rx math
Region: US
Friday 18 July 2025 01:04:03 GMT
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Glory of God(Son of thunder ⛈) :
how about solve 2x2 case
assume it holds for k
2025-07-18 11:02:26
1
Butch Flowers :
This was a lot of fun to watch and helped make this level of math more approachable ❤️
2025-07-19 00:37:17
3
Mike :
The skill of reading math, as one of my old professors used to call it
2025-07-18 02:54:53
1
cromblerbomp :
Wow, I really didn’t see where this was going until 3:00, and then when you finally said to multiply everything by A, I just got this really beautiful wash of revelation. Lovely presentation.
2025-07-18 05:00:06
3
adamhenderson4523 :
I always check - unless I know Ive proved it before .. or if the author says “the proof is complicated - see this reference”.
2025-07-23 14:36:55
1
Horseshoe for luck :
Like, it makes sense. But what did yiu say Omega was supposed to be?
2025-07-18 10:19:08
0
Pablo :
Great
2025-07-18 04:49:12
0
Enzo's Selections :
thank you this was cool. i need to get back into math. taught myself calc from the ground up with a Dover textbook and it made a huge difference
2025-07-18 03:38:23
2
dmcdanie86 :
Reading math past a certain level is either 500 words per minute, or 500 minutes per word. There's no in between.
2025-07-18 03:31:49
133
fra 🥀 :
personally, i really like the proof using minimal polynomials, because it shows how linear algebra can actually be /algebra/, as in abstract algebra, and not just computations with matrices. it uses a beautiful theorem that generalizes the chinese remainder theorem, saying that if a polynomial p can be factored into pairwise coprime factors q_1,...,q_r, and A is a square matrix such that p(A)=0, then V is the direct sum of all the Ker(q_i(A)). in this case, if A^k=id, it means that its minimal polynomial divides x^k-1. if k is prime to the characteristic of the field, this implies that the minimal polynomial is separable, so (assuming the field is algebraically closed) the minimal polynomial looks like (x-a_1)*...*(x-a_r). by the previous theorem, this means that V is the direct sum of the eigenspaces of A, meaning that A is diagonalizable.
2025-07-22 09:07:38
1
Juan :
I don't.
2025-07-18 01:48:11
17
al :
love this
2025-07-18 01:54:18
3
user144931 :
That’s so interesting, I guess the underlying motivation here is that A itself is acting as a “6th root of unity” so of course it’s eigenvalues will behave that way, and that cyclic structure motivates setting up the eigenvectors like that
2025-07-18 04:52:37
8
pythagoras_other_theorem :
You don't have to dumb it down for me. Just say what you mean.
2025-07-19 08:52:19
0
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