What does isomorphic mean? What is an isomorphism? 

Exercise for the viewer. Repeat this problem in the case where 2×2 matrices are replaced with general n×n matrices. Of course n in Z+ is fixed.

@@Grassmpl x|------> x/n * J where J is the all 1s matrix of order n?

Hey man, hope you liked it! Congrats on your big integral win. 💪

We may be the most significant isomorphic sets

Welcome :)

Yes I agree. The trick here is the standard identity matrix is NOT the identity of this group. Since none of the matrices have full rank, we are off the hook.

+Epic Math Time could you say that The integral of f(x) dx and f(y)dy is isomorphic if the only difference between the two is the symbol?

@@poutineausyropderable7108 I mean, those two things are really one-the-nose equal.

The term "isomorphism" is ambiguous unless the intended underlying equivalence relation is well defined

Maybe I'm not getting the irony but how is this a group?

Lol I came to the comments to see if there was one on the music, I thought it was runescape music

Great! I'm glad you were able to get something out of it. That's very motivating. Thanks for the comment!

Have you used character tables or learned about group representation theory in your studies? These particular fields are strongly linked. The fact that you are able to see a connection like that at such a foundational level is very impressive.

@@EpicMathTime Yessss! You determine the point group, represent them as matrices, and every time you perform an operation, the molecular structure is still the same (i.e. geometry, properties, etc) but you reduce the irreducible representations of the molecule. It helps us predict the translational, vibrational, and rotational motions of the molecule.

Yeah I agree. Maybe it would help people understand math concepts better

In an abstract categorical sense, certainly, but that isn't enlightening because category theory puts an abstraction on isomorphisms to the point that it may not capture the intuition that I wanted to address here. But in a more concrete mathematical sense, I think a good example is as follows: instead of vector spaces over a field F, with isomorphisms in the usual way (bijective linear transformations), we can look at all vector spaces over any field isomorphic to F, with isomorphisms taken to be bijective semilinear maps. Semilinear maps are more than just a map on the underlying sets because it must not only translate the vectors in the vector space, but also the scalars to the new field. So it effectively contains two maps, translating the vectors, and translating the field elements in that external field acting on the vector space. In the standard study of linear algebra, this is a nonissue and not needed because we lock interacting vector spaces into being over the same field. Once we broaden the class of objects in the way described, one map on the underlying vectors is not enough to preserve what we want.

@@EpicMathTime Nice one! If we cheat a little bit, though, then we could see such an object as an ordered pair (F,V), where F is a field and V is a vector space over F. Then a morphism could still be seen a map f: (F_1,V_1) -> (F_2,V_2).

@@supermarc There's nothing wrong with describing V over F as a single object (which it is) and denoting it (V,F), but the set-wise meaning of this is unclear. If we want to have a map f:(V,F)->(V',F'), we have to know what (V,F) means as a set. Are elements of (V,F) ordered pairs (v,c)? If so, it's hard to make this agree with the notion of our vector space (though I'd have to think about that more). If the elements of (V,F) are just vectors, then we of course have the same issue as before. Now, if we instead are associating (V,F) to (V',F') not by a map between them as sets, but in a way that views them as single mathematical objects in which (V,F) is the "input" and (V',F') is the "output" (as opposed to the domain and target) then we aren't describing a morphism at all, but a functor.

Haha nope, although someone else did mention that recently.

Except Paul speaks with a Canadian accent and Epic sounds southern U.S., probably Louisiana based on his LSU hat :-)

Same as itself is supposed to be an obvious fact…like for example say I took you into a cloning machine and made a perfect clone of you…you and this clone are isomorphic because there is a mapping you can do, atom by atom where that mapping is bijective. Say now that I took your clone and I smushed it into just a lumpy pile of goo. This lumpy pile of goo is still isomorphic to you because there is still a bijective mapping you can make with every atom of your smushed clone and you. You can think of a bijective mapping as me taking pieces of your smushed clone, and placing them somewhere new…so a -1 from the clone and a +1 over here…and slowly but surely from the pile of mush I build you up again into a human being. These operations (+1 and -1) is the bijective function in that case…but the function can be any mathematical function, so long as there is an inverse operation to it.

These are all correct as morphisms. They are isomorphisms with the added stipulation they are also bijective, with inverses that are also morphisms. (that is, an isomorphism is a special kind of morphism). For topology, the morphisms are continuous functions. Sometimes the steps for a morphism to become an isomorphism get satisfied automatically, depending on category. For Euclidean geometry, all morphisms are automatically isomorphisms (every Euclidean transformation is bijective and has an inverse that is also a Euclidean transformation). For most algebraic categories, any bijective morphism is automatically an isomorphism (the inverse of any linear map is a linear map). This isn't true of topology though (there are bijective continuous functions whose inverses are not continuous). So the statement "bijective morphism whose inverse is a morphism" is meant to be an all-encompassing general description.

:) Proof: First, we show if A in G is invertible, then the identity of G is the identity matrix. Let E denote the identity matrix of G, let A* denote the inverse of A in G (let A^(-1) be its standard inverse, and I the standard identity matrix). Then we have that AA* = E and AA^(-1) = I, hence AA*AA^(-1) = EI = E => AEA^(-1) = E => AA^(-1) = E => E = I. Hence E = I. Now, we show if B in G is not invertible, then the identity of G cannot be the identity matrix. B is in G, therefore it has an inverse B* such that BB* = B*B = E. Since B is not invertible, E =/= I. This shows that A and B cannot lie in the same group (as A's membership requires E = I, while B's requires E =/= I).

@@EpicMathTime Well done. Your grade is 5/5. Here is another trivial exercise which I know won't stump you. Let G consist of invertible matrices and H consist of singular matrices. Is it possible that G is isomorphic to H? Either give an example or a disproof as appropriate.

@@EpicMathTime btw these questions are from my own head which I know the answer to. Ie they are NOT from textbooks, internet, home works, exams, etc.

@@Grassmpl Yes, it is possible, let E be an idempotent matrix, then {I} and {E} are isomorphic. Or do we want to ban that?

@@EpicMathTime this is good. As u can see very trivial. Although since I is idempotent u should mention that E is singular as well. Better yet let E=0

Ask Tim Pool.

No wait, you wrote like normally and mirrored the footage. That must be it.