Dual basis

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  • Опубліковано 4 жов 2024
  • Dual basis definition and proof that it's a basis
    In this video, given a basis beta of a vector space V, I define the dual basis beta* of V*, and show that it's indeed a basis. We'll see many more applications of this concept later on, but this video already shows that it's straightforward to construct one. Moreover, we get a very elegant representation formula of a functional in terms of the dual basis. Enjoy!
    Subscribe to my channel: / drpeyam
    Dual Space video: • Dual Space
    Example of dual basis: • Dual Basis Example
    Check out my Dual Space playlist: • Dual Spaces

КОМЕНТАРІ • 79

  • @chachomask
    @chachomask 4 роки тому +67

    Rumors has it that this guy has infinite amount of boards he can just switch between

  • @hansbaeker9769
    @hansbaeker9769 Рік тому +9

    I'm getting close to retirement and am determined to make relearning the math I've forgotten a retirement goal. So now I'm reviewing dual vector spaces which is one area I've pretty much forgotten. This is definitely helping.

  • @sepidehtje
    @sepidehtje 3 роки тому +10

    Thank you so much for this amazing explanation! My own professor just glossed over it by saying "there exists a basis" that's it. It was driving me wild. Now THIS is a proper proof. You are a lifesaver.

  • @FT029
    @FT029 4 роки тому +28

    The visualization of functions as pictures (with all the dots) was extremely helpful! After that the proofs of linear independence and spanning became more intuitive.

  • @blackpenredpen
    @blackpenredpen 5 років тому +75

    Peyam, can you do 100 dual basis in one take?

    • @drpeyam
      @drpeyam  5 років тому +29

      Hahahahahahahaha

  • @jubachoomba
    @jubachoomba Рік тому +1

    Reisz representation theorem pops right out: bravo, Prof!

  • @zehrademirhan600
    @zehrademirhan600 3 роки тому +9

    This lecture is awesome:) i really don't understand from my instructor but you made all of the things clear . Thank you...

  • @williamhogrider4136
    @williamhogrider4136 Рік тому +2

    This is very well explained, so now I have an idea on dual space and dual basis but I'd like to revisit this video again, it's awesome

  • @apappas4482
    @apappas4482 4 роки тому +4

    Thank you for explaining slowly and clearly.

  • @rohitkantipudi6605
    @rohitkantipudi6605 4 роки тому +4

    this deserves way more views
    great explanation

  • @diegoviejocarretero7629
    @diegoviejocarretero7629 3 роки тому +2

    thanks for sounding so happy, I really needed that to cheer me up studyng for this

  • @klam77
    @klam77 4 роки тому +10

    excellent; better than my textbook!

  • @adebayoemmanuel911
    @adebayoemmanuel911 Рік тому

    I've been looking for a way to picture a dual basis. Thank you so much for this

  • @speedbird7587
    @speedbird7587 8 місяців тому

    Thanks for the lecture. It was full of excitement and energy. I really liked it.

  • @owen5106
    @owen5106 Рік тому

    just found your channel, im reading linear algebra done right on my own, your a G my dude. so helpful tyvm

  • @basakatik4770
    @basakatik4770 4 роки тому

    You are a hero Dr. Peyam! You made again everything is visible! Are you sure you are human!!!

  • @PascalsLaptop
    @PascalsLaptop 9 місяців тому

    thank you very much for making these videos

  • @sundayscrafter1779
    @sundayscrafter1779 5 років тому +4

    Now is time for the double dual teacher PiM :) It’s got pretty interesting properties :)

    • @drpeyam
      @drpeyam  5 років тому +1

      Already up!

  • @gorkifreire1380
    @gorkifreire1380 3 роки тому +1

    Amazing explanation thank you very much

  • @arseniikvachan
    @arseniikvachan 3 роки тому +1

    What the hell? Why should I go to university, if I have Internet with Dr. Peyam. Infinite thanks to you.

