Dual basis

[@chachomask]

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

[@hansbaeker9769]

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]

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]

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]

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

[@drpeyam]

Hahahahahahahaha

[@zehrademirhan600]

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

[@jubachoomba]

Reisz representation theorem pops right out: bravo, Prof!

[@rohitkantipudi6605]

this deserves way more views great explanation

[@apappas4482]

Thank you for explaining slowly and clearly.

[@williamhogrider4136]

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]

Thank you!!!

[@diegoviejocarretero7629]

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

[@adebayoemmanuel911]

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

[@klam77]

excellent; better than my textbook!

[@owen5106]

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

[@drpeyam]

Happy to help!

[@basakatik4770]

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

[@speedbird7587]

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

[@hojungkim241]

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.

[@Ruvine_]

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]

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

[@arnavarora7023]

These videos are really helpful. Thanks a lot.

[@gorkifreire1380]

Amazing explanation thank you very much

[@arseniikvachan]

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

[@sundayscrafter1779]

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

[@drpeyam]

Already up!

[@donqe6439]

better than my professor. you should become a professor man

[@drpeyam]

Awwwww I am actually 😄

[@PascalsLaptop]

thank you very much for making these videos

[@estebangimenez3714]

u saved my life, congrats from Spain.

[@thehsuisbhsulsh]

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]

By definition

[@thehsuisbhsulsh]

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.

[@amergoel3117]

thank you so much for making this video you saved me

[@thecarlostheory]

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

[@margueritedepompadour7031]

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

[@MoonLight-sw6pc]

U r right! linear algebra is beautiful :) شكرا

[@QT-yt4db]

Very helpful... Thank you...

[@lazarsavic6613]

Extremely helpful! :)

[@dgrandlapinblanc]

Pretty cool. Thank you very much.

[@AbdelrahmanAnbar]

this is amazing

[@sekhar018]

Excellent ✌️

[@justinbond7456]

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?

[@arechilasalvia8331]

Do one video on quotient space

[@112BALAGE112]

Are you planning to discuss differential forms?

[@drpeyam]

Not really

[@TheNachoesuncapo]

This would be my summer fun!!!

[@111abdurrahman]

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]

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

[@gordonchan4801]

Is it like dual citizenship?

[@drpeyam]

Hahaha

[@blackpenredpen]

LOLLL

[@mario1415]

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

[@tom13king]

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]

Second year :)

[@tom13king]

@@drpeyam Looking forward to it then.

[@chiruvolusridevi8163]

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]

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

[@climitod8524]

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]

Check out the playlist

[@thepositron5676]

Great explanation! Thank you :D

[@sandorszabo2470]

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

[@drpeyam]

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

[@r75shell]

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

[@drpeyam]

At which point?

[@r75shell]

@@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]

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

[@r75shell]

@@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.

[@schlechtestergtaspielerdek3851]

This is electroboom for mathematicians or cs students XD. THx

[@lynny7868]

2:30, 6:05

[@rodrigodiazarancibia5486]

Thanks man!

[@AkshayKumar-pt6fz]

This is so good!! 👍👍

[@snehadwivedi5894]

Much needed

[@bhargav7478]

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

[@Victual88]

Thanks Peyam! [+1 sub] :)

[@drpeyam]

Thank you!!! 😊

[@ziminfan1664]

You r awesome

[@drpeyam]

❤️

[@youssefbenhachem993]

Thanks . That's awesome

[@qrubmeeaz]

Yipeee!!