if all pairs of distinct vectors in the set are orthogonal." So we also should concluded that set a and d are orthogonal, a and b are orthogonal, b and d are orthogonal to each other.

    • 2 years ago
  15. KonradZuse Group Title
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    Now it also says that orthogonal sets have to have a norm = 1 to be orthogonal, so should I check for that also? a's norm is 2, but by normalizing it you get 1 as the new norm.

    • 2 years ago
  16. KonradZuse Group Title
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    @jim_thompson5910 save me ;)

    • 2 years ago
  17. UnkleRhaukus Group Title
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    you could plot the point on a graph

    • 2 years ago
  18. UnkleRhaukus Group Title
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    orthogonal points will be separated by a 90° angle at the origin

    • 2 years ago
  19. UnkleRhaukus Group Title
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    orthogonal sets do not have to have a norm = 1. orthonormal sets do

    • 2 years ago
  20. KonradZuse Group Title
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    DEFINITION 1 A set of two or more vectors in a real inner product space is said to be orthogonal if all pairs of distinct vectors in the set are orthogonal. An orthogonal set in which each vector has norm 1 is said to be orthogonal. Wtf is ortonormal? @UnkleRhaukus

    • 2 years ago
  21. KonradZuse Group Title
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    @tkhunny

    • 2 years ago
  22. tkhunny Group Title
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    OrthoNormal = Orthogonal + Normal = Orthogonal + (Unit Length) Orthogonal -- Calculate the Inner Product. If you get "0", you are done. They are Orthogonal. Normal -- Calculate the Length of each Basis Vector. If all lengths are "1", you are done - Normal it is.

    • 2 years ago
  23. KonradZuse Group Title
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    so did I do this correctly since the vectors a b and d are orthogonal, that the sets a and b, b and d, d and a are arthogonal?

    • 2 years ago
  24. tkhunny Group Title
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    You will need to test all basis vectors, pairwise, for orthogonality. You will need to test all basis vecors individually for length. Just question on notation: a = (0, 1), 2, 0; Is that one basis vector (0,1) and another basis vector (2,0), or is it something else. I'm hoping you just failed to use the parentheses on the second vector. Otherwise, I don't know what it means.

    • 2 years ago
  25. KonradZuse Group Title
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    yeah I failed haha.

    • 2 years ago
  26. KonradZuse Group Title
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    well I tested each pair of vectors to see if they were orthogonal to each other, and A, B, and D are the ones that are orthogonal.

    • 2 years ago
  27. KonradZuse Group Title
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    definition 1 states that if each are orthogonal sets, then they will be orthogonal to each other.

    • 2 years ago
  28. tkhunny Group Title
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    a = (0, 1), 2, 0; b = (-1/sqrt(2), 1/sqrt(2)), 1/sqrt(2), 1/sqrt(2); c = (-1/sqrt(2), -1/sqrt(2)), 1/sqrt(2), 1/sqrt(2); d = (0, 0), 0, 1; a --- Orthogonal, but not Normal d --- Definition Question,is EVERTHING considered Orthogonal to the zero-vector? Are you SURE that using the zeo-vector constitutes a Basis?

    • 2 years ago
  29. KonradZuse Group Title
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    Idk I forgot :P

    • 2 years ago
  30. tkhunny Group Title
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    Well, check that definition and be sure. You'll be the only one in your class to notice the technicality.

    • 2 years ago
  31. KonradZuse Group Title
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    BLEHEHEH

    • 2 years ago
  32. KonradZuse Group Title
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    This one?? I"t is not true that every vector space has a basis in the sense of Definition 1. The simplest example is the zero vector space, which contains no linearly independent sets and hence no basis. The following is an example of a nonzero vector space that has no basis in the sense of Definition 1 because it cannot be spanned by finitely many vectors."

    • 2 years ago
  33. KonradZuse Group Title
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    so does that mean the only orthagonal sets are a and b?

    • 2 years ago
  34. tkhunny Group Title
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    Okay, we threw out 'c' as not being orthogonal. That leaves ONLY the length of the b-vectors to consider. Are they length one (1)? I think we have a winner.

