Nice! I feel like actual justification should've been given to the fact that the determinant is invariant with respect to the ordering of the subsets, though. We cannot conclude that just because of the problem statement. It's quite possible that the solution involves breaking it up into cases of different orderings or classes of orderings. Obviously that didn't happen here, but still. Also, the reason for the determinant being invariant is simple anyway. If we pick one ordering and then switch two sets, it is fairly simple to see that that corresponds to switching two rows and then two columns of M_n. Switching rows and columns multiplies the determinant by -1 and we do that twice, so the net effect is just multiplying by 1. Therefore the determinant is invariant. (My suggestion to those who don't see why switching sets has this effect is to play around with the n=2 and n=3 cases a bit. The intuition will become obvious and then you can come up with an effective argument.)
Rather than using that obscure “fact” about determinants, pulled out of thin air at 13:46 , just use row operations to subtract the bottom half of the big matrix from the top half. This gives you a matrix of the form ((0,M)(M,0)) which is block diagonal with two copies of M. The result follows.
So, why induction? Apart from the fact that it is usually used for calculating determinants, one might have experimented with a few small cases (the "first" determinants, like for n=2 and n=3) and observe that they are -1. Then a natural guess is that they are all -1 and it is here where someone might think of solving the problem via induction. Furthemore, at the last step, when we swap columns, the resulting upper-right matrix is not necessarily M_(n-1), but the solution still works.
For the "well-known" fact": Each element of the set either belongs to the set or not. There are two ways for each element of the set. All these choices are independent. Thus, 2x2x2...x2=2^n
Nice! I feel like actual justification should've been given to the fact that the determinant is invariant with respect to the ordering of the subsets, though. We cannot conclude that just because of the problem statement. It's quite possible that the solution involves breaking it up into cases of different orderings or classes of orderings. Obviously that didn't happen here, but still.
Also, the reason for the determinant being invariant is simple anyway. If we pick one ordering and then switch two sets, it is fairly simple to see that that corresponds to switching two rows and then two columns of M_n. Switching rows and columns multiplies the determinant by -1 and we do that twice, so the net effect is just multiplying by 1. Therefore the determinant is invariant. (My suggestion to those who don't see why switching sets has this effect is to play around with the n=2 and n=3 cases a bit. The intuition will become obvious and then you can come up with an effective argument.)
Thanks for the reply. I will pin this comment so that it can help everyone!
another way to conclude this is each ordering correspond with a permutation matrix, that is M = P.M*.P^(-1), so that det(M) = det(M*)
I love the snaps :-)
Great feature of these talks
Brilliant, I almost had it, at least started in the right direction :)
Rather than using that obscure “fact” about determinants, pulled out of thin air at 13:46 , just use row operations to subtract the bottom half of the big matrix from the top half. This gives you a matrix of the form ((0,M)(M,0)) which is block diagonal with two copies of M. The result follows.
fantastic! happy new year!
Didn't know that the row operation (affine combination? is there a term for it?) didn't change the determinant. Cool!
So, why induction? Apart from the fact that it is usually used for calculating determinants, one might have experimented with a few small cases (the "first" determinants, like for n=2 and n=3) and observe that they are -1. Then a natural guess is that they are all -1 and it is here where someone might think of solving the problem via induction.
Furthemore, at the last step, when we swap columns, the resulting upper-right matrix is not necessarily M_(n-1), but the solution still works.
For the "well-known" fact":
Each element of the set either belongs to the set or not. There are two ways for each element of the set. All these choices are independent. Thus, 2x2x2...x2=2^n
I think you have to prove that re-ordering does not change deteminant and should not depend on the question itself for that!
Nice, really nice
Ah, I broke it into 2 cases instead of 3. That was my mistake.
But the question did not state whether the sets are different.
The question uses the word "The", which means that we're talking about THE non-empty sets of {1,2,...,n}.
Nice
Can you make it the graph of function f(x)={0, x€Q and 1, x€I}?! It's impossible for you?! 😀😉
Can someone link me A3 of the same year I'm not able to find any video solution
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I boycott all products advertised on RUclips. Great!