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.
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.
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.
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.
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.
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?
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.
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??
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 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 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.
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.
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.
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
Reisz representation theorem pops right out: bravo, Prof!
This lecture is awesome:) i really don't understand from my instructor but you made all of the things clear . Thank you...
this deserves way more views
great explanation
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!!!
thanks for sounding so happy, I really needed that to cheer me up studyng for this
You are a hero Dr. Peyam! You made again everything is visible! Are you sure you are human!!!
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!
Thanks for the lecture. It was full of excitement and energy. I really liked it.
excellent; better than my textbook!
Now is time for the double dual teacher PiM :) It’s got pretty interesting properties :)
Already up!
What the hell? Why should I go to university, if I have Internet with Dr. Peyam. Infinite thanks to you.
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
Amazing explanation thank you very much
These videos are really helpful. Thanks a lot.
Thanks a lot! I never got what those notations really meant
thank you very much for making these videos
thank you so much for making this video you saved me
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.
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?
Dr. Peyam! Now make a video about the Mackey topology! =)
Do one video on quotient space
Are you planning to discuss differential forms?
Not really
This would be my summer fun!!!
U r right!
linear algebra is beautiful :)
شكرا
Vey awesome video. Thx a lot, ur helping me so much :D
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
Very helpful... Thank you...
Pretty cool. Thank you very much.
u saved my life, congrats from Spain.
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.
Thanks man!
Excellent ✌️
Extremely helpful! :)
Hi Peyam, Do you plan to talk about multilinear algebra? I hope so :-)
I’ll talk about multilinearity of the Determinant at some point!
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
Is it like dual citizenship?
Hahaha
LOLLL
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.
Great explanation! Thank you :D
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
This is electroboom for mathematicians or cs students XD. THx
Much needed
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.
Thanks Peyam! [+1 sub] :)
Thank you!!! 😊
This is so good!! 👍👍
Thanks . That's awesome
so dual space of vector space V is basically adam world of origin world a.k.a. vector space V.
Yipeee!!
You r awesome
❤️
2:30, 6:05