I hadn't the slightest clue what my professor was explaining when he went over this, and in 11 minutes you explained it in a way I completely understood. Thank you so much.
Right? These rules for building determinants seemed to come out of nowhere?! I always wondered how this seemingly random combination of matrix-elements came to be - now I finally found the real consistent definition of it that is applicable to any square matrix of order n.
That is a great video, yet I believe that you could make it even better. The transition from perceiving a permutation as a tuple to perceiving it as a function could have been made explicit. It is still absolutely amazing, though!
Why does the labelling of terms begin with 1? Is this standard procedure, or do some people start with 0, going a00, a01, a02 etc? I thought of this because its equivolence with counting in ternary. (Perhaps being able to position an element of a matrix by calling out a base 3 number)
Hi Professor, based on the theorem that det(A) = det(A-transpose), I'm wondering whether it is also true to express the algebraic definition in such a way that the columns are in order from 1 to N, and the rows are from ɛ(1) to ɛ(N). Would that also work as a definition of the determinant? Thanks!
For the sign, you could also trace diagonals in the matrix, the ones that go descending from left to right are positive and the ascending ones are negative, like being the inverse of the slope you get if it was a cartesian plane.
@@MathTheBeautiful really? I did spot the clockwise order for the positive arrangements [n=3] (1,2,3), (2,3,1), and (3,1,2) and negative for ccw (1,3,2), (2,1,3), and (3,2,1). Your parity of the permutation could I guess be simply put into a formula. I wonder if that would then look like Mirsky's notation in (1955: 3) An Introduction to Linear Algebra? (which is proving difficult to follow by itself)
Mirsky's notation is now understood [e.g. (1,2,3) = sgn(2-1)(3-2)(3-1) = 1]; and I've the circular notation clarified (see Socratica: Cycle notation of permutations) - including transpositions [(135) = (13) (15)] . Elliot Nicholson was also ok on parity of permutations. Michel van Biezen I'd use for the simple alternative type of formula for the sign. Enjoyed getting to this point but a frustrating search , at times.
Near the end, I can't understand how you are calculating five swaps. I can see that the given permutation can be expressed as a 4-cycle and a k-cycle is a composite of k-1 transpositions, so the given permutation is a composite of 3 transpositions, this implies that the permutation is odd and hence it's sign is -1.
The number of swaps does not matter, as long as you know if it is even or odd. An odd permutation can never be written as a composition of an even number of steps, and vice versa.
It can be proven by induction rather easily that our method of deriving determinant by gaussian elimination conforms to this general formulae for any NxN matrix
I still can't understand what the fuck does this have to do with the determinant by linear transformation definition. Like it's just an awesome coincidence of mathematics or what? i don't understand :'(
Hello, I've tried your procedure of counting column permutations to attribute the sign of each term of a 4x4 matrix and compared them to the terms obtained by reducing the 4x4 determinant to 3x3 determinants times the appropriate value of the first row (Indian method). Here a picture of my work : imgur.com/a/tGzpm . The result is not satisfying. I obtain opposite signs for some terms (-a12 a23 a34 a41 by permutation (same as you at 8:44) and +a12 a23 a34 a41 by doing the reduction of the determinant). Could you please take a look at my work and give me a hint of why this is not consistent depending of the procedure used? Thank you for your help and many thanks for all these great videos.
I just found the reason why it wasn't consistent by looking how you choose the points on your 5x5 matrix. They're not align and so combining the Indian method with the American method doesn't work. To have the good sign it's necessary to reduce the matrix to a 2x2 matrix and then, the Indian crossing give us the correct sign for each value. I know you gave us a better way to find any values and the correct sign by counting the permutations but each method should be equivalent in my mind.