  • @arnavarora7023
    @arnavarora7023 Рік тому

    These videos are really helpful. Thanks a lot.

  • @Ruvine_
    @Ruvine_ 2 роки тому

    AWESOME GRAPH! I had the topic of duals on class and didn't understand a thing. Then I watched 4 other youtube videos and understand nothing to see your graph and understand everything instanlty.

    • @drpeyam
      @drpeyam  2 роки тому

      Thanks so much!! Yes, the graph really cleared it up for me

  • @margueritedepompadour7031
    @margueritedepompadour7031 6 місяців тому

    Thanks a lot! I never got what those notations really meant

  • @TheNachoesuncapo
    @TheNachoesuncapo 5 років тому +5

    This would be my summer fun!!!

  • @amergoel3117
    @amergoel3117 Рік тому

    thank you so much for making this video you saved me

  • @MoonLight-sw6pc
    @MoonLight-sw6pc 5 років тому +3

    U r right!
    linear algebra is beautiful :)
    شكرا

  • @dgrandlapinblanc
    @dgrandlapinblanc 4 роки тому +1

    Pretty cool. Thank you very much.

  • @estebangimenez3714
    @estebangimenez3714 5 років тому

    u saved my life, congrats from Spain.

  • @QT-yt4db
    @QT-yt4db 2 роки тому

    Very helpful... Thank you...

  • @lazarsavic6613
    @lazarsavic6613 2 роки тому

    Extremely helpful! :)

  • @thecarlostheory
    @thecarlostheory Рік тому

    Vey awesome video. Thx a lot, ur helping me so much :D

  • @rodrigodiazarancibia5486
    @rodrigodiazarancibia5486 Рік тому

    Thanks man!

  • @thepositron5676
    @thepositron5676 5 років тому +1

    Great explanation! Thank you :D

  • @arechilasalvia8331
    @arechilasalvia8331 3 роки тому

    Do one video on quotient space

  • @hojungkim241
    @hojungkim241 4 роки тому +1

    I have a question in proving a1=f(v1) in 17:04. Can we use f(v1)=a1f1(v1)+…+anfn(vn) before we prove any element in V* can be expressed by linear combination of ß*? I mean, we were on the way to prove f=a1f1+…anfn but we used that before finish the proof.

  • @climitod8524
    @climitod8524 4 місяці тому

    AH dangit I wish you did a video however on a dual map as that's the definition Im having trouble understanding how its all mapped and stuff.

    • @drpeyam
      @drpeyam  4 місяці тому

      Check out the playlist

  • @schlechtestergtaspielerdek3851

    This is electroboom for mathematicians or cs students XD. THx

  • @mario1415
    @mario1415 5 років тому

    Dr. Peyam! Now make a video about the Mackey topology! =)

  • @sekhar018
    @sekhar018 2 роки тому

    Excellent ✌️

  • @112BALAGE112
    @112BALAGE112 5 років тому +1

    Are you planning to discuss differential forms?

  • @justinbond7456
    @justinbond7456 Рік тому

    awesome lecture, thanks! however i have difficulty seeing that it is enough to show f(vi) = g(vi) for the spanning proof. is this specifically because of the linearity of the functionals?

  • @snehadwivedi5894
    @snehadwivedi5894 4 роки тому

    Much needed

  • @gordonchan4801
    @gordonchan4801 5 років тому +8

    Is it like dual citizenship?

  • @Victual88
    @Victual88 7 днів тому

    Thanks Peyam! [+1 sub] :)

    • @drpeyam
      @drpeyam  7 днів тому

      Thank you!!! 😊

  • @youssefbenhachem993
    @youssefbenhachem993 5 років тому +1

    Thanks . That's awesome

  • @sandorszabo2470
    @sandorszabo2470 5 років тому +1

    Hi Peyam, Do you plan to talk about multilinear algebra? I hope so :-)

    • @drpeyam
      @drpeyam  5 років тому +2

      I’ll talk about multilinearity of the Determinant at some point!

