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How have they calculated these pseudoinverses?

  1. Dec 10, 2015 #1
    • Moved from a technical forum, so homework template missing.
    fieHb7f.png

    imgur link: http://i.imgur.com/fieHb7f.png

    UpDPZ6K.jpg

    imgur link: http://i.imgur.com/UpDPZ6K.jpg

    Solution:

    gUHCb1u.png

    imgur link: http://i.imgur.com/gUHCb1u.png

    I can see that [itex]A^+A[/itex] and [itex]AA^+[/itex] are matrices that project onto the column and row space respectively.

    But

    (i) where does the [itex]A^+[/itex] figure into the calculation, because those answers are not A matrix-multiplied by [itex]A^+[/itex] or [itex]A^+[/itex] matrix-multiplied by A?

    (ii) if [itex]A^+[/itex] isn't used to find the last two answers, why calculate it in the first place, what purpose does it serve? I could just find the column and row space basis vectors by other methods and construct those last two answers accordingly.

    (iii) why are those answers given in the form that they are? wouldn't [itex]A^+A[/itex] be just as good given as

    [tex]A^+A = \begin{bmatrix}1 & 2\\2 & 4\end{bmatrix}[/tex] ?
     
  2. jcsd
  3. Dec 15, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Dec 16, 2015 #3
    Matrix ##A^+## was computed using singular value decomposition.
    You are asked how the fundamental subspaces of ##A## and of ##A^+## are related.
    Concerning your questions:
    (i) Check your calculations, the given matrices are exactly ##A^+A## and ##AA^+##;
    (ii) This is a way to check your results. According to the theory ##A^+A## and ##AA^+## give you orthogonal projections onto row and column spaces of ##A## respectively (in fact, these conditions define ##A^+## uniquely), so computing these projections using other methods you can check your answer.
    (iii) your formula for ##A^+A## is wrong, how did you get it?
     
  5. Dec 16, 2015 #4
    I thought A dagger was just 1/50 of A, but it's not, it's 1/50 of A transpose. Even after doing the SVD calculations, somehow this misapprehension snuck in.

    Just a bit of blindness Matlab confirms.
     
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