Gram-Schmidt Orthogonalization

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  • Опубликовано: 21 дек 2024

Комментарии • 22

  • @monatoshbarman11
    @monatoshbarman11 5 месяцев назад +3

    The ending explanation about how the entries in R can be seen/found in previous calculations was very helpful..
    Thank you so much.

  • @dinuthomas4531
    @dinuthomas4531 3 года назад +9

    Excellent explanation and conclusion on QR relations

  • @Y2B123
    @Y2B123 2 года назад +2

    These videos are incredibly helpful in refreshing those ideas. It makes these concepts really stick in my mind.

  • @AbdullahArshad-u2w
    @AbdullahArshad-u2w 2 месяца назад +1

    Good understanding for R

  • @amratia99
    @amratia99 6 лет назад +14

    Well explained ! But the professor said in the lecture that :
    C = c - (aTc/aTa)*a - (bTc/bTb)*b
    This will give me another answer which is wrong .. but I don't know why .. If we use this formula that means that we want a vector prep to a .. then take this vector and make it prep to b .. don't know why didn't work

    • @rooftop1510
      @rooftop1510 6 лет назад +8

      the video uses the new vector a and b! if u use the new vector aTa and bTb will be 1.

    • @sudharshantr8757
      @sudharshantr8757 Год назад

      since the short answer is already given, here is a deeper one. Let p and q be 2 dimensional vectors which are independent (but not orthogonal!). We know that p, q span the entire 2d space. Suppose, we want to express another 2d vector r = ap + bq. Can we say a = rTp/pTp and b = rTq/qTq? Draw it in a piece of paper and you will see we can't!
      we can write is such a fashion only if p and q are perpendicular.

  • @quirkyquester
    @quirkyquester 4 года назад +3

    great eg, thank you Ana!

  • @tanyabarnwal2061
    @tanyabarnwal2061 6 лет назад +4

    Excellent explanation

  • @matheuscardoso1110
    @matheuscardoso1110 5 лет назад +3

    great great video. thank you Ana!

  • @AnupKumar-wk8ed
    @AnupKumar-wk8ed 6 лет назад +5

    You are awesome!

  • @johannesmariomeissner7262
    @johannesmariomeissner7262 4 года назад +4

    2:21 "When you do the projection of a vector onto another, you have to divide by the length of in this case q1."
    Wouldn't it be dividing by the length squared? Or in other words, q1 dot q1?
    That's what the formula for Gram-Schmidt tells us: q2' = (q1 q1T b) / (q1T q1)
    where q1T q1 happens to be the length squared of q1.

    • @josephlevine2836
      @josephlevine2836 4 года назад +5

      You almost got it! For any vector a, aT a = |a|^2. In the case where the vector is q it is defined to have length 1. qT q = 1^2 = 1
      An example might help. Look at the second term in Gram-Schmidt:
      a (aT b) / (aT a)
      = a(aT b) / |a|^2
      = a/|a| * aT/|a| b
      but a/|a| = q1 and aT/|a| = q1T
      so
      a (aT b) / (aT a) = q1 (q1T b)
      I used the parentheses in the final line to show the meaning: project b along the direction of a to get a scalar. Multiply that by the direction of a.
      Hope this helps!

  • @KG16888
    @KG16888 3 года назад +2

    very clear!

  • @user-qj6hl5xb8q
    @user-qj6hl5xb8q 3 года назад

    Around 7:40, how do we go from 2 3 0 in the second column of R
    to
    b=2q1 + 3q2?

    • @davidondev7382
      @davidondev7382 Год назад +1

      You might have found it out by yourself, but what she was doing is checking if A=QR, where Q came from the Gram-Schmidt, and R came from observation, (but she explained a general way to get R at the end).
      For b=2q1 + 3q2, it checked the second column of A (which is b) = the matrix Q multiply the second column of R (which is [2, 3, 0]), so it's b = R_{12}q_1 + R_{22}q_2 + R_{32}q_3.
      Hope this would be useful.

  • @thedailyepochs338
    @thedailyepochs338 4 года назад

    good one ana

  • @cheekyiscool
    @cheekyiscool 5 лет назад +1

    If you cannot follow the formula used here, I'd recommend going through this intro first: ruclips.net/video/rHonltF77zI/видео.html (Gram-Schmidt process by Khan Academy)

    • @iharsh386
      @iharsh386 Год назад +1

      thanks man, that really helps

  • @yetaowu2658
    @yetaowu2658 5 лет назад

    great!

  • @rickshawty
    @rickshawty 7 месяцев назад

  • @uffzymoron2407
    @uffzymoron2407 Год назад +3

    only she can fix me