Least squares using matrices | Lecture 26 | Matrix Algebra for Engineers

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  • Опубликовано: 4 ноя 2024

Комментарии • 28

  • @ProfJeffreyChasnov
    @ProfJeffreyChasnov  4 года назад +3

    Find other Matrix Algebra videos in my playlist ruclips.net/p/PLkZjai-2Jcxlg-Z1roB0pUwFU-P58tvOx

  • @johnadams6249
    @johnadams6249 Год назад +4

    This is a great overview of the problem! A teacher of mine went over the steps without all the justification, so this was very helpful!

  • @area51xi
    @area51xi 3 года назад +3

    Wow. I was initially skeptical when I started watching thinking this is just another video on this topic. I already knew this material but now I know it much better. Very clear explanation. Thank you.

  • @harshvardhanranvirsingh9473
    @harshvardhanranvirsingh9473 3 года назад +2

    Sir, you are amazing! This is by far the best at least according to me a best explanation for least squares optimization problem.

  • @sanjogkarki7999
    @sanjogkarki7999 5 лет назад +2

    Great video to quickly review the concepts. Thanks Jeffrey.

  • @ElizaberthUndEugen
    @ElizaberthUndEugen 5 лет назад +8

    Ax = b in general has no solution here, since b is in an n-dimensional space and Ax is in a 2-dimensional (at most 2) subspace of that space, since Ax is a linear combination of 2 n-dimensional vectors.
    Just for anyone else that got stuck there for a moment.

  • @ArmstrongNigere
    @ArmstrongNigere Год назад

    great video , finally understand clearly the topic

  • @area51xi
    @area51xi 3 года назад +3

    I think what makes this topic confusing for beginners is that the x vector in Ax=b does not represent the "x's" but rather represents the betas. And the b in Ax=b does not represent the betas but rather the y's. It is an unfortunate accident of the nomenclature.

    • @ProfJeffreyChasnov
      @ProfJeffreyChasnov  3 года назад +1

      That would be a beginning beginner.

    • @jackieliu1472
      @jackieliu1472 2 года назад

      @@ProfJeffreyChasnov me is a beginning beginner ^o^ ty for the great lectures

  • @weisanpang7173
    @weisanpang7173 Год назад

    At 7:17, you said "if AX equals B is overdetermined, that means B is not in the column space of A, there's no solution to X". Where does that come from ?

  • @max.hastings
    @max.hastings 4 года назад +34

    Is this dude really out here writing backwards

    • @10bokaj
      @10bokaj 4 года назад +9

      or maybe some tech has been made that can flip a video (omg what has the world come to)...

    • @Peter_1986
      @Peter_1986 4 года назад +2

      I think that he is recording himself behind a transparent wall, and then he reflects that recording around the vertical axis.

    • @cathyfeng3385
      @cathyfeng3385 4 года назад

      or he's writing on a mirror and the video might be everything in the mirror

    • @max.hastings
      @max.hastings 4 года назад +1

      @@cathyfeng3385 writing on glass and then flipping video horizontally

    • @charliehou9553
      @charliehou9553 3 года назад +2

      TENET

  • @CRISTIANO18786
    @CRISTIANO18786 2 года назад

    it's not necessary that an overdetermined system has no solutions, it is unlikely in a real world scenario but it's not necessarily the case. If the columns of an overdetermined matrix are dependant for example in the 2D case then they lie on a line. my point is that the fact that there is no solution is not because the system is overdetermined but because b happens to not lie in the column space

  • @Flannel_girly
    @Flannel_girly Год назад

    you are the best!!!

  • @benjellounyassine8646
    @benjellounyassine8646 2 года назад

    Very helpful, thank you

  • @michaelumitbozdemir3185
    @michaelumitbozdemir3185 3 года назад

    You are a legend 👍

  • @wallyduboss9464
    @wallyduboss9464 2 месяца назад

    you lost me at Ax =b. . you have x under B and B1, A under X and b under the y's.
    If y=Ax +b, isn't what you wrote confusing?

  • @zhou6075
    @zhou6075 3 года назад

    thank you

  • @martinsanchez-hw4fi
    @martinsanchez-hw4fi 3 года назад

    In 7:15. If Ax=b it is not true that this implies that b does not lie in the collumnspace of A. It may be overdetermined and have a solution.

  • @anselmo2194
    @anselmo2194 4 года назад

    Nice sir

  • @sollinw
    @sollinw 4 года назад

    great!