Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

KonradZuse

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;

  • one year ago
  • one year ago

  • This Question is Closed
  1. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  2. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  3. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    @mahmit2012 @kropot72 @asnaseer

    • one year ago
  4. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    • one year ago
    1 Attachment
  5. mahmit2012
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
  6. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    so I basically compare 1 at a time to each other...?

    • one year ago
  7. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    in the book's example they compare each of them to each other it seems.(u1,u3) = (u2,u1) = (u2,u3) = 0

    • one year ago
  8. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    I guess they do 2 at a time though.....

    • one year ago
  9. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    @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?

    • one year ago
  10. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    Also wouldn't a b and D be an orthogonal set since each individual is orthogonal?

    • one year ago
  11. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    @UnkleRhaukus

    • one year ago
  12. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  13. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    @Outkast3r09 save me :)

    • one year ago
  14. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  15. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  16. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    @jim_thompson5910 save me ;)

    • one year ago
  17. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 0

    you could plot the point on a graph

    • one year ago
  18. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 0

    orthogonal points will be separated by a 90° angle at the origin

    • one year ago
  19. UnkleRhaukus
    Best Response
    You've already chosen the best response.
    Medals 0

    orthogonal sets do not have to have a norm = 1. orthonormal sets do

    • one year ago
  20. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  21. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    @tkhunny

    • one year ago
  22. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
  23. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

    • one year ago
  24. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
  25. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah I failed haha.

    • one year ago
  26. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  27. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    definition 1 states that if each are orthogonal sets, then they will be orthogonal to each other.

    • one year ago
  28. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    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?

    • one year ago
  29. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    Idk I forgot :P

    • one year ago
  30. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    Well, check that definition and be sure. You'll be the only one in your class to notice the technicality.

    • one year ago
  31. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    BLEHEHEH

    • one year ago
  32. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    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."

    • one year ago
  33. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    so does that mean the only orthagonal sets are a and b?

    • one year ago
  34. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
  35. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    I did't think we needed to use basis on this one... :(

    • one year ago
  36. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
  37. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    mhm

    • one year ago
  38. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    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).

    • one year ago
  39. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    so a is gone too?

    • one year ago
  40. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    Well, how are we doing? Are we getting to the bottom of this one, too?

    • one year ago
  41. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    I thought we were comparing based on the definition.. Guess not :P

    • one year ago
  42. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    'a' is gone by inspections. (2,0) has length 2, not 1.

    • one year ago
  43. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    kk

    • one year ago
  44. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  45. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    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!

    • one year ago
  46. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    Which of the sets in Exercise 1 are orthonormal with respect to the Euclidean inner product on R^2

    • one year ago
  47. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    which from the book def 1 says what i've been saying above.

    • one year ago
  48. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  49. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    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!

    • one year ago
  50. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    maybe that means something else.... But liek you were saying above norm = 1.

    • one year ago
  51. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm?

    • one year ago
  52. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    not sure what you mean...?

    • one year ago
  53. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    I think the last word in Definition 1 should be "orthonormal".

    • one year ago
  54. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    meh stupid book I hate thee....

    • one year ago
  55. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    This is what I was saying about normalizing.... It shows v2 and v2 = sqrt(2) but then normalizing it they got it = 1?

    • one year ago
    1 Attachment
  56. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    it says we should verify it tho.

    • one year ago
  57. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
  58. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    by the way I get an inner product of 1 for c.

    • one year ago
  59. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    Verify what? We did all the verification in order to answer the question. Show the inner product (0) and the Norms (1). Done.

    • one year ago
  60. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    look at the picture i posted.....

    • one year ago
  61. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    it takes the norm of 3 vectors, and then normalizes the vectors which then sets them = 1?

    • one year ago
  62. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    a also = 1 if done this way....

    • one year ago
  63. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    ?? Both pieces are negative. Can't be +1.

    • one year ago
  64. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    neg * neg + pos * pos...

    • one year ago
  65. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    ?? \((<a,b><c,d>) = a*c + b*d\) You seem to be doing a*b + c*d.

    • one year ago
  66. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh yeah :P good call.

    • one year ago
  67. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
  68. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    so for the norm of a do we just do each individual section? sqrt(0^2 + 2^2) = 2

    • one year ago
  69. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    kk

    • one year ago
  70. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    norm =length?

    • one year ago
  71. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    so we take each individual vector than to see? So all vectors have to = 1 then?

    • one year ago
  72. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
  73. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    I just wanna write it out all nice :).

    • one year ago
  74. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    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;

    • one year ago
  75. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    This is what I wrote out... Should be good :). Thanks for the help!

    • one year ago
    1 Attachment
  76. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
  77. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    So it looks good? :) "An orthogonal set in which each vector has norm 1 is said to be orthognormal????"

    • one year ago
  78. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    I added that to my homework, that isn't in that pic :P

    • one year ago
  79. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    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!!)

    • one year ago
  80. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  81. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm not sure if that's a good example tho....

    • one year ago
  82. KonradZuse
    Best Response
    You've already chosen the best response.
    Medals 0

    Someone posted water flow, health care cost, population, NFL wins... :(

    • one year ago
  83. tkhunny
    Best Response
    You've already chosen the best response.
    Medals 1

    Interest Rate / Investment Earnings Prognostication.

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.