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Help with simple matrix algebra

  1. Jan 21, 2012 #1
    Hi all,

    I'm having trouble solving this matrix problem, basically I have,

    s=r*H

    where s = [ 1 1 0] and,

    H=
    1 0 1 1 0 0
    0 1 1 0 1 0
    1 1 0 0 0 1

    I am trying to find out what the matrix r is but it won't work in matlab. I have tried inverting H but it isn't a square matrix. So I tried pseudoinverse psinv and it still doesn't seem to be working. Any help would be greatly appreciated.
     
  2. jcsd
  3. Jan 21, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Your problem is that this is impossible. H has three rows and 6 columns so r must have three columns. If it also has n rows, s would have to have n rows and 6 columns. Since s has one row, n can be 1 but s has 3 columns, not 6.
     
  4. Jan 21, 2012 #3
    Your equation is of the form :
    [tex]
    \vec{a} = \vec{b}A
    [/tex]

    Where b is the 1x3 vector we are looking for and a is a 1x6 vector . As you mentioned, your matrix isn't square an thus not normally invertible. By the Moore-Penrose Pseudoinverse, there does exist a solution (infinitely many actually).
    Your matrix A is an mxn matrix where m<n and is also full-rank, so:
    [tex]
    A^\dagger = (A^TA)^{-1}A^T
    [/tex]
    Where A-dagger is the Moore-Penrose Pseudoinverse. Notice that only because A was full rank could you use this formula, otherwise you have to use the singular value decomposition .
    I've tried [tex] A^\dagger = (A^TA)^{-1}A^T [/tex] with your matrix but [tex] A^TA [/tex] turns out to be singular. I suppose you could try to take the pseudoinverse of that too, but then I'm not sure if everything will work out as you had wanted.

    Edit: I made a silly mistake with the size of r :redface: (as HallsofIvy pointed out). The above would have been useful for consistent sizes and a full rank matrix.
     
    Last edited: Jan 21, 2012
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