Prove that det(ATA) = det(AAT)

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  • Опубліковано 22 вер 2023
  • In this video, I showed the proof that the determinnat of the product of a matrix and its transpose is the determinant of the product of the transpose and the matrix.

КОМЕНТАРІ • 8

  • @holyshit922
    @holyshit922 10 місяців тому +6

    One assumption A must be the square matrix your proof to be valid
    but it is not implied from A^{T}A nor AA^{T}
    A can be nonsquare and both A^{T}A and AA^{T} can be square
    You should add assumption or split you proof to cases

  • @punditgi
    @punditgi 10 місяців тому +5

    Prime Newtons is the channel to watch! 🎉😊

  • @kingbeauregard
    @kingbeauregard 10 місяців тому +2

    Speaking of determinants, one time I was asking if you had any advice on efficient determinant algorithms that don't lend themselves to rounding errors. You'll be pleased to know that today I learned of the Bareiss Algorithm, and it's incredibly easy to program, it's fast, and it even guarantees integer results if all terms are integers. On top of all that, it's a little faster than Gaussian elimination.
    Let's say I have a matrix like so:
    a b c
    d e f
    g h i
    First thing I have to do is make sure "a" doesn't equal 0. If it does, I swap that first row with one of its successors, and then I make a note somewhere that I will have to multiply the final result by -1.
    Now I can get to work. This is the first pass, and so I am not going to do any arithmetic on the first row or the first column. But the values in the second/third rows and second/third columns, I will do math on. In particular, I will replace e with e*a - b*d, f with f*a - c*d, h with h*a - b*g, and i with i*a - c*g.
    Then, on the next pass ... well, first let me re-label all the elements of the matrix now that they have new values:
    A B C
    D E F
    G H I
    Okay, so on the next pass, first thing I do is make sure E isn't 0, and if it is I swap the row with a successor, and I make a note to multiply the final result by -1. Now that that's done, I ignore the first and second rows, and also the first and second columns. That leaves only I, and I will replace it with (I*E - F*H) / A. Note how there's a division there, and the value to divide by will move along the diagonal with each pass.
    But in this example, there are no more passes; we can just look at what's in the lower right cell, and that is our determinant. It will be (I*E - F*H) / A, times any -1 factors we noted along the way.

    • @PrimeNewtons
      @PrimeNewtons  10 місяців тому +3

      Wow! I am actually looking into rounding errors now and your comment is timely. I will take some time to go through this algorithm probably show this to some friends. Thank you.

    • @kingbeauregard
      @kingbeauregard 10 місяців тому +1

      @@PrimeNewtonsNeat! I don't know that anyone teaches this algorithm to students, but I bet you could turn the algorithm into an easy process to do on paper, like Sarrus and those arrows. I took a stab at it myself - drawing boxes around cells that interest us in a given iteration, that kind of thing - but I bet you can come up with a much smoother approach.

    • @kingbeauregard
      @kingbeauregard 10 місяців тому

      @@PrimeNewtons For what it's worth, I wrote a few different algorithms for calculating determinants in PHP: one that uses Laplace expansion, one that figures out all the permutations of rows and columns, and one that uses Bareiss. I used each to calculate the determinant of a 10x10 matrix that consisted of integers from 0 to 9. Here is the timing on the three:
      Laplace expansion: 1.9 seconds
      Permutations: 10.8 seconds
      Bareiss: 0.004 seconds
      Great Caesar's ghost!

  • @NURSAHKUBRAEKMEKCI
    @NURSAHKUBRAEKMEKCI 3 місяці тому

    Could you possible to show (AB)T=BTAT

  • @jonathanv.hoffmann3089
    @jonathanv.hoffmann3089 7 місяців тому +2

    ❤🎉