Dual basis
ฝัง
- เผยแพร่เมื่อ 21 ธ.ค. 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!
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Dual Space video: • Dual Space
Example of dual basis: • Dual Basis Example
Check out my Dual Space playlist: • Dual Spaces
Rumors has it that this guy has infinite amount of boards he can just switch between
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.
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.
Peyam, can you do 100 dual basis in one take?
Hahahahahahahaha
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.
Reisz representation theorem pops right out: bravo, Prof!
this deserves way more views
great explanation
This lecture is awesome:) i really don't understand from my instructor but you made all of the things clear . Thank you...
Thank you for explaining slowly and clearly.
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
Thank you!!!
excellent; better than my textbook!
thanks for sounding so happy, I really needed that to cheer me up studyng for this
I've been looking for a way to picture a dual basis. Thank you so much for this
just found your channel, im reading linear algebra done right on my own, your a G my dude. so helpful tyvm
Happy to help!
You are a hero Dr. Peyam! You made again everything is visible! Are you sure you are human!!!
Thanks for the lecture. It was full of excitement and energy. I really liked it.
What the hell? Why should I go to university, if I have Internet with Dr. Peyam. Infinite thanks to you.
Amazing explanation thank you very much
Now is time for the double dual teacher PiM :) It’s got pretty interesting properties :)
Already up!
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.
These videos are really helpful. Thanks a lot.
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.
Thanks so much!! Yes, the graph really cleared it up for me
better than my professor. you should become a professor man
Awwwww I am actually 😄
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?
By definition
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.
thank you very much for making these videos
thank you so much for making this video you saved me
Are you planning to discuss differential forms?
Not really
Thanks a lot! I never got what those notations really meant
Do one video on quotient space
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?
This would be my summer fun!!!
Pretty cool. Thank you very much.
u saved my life, congrats from Spain.
Vey awesome video. Thx a lot, ur helping me so much :D
this is amazing
Very helpful... Thank you...
Is it like dual citizenship?
Hahaha
LOLLL
U r right!
linear algebra is beautiful :)
شكرا
When would you do this at university? We didn't do it in first year Linear Algebra, would it be done in second year?
Second year :)
@@drpeyam Looking forward to it then.
Extremely helpful! :)
Excellent ✌️
Thanks man!
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.
Check out the playlist
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??
Any two norms on R^n are equivalent, so it doesn’t matter which one we use
Hi Peyam, Do you plan to talk about multilinear algebra? I hope so :-)
I’ll talk about multilinearity of the Determinant at some point!
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 ?
No f1 becomes kronecker. The first step is for general f
Great explanation! Thank you :D
Dr. Peyam! Now make a video about the Mackey topology! =)
This is electroboom for mathematicians or cs students XD. THx
Much needed
This is so good!! 👍👍
shouldn't f(2v) = 2f(v)? this means that f(2v1) = 2f(v1), thus not equal to zero.
At which point?
@@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.
By everywhere else I mean zero on all the basis vectors other than v1
@@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.
2:30, 6:05
Thanks . That's awesome
Thanks Peyam! [+1 sub] :)
Thank you!!! 😊
so dual space of vector space V is basically adam world of origin world a.k.a. vector space V.
You r awesome
❤️
Yipeee!!