good! but the most natural way to find the determinant is to realize that if we do a substitution, we can get 0 determinant, namely x→y, x→z, x→t, y→z, y→t, or z→t. Since you get 0 determinant after the substitution, you must have the original determinant in the form P(x-y)(x-z)(x-t)(y-z)(y-t)(z-t), and P has to be a constant for the degree must be 6, and P has to be 1 (or -1, if you manage to flip some order of subtraction) by using term comparison
Here is the proof what you mentioned when I asked about the "usual proof". Thanks! (My taste in linear algebra is very similar to yours. Maybe we learnt from the same book :-) )
is it determinant but when you are taking common you are taking it as we do in a matrix what I want to say is when we multiply a determinant is multiply by a scalar we only multiply it to a single row but in the matrix we multiply it to all the rows
You should make some miscellaneous vids on how the Jacobian and linear algebra mix, as well as the covariance matrix and its properties. Also if you can, some videos on Numerical methods involving systems of linear equations like Gauss-Seidel etc. Thanks!
11:27 shouldn't it be t-z?
Yep
How is your comment 1 week older if the video has been published just a few minutes ago?
@@yaaryany the video was prolly unlisted
@@gesucristo0 oh I see
yes
Thanks Dr. Peyam, I loved your enthusiasm! Makes math extra fun to see someone excited to teach :)
God bless u Dr peyam i am jamal a math teacher from palestine
Dr.peyam,thank you for your explanation❤,you explain very well,have a nice day!
Hey, that was one of my assignments! :D
I don't care if it's trivial or obvious, where did the first row and column go?
It vanished as we do determinant along coloumn 1 !
You can actually use row/collum reduction to find the determinant of a 3x2 matrix(3x2 meaning 2d space as the input and 3d space as the output).
Wow you've helped me learn a nice concept...thanks so much
Thank you so much!
good! but the most natural way to find the determinant is to realize that if we do a substitution, we can get 0 determinant, namely x→y, x→z, x→t, y→z, y→t, or z→t. Since you get 0 determinant after the substitution, you must have the original determinant in the form P(x-y)(x-z)(x-t)(y-z)(y-t)(z-t), and P has to be a constant for the degree must be 6, and P has to be 1 (or -1, if you manage to flip some order of subtraction) by using term comparison
That’s the cool way I did in the first Vandermonde video (the one with 14k views)
Here is the proof what you mentioned when I asked about the "usual proof". Thanks! (My taste in linear algebra is very similar to yours. Maybe we learnt from the same book :-) )
is it determinant but when you are taking common you are taking it as we do in a matrix what I want to say is when we multiply a determinant is multiply by a scalar we only multiply it to a single row but in the matrix we multiply it to all the rows
I wonder what this sort of thing would be used for. Perhaps a discrete Fourier transform?
It’s used to show that there is a nth degree polynomial going through n points
Thanks for your nice tutoring, greetings dear Dr Peyam, the best to you.
Very nice explanation👌
Now how to find the adjoint of vandermonde :(
You should make some miscellaneous vids on how the Jacobian and linear algebra mix, as well as the covariance matrix and its properties. Also if you can, some videos on Numerical methods involving systems of linear equations like Gauss-Seidel etc. Thanks!
There are 2-3 videos on the Jacobian where I explain that
Wolfram Alpha: Input
Factor[ Det[{ {1,x,x^2,x^3}, {1,y,y^2,y^3}, {1,z,z^2,z^3}, {1,t,t^2,t^3} }]]
Output
-(t - x) (t - y) (t - z) (x - y) (x - z) (y - z)
Yep 👍
I wonder if Voldemort determinant will be invented someday :)
And i wanna see glass numbers
Nice
Hm, why am I having a deja vu?
I presented a different proof a year ago