    • 2 years ago
  35. KonradZuse Group Title
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    I did't think we needed to use basis on this one... :(

    • 2 years ago
  36. tkhunny Group Title
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    As far as the zero vector goes, it cannot be a Basis Vector, since it cannot EVER be linerly independent from other Basis Vectors. Example, If I have two basis vectors (1,0) and (0,1), I can get (0,0) from 0*(1,0) or from 0*(0,1). Definitely not a valid basis vector, zero.

    • 2 years ago
  37. KonradZuse Group Title
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    mhm

    • 2 years ago
  38. tkhunny Group Title
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    You always need to consider Basis Vectors, whether it is stated explicitly in the problem statement or not. The correct answer to 'd' is , "Cannot be an orthonormal basis, since it isn't a basis at all." We could also argue that the zero vector has length zero, which clearly is not one (1).

    • 2 years ago
  39. KonradZuse Group Title
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    so a is gone too?

    • 2 years ago
  40. tkhunny Group Title
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    Well, how are we doing? Are we getting to the bottom of this one, too?

    • 2 years ago
  41. KonradZuse Group Title
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    I thought we were comparing based on the definition.. Guess not :P

    • 2 years ago
  42. tkhunny Group Title
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    'a' is gone by inspections. (2,0) has length 2, not 1.

    • 2 years ago
  43. KonradZuse Group Title
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    kk

    • 2 years ago
  44. KonradZuse Group Title
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    my book says that you can first take the norm of a vector, then normalize it based on that norm, then take the norm again to see.... In that case a would = 1.

    • 2 years ago
  45. tkhunny Group Title
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    Maybe you're right. It does not ask, "Which form an orthonormal basis?" That is how I was thinking about the problem. It never hurts to go back and read the problem statement. Recap: a) No good. (2,0) has length 2, not 1 d) No good. (0,0) has length 0, not 1 c) No Good. Inner Product is -1, not 0 b) Got it!

    • 2 years ago
  46. KonradZuse Group Title
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    Which of the sets in Exercise 1 are orthonormal with respect to the Euclidean inner product on R^2

    • 2 years ago
  47. KonradZuse Group Title
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    which from the book def 1 says what i've been saying above.

    • 2 years ago
  48. KonradZuse Group Title
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    EFINITION 1 A set of two or more vectors in a real inner product space is said to be orthogonal if all pairs of distinct vectors in the set are orthogonal. An orthogonal set in which each vector has norm 1 is said to be orthogonal.

    • 2 years ago
  49. tkhunny Group Title
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    a) That makes no sense. It's not a very interesting question if we can just normalize the ones that aren't normalized. d) Of course, I dare you to normalize this one!

    • 2 years ago
  50. KonradZuse Group Title
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    maybe that means something else.... But liek you were saying above norm = 1.

    • 2 years ago
  51. KonradZuse Group Title
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    hmm?

    • 2 years ago
  52. KonradZuse Group Title
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    not sure what you mean...?

    • 2 years ago
  53. tkhunny Group Title
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    I think the last word in Definition 1 should be "orthonormal".

    • 2 years ago
  54. KonradZuse Group Title
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    meh stupid book I hate thee....

    • 2 years ago
  55. KonradZuse Group Title
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    This is what I was saying about normalizing.... It shows v2 and v2 = sqrt(2) but then normalizing it they got it = 1?

    • 2 years ago
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  56. KonradZuse Group Title
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    it says we should verify it tho.

    • 2 years ago
  57. tkhunny Group Title
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    DEFINITION 1 A set of two or more vectors in a real inner product space is said to be orthogonal if all pairs of distinct vectors in the set are orthogonal. An orthogonal set in which each vector has norm 1 is said to be orthonormal. I stand by this answer: a) No good. (2,0) has length 2, not 1 d) No good. (0,0) has length 0, not 1 c) No Good. Inner Product is -1, not 0 b) Got it! This seems to be most consistent with the problem statement and Definition #1.

    • 2 years ago
  58. KonradZuse Group Title
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    by the way I get an inner product of 1 for c.

    • 2 years ago
  59. tkhunny Group Title
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    Verify what? We did all the verification in order to answer the question. Show the inner product (0) and the Norms (1). Done.

    • 2 years ago
  60. KonradZuse Group Title
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    look at the picture i posted.....

    • 2 years ago
  61. KonradZuse Group Title
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    it takes the norm of 3 vectors, and then normalizes the vectors which then sets them = 1?

