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I Solving for a 3x3 matrix

  1. Apr 6, 2016 #1
    Hi all,

    I have this data that can be described by M*A = B, where M is a 3x3 matrix and A and B are 3x1 vectors.

    Since I know and can collect A and B data, and I have 9 unknowns in the 3x3 matrix, I thought that collecting 9 pairs of A and B vectors would yield the matrix M's coefficients via 9 equations.

    However, it has not been the case. For whatever reason, I'm not getting those coefficients. Is there something you recommend looking into?

    I have both Mathematica and Matlab. It's very possible I'm not considering or overlooking a linear algebra principle. I just tried Solve in Mathematica, and I get { }, which is a meaningless, or better :confused: response!

    Thank you
     
  2. jcsd
  3. Apr 6, 2016 #2

    Orodruin

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    How are you trying to reconstruct the matrix? As long as your choices of the vectors the matrix act on is linearly independent you should be able to make a full reconstruction.

    Another possibility is that your Mathematica code is wrong.
     
  4. Apr 6, 2016 #3
    Hi,

    I'm trying to reconstruct the matrix M with Mathematica's Solve[ ]:

    The vectors are basically (x, y ,z) and (x' , y' , z') coordinates

    Screen_Shot_2016_04_06_at_9_47_43_PM.png

    And the data behind this mess is here - shown in (x,y,z) row sets - where the "M4" and "IF" would be like A and B in M*A = B. The data looks linearly independent:

    Screen_Shot_2016_04_06_at_9_48_15_PM.png

    So, I don't know why I'm getting those empty brackets in Mathematica. Maybe I should explicitly write each equation, but I'm assuming Mathematica can handle that.
     
  5. Apr 7, 2016 #4
    I managed to find a number of the elements by rewriting my Solve[ ] call:

    Screen_Shot_2016_04_06_at_10_08_08_PM.png

    And, as you can see above, I'm left with the first column vector within M, i.e. [m11; m21; m31], left to be determined.

    To be honest, I don't understand why I have those 3 elements left to be determined. Using vector rows beyond row 2 (i.e. k > 2) are giving open brackets solutions.

    Mathematica! :nb)
     
  6. Apr 7, 2016 #5

    Orodruin

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    You are using too many vectors and so they are not linearly independent. Since you have numerical values you are likely getting rounding errors, making the system undetermined. Since you have only used vectors with x value zero, you are getting an undetermined first column.

    You need to use three linearly independent vectors.
     
  7. Apr 7, 2016 #6
    I get you; makes sense now! I also see what you mean by the three linear independent vectors as [1; 0; 0], [0; 1; 0], and [0;0;1]. They give away 3 columns of M.

    Too bad the system I'm crunching this data through doesn't have that level of direct control, so I need to get as close as possible to those 3 vectors for "A" in M*A = B. Hard to predict I will get those vectors, but I'll try. If I do, I will have the "B" vectors as well, and it will be the end of it.

    Do you happen to know if there is a hack "brute force" method :smile: with linearly dependent vectors? I would guess MATLAB has something like this. Reminds me of pseudo-inverse techniques, but I'm not looking for an inverse.
     
  8. Apr 7, 2016 #7

    Orodruin

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    No, if you do not have three linearly independent vectors you do not have enough information to reconstruct the matrix.

    Note: They do not need to be orthogonal, just linearly independent.
     
  9. Apr 12, 2016 #8
    Oroduin is correct. There is no unique answer. For example taking the cross product between the two vector yields another vector. A orthogonal rotation about this vector axis by 90 degrees will transform one vector into the other. That is one way to get a matrix. However, you could equally well take a unit vector in the direction a + b. This yields another vector c. Now a rotation of 180 degrees about c transforms a into a multiple of b. Now rescale the matrix to transform a into b.
     
  10. Apr 12, 2016 #9
    Sorry, I misunderstood
     
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