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KonradZuse

  • 2 years ago

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;

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  1. KonradZuse
    • 2 years ago
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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. KonradZuse
    • 2 years ago
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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.

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

  4. KonradZuse
    • 2 years ago
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  5. mahmit2012
    • 2 years ago
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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.

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

  7. KonradZuse
    • 2 years ago
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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

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

  9. KonradZuse
    • 2 years ago
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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?

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

  11. KonradZuse
    • 2 years ago
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    @UnkleRhaukus

  12. KonradZuse
    • 2 years ago
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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.

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

  14. KonradZuse
    • 2 years ago
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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.

  15. KonradZuse
    • 2 years ago
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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.

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

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

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

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

  20. KonradZuse
    • 2 years ago
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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

  21. KonradZuse
    • 2 years ago
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    @tkhunny

  22. tkhunny
    • 2 years ago
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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.

  23. KonradZuse
    • 2 years ago
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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?

  24. tkhunny
    • 2 years ago
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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.

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

  26. KonradZuse
    • 2 years ago
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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.

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

  28. tkhunny
    • 2 years ago
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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?

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

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

  31. KonradZuse
    • 2 years ago
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    BLEHEHEH

  32. KonradZuse
    • 2 years ago
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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."

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

  34. tkhunny
    • 2 years ago
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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.

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

  36. tkhunny
    • 2 years ago
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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.

  37. KonradZuse
    • 2 years ago
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    mhm

  38. tkhunny
    • 2 years ago
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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).

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

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

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

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

  43. KonradZuse
    • 2 years ago
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    kk

  44. KonradZuse
    • 2 years ago
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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.

  45. tkhunny
    • 2 years ago
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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!

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

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

  48. KonradZuse
    • 2 years ago
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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.

  49. tkhunny
    • 2 years ago
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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!

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

  51. KonradZuse
    • 2 years ago
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    hmm?

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

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

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

  55. KonradZuse
    • 2 years ago
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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?

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

  57. tkhunny
    • 2 years ago
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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.

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

  59. tkhunny
    • 2 years ago
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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.

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

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

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

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

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

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

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

  67. tkhunny
    • 2 years ago
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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.

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

  69. KonradZuse
    • 2 years ago
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    kk

  70. KonradZuse
    • 2 years ago
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    norm =length?

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

  72. tkhunny
    • 2 years ago
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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.

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

  74. tkhunny
    • 2 years ago
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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;

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

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  76. tkhunny
    • 2 years ago
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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.

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

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

  79. tkhunny
    • 2 years ago
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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!!)

  80. KonradZuse
    • 2 years ago
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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.

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

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

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

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