What the requirements for thus course

I think this video quite literally changed my life

After 49:00 he said "associating" and that was it. I started dissociating.

Be careful though, the reflection that is meant in glide reflection is not an elementof O(n), it is geometrically a reflection through some line, not as qn element of O(n) .

Why are we proving again the dimension theorem?

17:14 I laughed out loud when he made the capital W into extra-capital W's by extending the "legs" an absurd amount 😂

6:26 The representation of the hopf decomposition of S^3, in the tangent space at the South pole of S^3, by means of the stereographic projection from the North pole of S^3.

3:16 This time the z2 axis is shown in details. And the number 1-2i is drawn.

2:47 The z1 axis is being developed. And the number 2-i is depicted.

I guess the "clear" way of the main method would be to(I didn't understand it ,there is a tricky part that I didn't understand probably, all the coefficients already 0 at the start , you cannot assume it is not and continue, even though you continue , all the first coefficients being zero in the end is a solution to that Sy. Of Lin. Eq) proove two basis has the same number of elements , they are almost ewuivalent with the use of little theorems we've prooved. And suppose we have w and v set as basis . Sum wi with coefficients 1 , it is not zero since wi constitutes basis by assumption. Change every wi in the summation with cijvj, again total summation is not zero ; change the order of summation(since it is finite , and summation is commutative in vec. spaces)the statement that they are all zero is false , so the system of n linear equations(since all of them has one of the vi's , they are linearly independent) with m unknowns has no solution(set them all equal to zero , "this is false" means we have no nontrivial solution for it); and this is true iff n>=m ; do the same with sumvi≠0 , we get m≥n ; so m=n , very resemblent to the video's and book's proof(they are same) but this makes sense to me .

It is interesting to see so many middle aged people in the audience.

really helpful

The mapping between G1 and G2 preserves the multiplication table: here we go, for the first time I finally really understand group isomorphisms 😊

Bro these guy’s lecturers are unavailable and mfing richard taylor himself is the replacement

❤

Guys I think Harvard might be hard....

This channel is cool! Thank you! I would like to mention that in the beginning of the imagination exercise I was a bit confused by the lizard metaphor because I thought that if their vision sense is the 2D analogous of 3D lizards, they'd have two receivers in the front and they would only perceive lines and dots. So if thew were to perceive shapes in the plane, they'd have to have sensors that encapsulate the shapes they interact with. Am I missing something?

can I share this on my channel?

Neither of the links posted work for me, was able to find the lecture notes by pasting the url into google and selecting the closest result but that trick didnt work for the exercises sadly. Lmk where I could find this information

Hmmm,brilliant! improve my tech art skills(in gamedev).

23:15 was that a fart?

Çok değişik bir şey ne anlatmaya çalışıyor Türkçe altyazı kullansanız

Dear KZbin gods, please suggest more of these videos, Amen.

Morphism = Map Homomorphism = A one to one map that preserves structure(multiplication) Isomorphism = A homomorphism that is bijective Automorphism = An isomorphism to itself

37:28 talking about Abel and Galois.

When he is making a group homomorphism from G to G prime and he claims that the kernel H divides the group G into equivalence classes, each class having the same number of elements, and then he did the same with H NOT being a normal subgroup and just a subgroup, and he said that the equivalence classes from this H are still the same size and the corollary still holds, why did he start with explaining this for when H is normal? If the subgroup being normal isn't important, why not first explain the general case and then maybe give the specific case where H is the kernel? Or am I missing something

There are 2 rings of order 4 ? 15:06 Is that true?

3:27 No time to waste in greetings and thanks.

HW: 2.1.5, 2.1.7, 2.2.1, 2.2.15, 2.2.17

Assignment: 1.1.7, 1.1.16, and 1.1.17

It's great information american

It was very clever to name all the videos in the playlist with one name

Great lecture

What am I missing in min 40:12? Find invertible matrices such that AB is not equal to BA? Here: let A = (1, 2; 3, 4) and let B = (3, 4; 1, 2). They are both invertible. If you take A*B row 1 column 1 entry, you get 5. If you take B*A row 1 col 1 entry, you get 15. The products are not equal. Other than that, great lecture. I just completed the first course in Abstract Algebra at my local college, and am moving on to other courses. I thought it'd be a shame to forget what I've learned because it was a fascinating topic. My prof closed the course in D2L with her video lectures, so I took to KZbin to see what's available there. Really glad I found this channel, and I hope the content is available for a long time. I really like that the first lesson reviewed linear algebra because I took that course 25 years ago when I was a "proper" college student. My intro course spend more time on Sn permutations, and I hope I learn more about matrices and vector spaces as groups here.

lecture ends at 39:00

🤯🤯🤯🤯🤯🤯

For the last part about gaussian integers and the related ideals to have isomorphism with Z/pZ, where p = 4m+1, the part about f(i) has order 4 is slightly misleading I believe, because although we necessarily have (f(i))^4=1, it does not imply f(i) has order 4, it could have order 2, or 1, because homomorphism does not necessarily preserves the ring structure. Nonetheless it does not affect the flow of the proof overall, can directly start by observing the existence of an element of order 4 in the multiplicative group of Z/pZ, and proceed to use that to construct an ideal to satisfy the isomorphism.

The guy may be a genius, but he's boring, dry and unpleasant. Maybe he's simply shy, but they are characters like him. that make you abandon the desire to study mathematics.

Thank you for the resource. I wish the titles were descriptive. It is not convenient to open every video to understand specific concept.

Ridiculously hard as scores show. I surprised that the prof., with all his experience , handed this out. Oh, he does say it's too hard. What a guy. Some confidence builder!!

hey I m here for the semiring definitions, which is need for me to understand signed measures and bounded variations...but I m an economist so I have absolutely no idea what I am saying

These lectures are marvellous. I studied group theory many years ago but fro a more abstract point of view: axiom, theorem, proof. Here we have discovery driven by example where the theorems just pop out, well motivated. Enjoying this immensely, thank you. Great lecturing style combining talk and text with real enthusiasm for the subject.

25:50

17:55 parameterizations/coordinate patch; 23:00 mobius

it looks frightening.

Here we clearly see what ''expert'' is.

thanks

Thanks

Thanks.

Another movie example: Ron Lola Run. We see essentially the same events several times, each with minor changes that lead to vastly different outcomes.