Pseudoinverses

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  • Опубліковано 8 вер 2024

КОМЕНТАРІ • 15

  • @YouJLT
    @YouJLT 4 роки тому +16

    Small mistake, at 14:07 it is said "if x is not in the nullspace of A, A+A gets x back" , No we get the projection of x onto the rowspace of A

    • @walterlevy5924
      @walterlevy5924 4 роки тому +1

      jean-luc thirion, thanks for pointing that out - it is easy to believe an arbitrary vector must be in one of the fundamental subspaces but that is simply not true. Take for example the column vector (1 1), not in N(A) and not in the row space. It gets projected to column vector 3/5 (1 2), now in the row space. The rest of the vector, 1/5 (2 -1), lies in the null space.

    • @TommyEVO3D
      @TommyEVO3D 2 роки тому +3

      I believe he means "x is in row space of A" instead of "x is not in the null space of A" .

    • @zokalyx
      @zokalyx 2 роки тому

      @@TommyEVO3D he meant that, but what he said was a bit confusing

  • @user-pd1sx9tx4q
    @user-pd1sx9tx4q 3 роки тому +2

    the best TA in the course

  • @iglidraci
    @iglidraci 25 днів тому

    Couldn't he have taken the right inverse immediately? It's full row rank matrix.

  • @AnupKumar-wk8ed
    @AnupKumar-wk8ed 6 років тому +1

    Nice video. Thanks.

  • @middlevoids
    @middlevoids 7 місяців тому

    Is there a mistake at 8:52? He wrote V^T instead of V (1, 2
    -2, 1)?

  • @quirkyquester
    @quirkyquester 4 роки тому

    thank you!

  • @chotirawee
    @chotirawee 4 роки тому +1

    Sigma is 1*2 but there are two eigenvalues. This means that only one eigenvalue will be written in Sigma metrix. What make him decide to use sqrt(5)? what not 0?

    • @felipealb
      @felipealb 4 роки тому +2

      I think is because the rank(Sigma) = rank(A) = rank(A^t*A) = r = 1. If Sigma = [ 0 0 ] the rank will be zero and the pseudoinverse of sigma matrix will be undefined ( 1/0 )

    • @zokalyx
      @zokalyx 2 роки тому +1

      in the sigma matrix we only care about positive eigenvalues; zero eigenvalues correspond to nullspaces which we don't really care about because we know how they behave already.

  • @william061490
    @william061490 Рік тому +1

    Question - at 11:24, why the vector (-2 1)?? Thanks!

    • @ickgloo
      @ickgloo Рік тому +1

      That's the null space vector X. if you calculate Ax=0 and solve for x you get (-2 1).

    • @tzxu
      @tzxu 4 місяці тому

      @@ickgloo you shall choose (2, -1) instead, otherwise you get A = (1, -2) which is wrong.