  • @chiruvolusridevi8163
    @chiruvolusridevi8163 4 роки тому

    only presentation I found , that goes into the details, step by step. I didnt follow this: In the first graph of f(v) vs v1,v2...vN the ordinates are points f(v1), .... . Later f(v1) becomes kronecker delta & f(v1) = 1. How ?

    • @drpeyam
      @drpeyam  4 роки тому

      No f1 becomes kronecker. The first step is for general f

  • @AkshayKumar-pt6fz
    @AkshayKumar-pt6fz 4 роки тому

    This is so good!! 👍👍

  • @r75shell
    @r75shell 5 років тому +1

    shouldn't f(2v) = 2f(v)? this means that f(2v1) = 2f(v1), thus not equal to zero.

    • @drpeyam
      @drpeyam  5 років тому

      At which point?

    • @r75shell
      @r75shell 5 років тому

      @@drpeyam let me rephrase, if f1, f2, f3.. (basis) all equal to zero everywhere except basis vectors, how could some LT be nonzero to anywhere else, like 2 times v1 - doubled basis first vector?
      I think it should be f1(k*v1) = k for any k, and zero everywhere else.

    • @drpeyam
      @drpeyam  5 років тому

      By everywhere else I mean zero on all the basis vectors other than v1

    • @r75shell
      @r75shell 5 років тому

      @@drpeyam so, you want to say that f1 is some LT such that it has 0 at any basis vector except v1, and it is equal to 1 at vector v1, and f1 of any other vector have a value as it fit? Then, would be nice to say why f1 exists and is it unique and so on.
      So f1(k v1) = kf1(v1) = k just because it's one of properties of LT.
      Oh, looks like f1 is non zero in many places, not only for vectors kv1, that's why you say like that.
      Anyway, I was confused and now I got it.

  • @thehsuisbhsulsh
    @thehsuisbhsulsh 2 роки тому

    How and why at 6:33 fi(vj) should be precisely equal to 1 when i=j.
    Why not some other value as I am unable to understand that?

    • @drpeyam
      @drpeyam  2 роки тому

      By definition

    • @thehsuisbhsulsh
      @thehsuisbhsulsh 2 роки тому

      Ok I got it now. Earlier I was thinking that what if we choose some function say f(x)=2x +3 then it won't guarantee that I will get fi(vi) = 1 precisely.
      But now I got it that we won't choose such a function as basis rather we would choose a function , to qualify as basis which would provide me with result fi(vi) =1 precisely.
      Thanks for it.
      But how can say with surity that we will always find such functions ( i.e. fi(vi)=1 ) in our dual space to qualify the criteria to be the basis.

  • @111abdurrahman
    @111abdurrahman 4 роки тому

    Dr Peyam, I have a question. They say that dual space of Rn is Rn ... but while doing proof we use CS inequality and we assume the standard euclidean norm on Rn. What if that norm on V is 1-norm, would then be the norm on dual space.... 1-norm? or sup-norm??

    • @drpeyam
      @drpeyam  4 роки тому

      Any two norms on R^n are equivalent, so it doesn’t matter which one we use

  • @bhargav7478
    @bhargav7478 4 роки тому

    so dual space of vector space V is basically adam world of origin world a.k.a. vector space V.

  • @tom13king
    @tom13king 5 років тому +1

    When would you do this at university? We didn't do it in first year Linear Algebra, would it be done in second year?

    • @drpeyam
      @drpeyam  5 років тому +1

      Second year :)

    • @tom13king
      @tom13king 5 років тому

      @@drpeyam Looking forward to it then.

  • @lynny7868
    @lynny7868 Рік тому

    2:30, 6:05

  • @ziminfan1664
    @ziminfan1664 5 років тому

    You r awesome

  • @qrubmeeaz
    @qrubmeeaz 3 роки тому

    Yipeee!!