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Matrix Problem

  1. Apr 16, 2008 #1
    the correlation between A and B is 0.8 and between B and C is 0.9. I have to calculate what the least correlation A and C can be.

    1 & 0.8 & x\\
    0.8 & 1 & 0.9\\
    x & 0.9 & 1

    using sthe standard equation for eigenvalues you get:

    [tex] -\lambda^3 + 3\lambda^2 + \lambda(x^2-1.55) - x^2 + 1.44x -0.45 [/tex]

    2 points have a used the correct methodology so far?
    Are my calculations correct?
    how do I proceed from here?

  2. jcsd
  3. Apr 16, 2008 #2


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    I'm not sure what you are doing so far. Which, of many definitions of "correlation", are you using here? What does the x in your matrix represent? I will point out that the standard "equation" for eigenvalues that you give is not an equation at all. I presume you meant to make that equal to 0. Why did you write that equation?
  4. Apr 16, 2008 #3
    I mean corelation betwen two random variables so the correlation matrix shouls be symmetric and positive defintie.

    So the x in the equation represents the unknown correlation between A and C.

    Well this is the problem:

    look at the equation circa half way down:


    (I try to reproduce it below but its probably clearer on Wikipeadia)
    det[A-lambda I3]= -lambda^3+lambda^2 Trace(A) + 0.5*lambda[Trace(A^2) -Trace^2(A)] + det [A

    when this is set equal to zero and solved you get the eigenvalues.

    If you sub in the values of my matrix including the values x you get the original equation I gave.

    Really I need to find the range of values x that give positive values of the eigenvalues of my correlation matrix (to keep it positive definite). Taking the lower bound to be the lowest possible value of x.

    I think the way I am proceeding is correct? (But I'm not sure)
  5. Apr 17, 2008 #4


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    That's the crucial point, then.

    Since the diagonal values are all 1, I would keep the equation in terms of [itex]1- \lambda[/itex] and then let [itex]u= 1-\lambda[/itex]. The equation then is
    [itex]u^3- (1.45+ x^2)u+ 1.44x= 0[/itex]
    In order that [itex]\lambda> 0[/itex], u must be less than 1.

    In order that there be no roots greater than 1, the expression on the left side of the equation must be always positive or always negative for u> 1. It's easy to see that, as u goes to infinity, that goes to positive infinity so it must be always positive for u> 1.
  6. Apr 17, 2008 #5
    Hi There,

    HallsofIvy thanks for your help with this. I gotta couple of quesitons see your amended quote and below.

    even given the last couple of points surely I am still stuck with one equation and two unknowns? u AND x.
    [itex]u^3- (1.45+ x^2)u+ 1.44x= 0[/itex]
    Last edited: Apr 17, 2008
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