Isomorphism

What does it mean for two spaces to be isomorphic? In this video, I define the notion of isomorphism of vector spaces, and show that P2 and R3 are isomorphic.

[coefficient]

Thanks a lot Dr peyam, really needed this

[AhmedIsam]

But, in group theory, they ask for more to declare an isomorphism. Specifically, what operation corresponds to what operation in each group.

[che_sta]

I love ur energy, thanks for helping me understand this! Finals in 2 days haha

[ethancheung]

Have you upgraded your camera? The frame rate seems higher and graphics is clearer. Also you have a nice angle, this lecture room is nice

[andreutormos]

I would love a video on Isomorphism Theorems

[MoonLight-swpc]

U r cool man!
Keep going!




Fighting!

[joeaverage]

Thank you. I really appreciate your videos Dr Peyam.


Btw, I just realized that it is Dr Peyam not Peynam lool

[maloriekasparian]

This has been so helpful for me today thank you!

[ShaolinMonkster]

I 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?

[Aviationlover-belugaxl]

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?

[rafaelhernandez]

Have you considered doing analysis videos?

[markpearson]

I had my group theory exam this morning, just missed out :(

[shayanmoosavi]

So 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?

[martinbranson]

When proving T to be 1-1 He proves by example,
Shouldn’t he generalize it to all p? What am I missing?

[dgrandlapinblanc]

Thanks. Question : an linear application is always onto and one to one ?

[rktiwa]

Everytime I try to view it I laugh and switch off. May be because I am not up to it.