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Curious for some insight: inverse of a random matrix is really ordered!

  1. Sep 25, 2010 #1

    [PLAIN]http://www.mathworks.co.uk/products/demos/shipping/matlab/inverter_01_thumbnail.png [Broken]
    ^^^ Random matrix, 100x100

    [PLAIN]http://www.mathworks.co.uk/products/demos/shipping/matlab/inverter_02_thumbnail.png [Broken]

    ^^^ Inverse of this random matrix.

    The given reason is; "each element in this matrix ("b") depends on every one of the ten thousand elements in the previous matrix ("a")." but that still makes it random... right?

    So... if each element is dependant on 10,000 other elements, it must be pretty random as well. My knowledge on linear algebra is pretty weak, but I'm pretty sure this inverse matrix is unique, so isn't there some sort of loss of information entropy going on here if we go from a very random looking picture to a very un-random looking picture through a one-to-one function?

    What's going on?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 25, 2010 #2


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    The discussion of how the entries are not random is simply saying that the construction of the inverse from the original matrix is a deterministic process (which is a good thing)

    This isn't really a linear algebra question. The one-to-one function doesn't have to preserve "randomness" (whatever that really means): The probability of getting A and A-1 by entering random entries into a matrix don't have to be the same. And all those colors mean is that

    Let's consider a slightly different tack: We pick a number between 0 and 100 at random uniformly. When we look at what kind of numbers we get, they'll be evenly spread between 0 and 100. Then consider looking at their inverses: if the random number is x, consider 1/x. Then 99% of the 1/x's will be between 0 and 1, and only 1% will be between 1 and infinity.

    So if we had some colored dots in a row where the color indicated what value we have for each random number, the colored dots for the 1/x's will be far more uniform than for the x's. This isn't a statement about randomness, just about how the range that the numbers fall between is smaller
  4. Sep 25, 2010 #3
    Ok, so some off the apparent "unrandomless" appears that way because the inverse matrix is coloured in according to the same rules as the other one (using the same colourmap) though its range is different.

    But there is more "unrandomness" than just the element values- there are clear gridlines. Even if the range of the numbers is changed to compensate for the effect you describe, we still see some order emerging from the plot of the matrix.

    Thanks for the reply, truth is I maybe don't know what I'm talking about but I do find it fascinating nevertheless.
  5. Sep 25, 2010 #4


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    Assuming this example is representative, it says that the inverse of a matrix with IID entries coming from the uniform distribution on [0,1] has some interesting features, e.g. that there seems to be some strong correlations between entries in the same row, or the same column.

    Trying to explain the distribution of the entries in the inverse matrix, I think, would count as linear algebra.
  6. Sep 25, 2010 #5


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    When I said "this isn't really a linear algebra question" I was referring to the idea of how applying a function can change the seemingly randomness of something.
  7. Sep 25, 2010 #6


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    Look at the formula for the inverse of a matrix. Each Cij is (-1)i+j times the determinant of the matrix you have left if you remove row i and column j from A. The correlations must be caused by the fact that a vast majority of the numbers that go into the calculations of Cij and Cik are the same.
  8. Sep 26, 2010 #7
    If you look at the reverse direction, it becomes more or less trivial i.e. inverse of a "structured" matrix has no apparent "structure". Quotations marks are meant for the reserved use of the word not the literal meaning.

    Thus, it is fair to say that structure is lost during inversion not created.
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