Dual basis

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Күн бұрын

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

Пікірлер: 79

@chachomask 4 жыл бұрын

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

@blackpenredpen 5 жыл бұрын

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

@drpeyam 5 жыл бұрын

Hahahahahahahaha

@FT029 4 жыл бұрын

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.

@hansbaeker9769 Жыл бұрын

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 3 жыл бұрын

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.

@zehrademirhan600 3 жыл бұрын

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

@apappas4482 4 жыл бұрын

Thank you for explaining slowly and clearly.

@rohitkantipudi6605 4 жыл бұрын

this deserves way more views great explanation

@diegoviejocarretero7629 3 жыл бұрын

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

@williamhogrider4136 Жыл бұрын

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

@drpeyam Жыл бұрын

Thank you!!!

@jubachoomba Жыл бұрын

Reisz representation theorem pops right out: bravo, Prof!

@sundayscrafter1779 5 жыл бұрын

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

@drpeyam 5 жыл бұрын

Already up!

@arseniikvachan 3 жыл бұрын

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

@hojungkim241 4 жыл бұрын

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.

@klam77 4 жыл бұрын

excellent; better than my textbook!

@schlechtestergtaspielerdek3851 Жыл бұрын

This is electroboom for mathematicians or cs students XD. THx

@PascalsLaptop 9 ай бұрын

thank you very much for making these videos

@gorkifreire1380 3 жыл бұрын

Amazing explanation thank you very much

@margueritedepompadour7031 6 ай бұрын

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

@climitod8524 4 ай бұрын

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 4 ай бұрын

Check out the playlist

@gordonchan4801 5 жыл бұрын

Is it like dual citizenship?

@drpeyam 5 жыл бұрын

Hahaha

@blackpenredpen 5 жыл бұрын

LOLLL

@thecarlostheory Жыл бұрын

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

@TheNachoesuncapo 5 жыл бұрын

This would be my summer fun!!!

@justinbond7456 Жыл бұрын

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?

@thehsuisbhsulsh 2 жыл бұрын

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 2 жыл бұрын

By definition

@thehsuisbhsulsh 2 жыл бұрын

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.

@dgrandlapinblanc 4 жыл бұрын

Pretty cool. Thank you very much.

@speedbird7587 7 ай бұрын

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

@bhargav7478 4 жыл бұрын

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

@adebayoemmanuel911 Жыл бұрын

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

@sandorszabo2470 5 жыл бұрын

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

@drpeyam 5 жыл бұрын

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

@amergoel3117 Жыл бұрын

thank you so much for making this video you saved me

@112BALAGE112 5 жыл бұрын

Are you planning to discuss differential forms?

@drpeyam 5 жыл бұрын

Not really

@111abdurrahman 4 жыл бұрын

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 4 жыл бұрын

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

@tom13king 5 жыл бұрын

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 5 жыл бұрын

Second year :)

@tom13king 5 жыл бұрын

@@drpeyam Looking forward to it then.

@rodrigodiazarancibia5486 Жыл бұрын

Thanks man!

@arnavarora7023 Жыл бұрын

These videos are really helpful. Thanks a lot.

@chiruvolusridevi8163 4 жыл бұрын

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 4 жыл бұрын

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

@lynny7868 Жыл бұрын

2:30, 6:05

@Ruvine_ 2 жыл бұрын

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 2 жыл бұрын

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

@r75shell 5 жыл бұрын

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

@drpeyam 5 жыл бұрын

At which point?

@r75shell 5 жыл бұрын

@@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 5 жыл бұрын

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

@r75shell 5 жыл бұрын

@@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.