Комментарии:
Casually reveals another whiteboard
ОтветитьWhen proving T to be 1-1 He proves by example,
Shouldn’t he generalize it to all p? What am I missing?
For the last method to determine isomorphism how do we know dim(P2) is 3?
ОтветитьThis guy is so delightful. Love it!
ОтветитьI love ur energy, thanks for helping me understand this! Finals in 2 days haha
ОтветитьEverytime I try to view it I laugh and switch off. May be because I am not up to it.
ОтветитьThis helped me alot ❣️
ОтветитьThis has been so helpful for me today thank you!
ОтветитьI would love a video on Isomorphism Theorems
ОтветитьFACT
Dr Peyam is AMAZING
You are a math god thank you so much
ОтветитьSo is the second condition the same as linear independence or is it a coincidence?
In other words can we say that a linear transformation is 1-1 if the transformed bases are linearly independent?
What an styel your teaching..❤
ОтветитьThanks. Question : an linear application is always onto and one to one ?
ОтветитьU r cool man!
Keep going!
Fighting!
You’re great 👍🏻
ОтветитьBut, in group theory, they ask for more to declare an isomorphism. Specifically, what operation corresponds to what operation in each group.
ОтветитьHave you considered doing analysis videos?
Ответитьthanks
ОтветитьThank you. I really appreciate your videos Dr Peyam.
Btw, I just realized that it is Dr Peyam not Peynam lool
Thanks a lot Dr peyam, really needed this
ОтветитьHave you upgraded your camera? The frame rate seems higher and graphics is clearer. Also you have a nice angle, this lecture room is nice
ОтветитьI didn't quite understand, is it enough to show T(p) = 0 => p = 0 for the 1-1 ?
Why so , is there some kind of theorem that I'm forgetting?
I have been wondering:
I’ve come up with a continuous function from [0,1] to (0,1). y=sin(1/(1-x))) if 0<=x<1 and 1/2 if x=1. (I know it’s not linear but these aren’t vector spaces so idk. :/) And there’s a continuous function from (0,1) to [0,1], y=0 if x<1/4, 2(x-(1/4)) if 1/4<x<3/4, and 1 if x>3/4. However, these are not inverses and I don’t know of any invertible function between these 2 sets. Are the sets ((0,1) and [0,1]) isomorphic?
I had my group theory exam this morning, just missed out :(
ОтветитьAnother fron Group Theory.
ОтветитьPeyam Boss
Ответить