    • 2 years ago
  62. KonradZuse Group Title
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    a also = 1 if done this way....

    • 2 years ago
  63. tkhunny Group Title
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    ?? Both pieces are negative. Can't be +1.

    • 2 years ago
  64. KonradZuse Group Title
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    neg * neg + pos * pos...

    • 2 years ago
  65. tkhunny Group Title
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    ?? \((<a,b><c,d>) = a*c + b*d\) You seem to be doing a*b + c*d.

    • 2 years ago
  66. KonradZuse Group Title
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    Oh yeah :P good call.

    • 2 years ago
  67. tkhunny Group Title
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    Your picture has two sections. One side is determining orthogonality. The other side is concerned with CONSTRUCTING an orthonormal set. We are not doing any CONSTRUCTING in this problem. We are just determining if it has already been constructed.

    • 2 years ago
  68. KonradZuse Group Title
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    so for the norm of a do we just do each individual section? sqrt(0^2 + 2^2) = 2

    • 2 years ago
  69. KonradZuse Group Title
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    kk

    • 2 years ago
  70. KonradZuse Group Title
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    norm =length?

    • 2 years ago
  71. KonradZuse Group Title
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    so we take each individual vector than to see? So all vectors have to = 1 then?

    • 2 years ago
  72. tkhunny Group Title
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    That's it. In this case, if we don't get one (1), we throw our hands in the air and exclaim, "Not Normal!" We can make it Normal, but that is not part of this problem statement.

    • 2 years ago
  73. KonradZuse Group Title
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    I just wanna write it out all nice :).

    • 2 years ago
  74. tkhunny Group Title
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    I'm all for exploration and playing around to see what works. This answer is my actual thought process for answering this problem specifically. First, I rule out the obvious ones: a) No good. (2,0) has length 2, not 1 d) No good. (0,0) has length 0, not 1 Then, with a little work, I narrow it down some more. c) No Good. Inner Product is -1, not 0 b) Got it! If you want to display it very well, I would do this a = (0, 1), (2, 0); \(\sqrt{0^{2} + 1^{2}} = \sqrt{1} = 1\) - Normal \(\sqrt{2^{2} + 0^{2}} = \sqrt{4} = 2\) - Not Normal \((<0,1>,<2,0>) = 0*2 + 1*0 = 0\) - Orthogonal This set is NOT orthonormal And then proceed similarly with the other three. b = (-1/sqrt(2), 1/sqrt(2)), 1/sqrt(2), 1/sqrt(2); c = (-1/sqrt(2), -1/sqrt(2)), 1/sqrt(2), 1/sqrt(2); d = (0, 0), 0, 1;

    • 2 years ago
  75. KonradZuse Group Title
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    This is what I wrote out... Should be good :). Thanks for the help!

    • 2 years ago
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  76. tkhunny Group Title
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    Always a pleasure to see a good student working hard to get it right! Let the definitions soak into your brain. Oh, and don't forget to fix that error in your book's Definition 1.

    • 2 years ago
  77. KonradZuse Group Title
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    So it looks good? :) "An orthogonal set in which each vector has norm 1 is said to be orthognormal????"

    • 2 years ago
  78. KonradZuse Group Title
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    I added that to my homework, that isn't in that pic :P

    • 2 years ago
  79. tkhunny Group Title
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    orthonormal. No extra 'g' in there. And yes, the "Norm" is essentially the Length. Ortholengthal just looks dumb, so we go with orthonormal. (Not really. I just made that up. MATH JOKES!!)

    • 2 years ago
  80. KonradZuse Group Title
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    By the way we have a discussion post where it asks us to use 6.5 to post an example of why we should use this... Normally I go by what I know, but I'm curious if this will wokr... They talk about least square regression lines, and I saw one y = a + bx which I've used in Physics for finding % error and such in our data.

    • 2 years ago
  81. KonradZuse Group Title
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    I'm not sure if that's a good example tho....

    • 2 years ago
  82. KonradZuse Group Title
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    Someone posted water flow, health care cost, population, NFL wins... :(

    • 2 years ago
  83. tkhunny Group Title
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    Interest Rate / Investment Earnings Prognostication.

    • one year ago
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