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Math Help - Generalized Rayleigh Quotient

  1. #1
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    Generalized Rayleigh Quotient

    Given the Generalized Rayleigh Quotient
    Q(x) = (Ax dot x)/(Bx dot x)
    and that A and B are both real n x n matrices, and B is positive definite.
    How would I go about proving that Q(x) has at least one critical point. In class we restricted ||x|| to 1 so that ||x||^2 also is 1, but we never actually see ||x||^2 in this equation, so it doesn't seem like that would be helpful. If someone can please help me with where to start that'd be awesome! I know that taking the gradient is the first step, but after that I'm utterly confused.
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    Re: Generalized Rayleigh Quotient

    Quote Originally Posted by tubetess123 View Post
    Given the Generalized Rayleigh Quotient Q(x) = (Ax dot x)/(Bx dot x) and that A and B are both real n x n matrices, and B is positive definite
    Surely you forgot to say that A,B are symmetric. Hint: Use the substitution x=Py where P\in \mathbb{R}^{n\times n} is a non singular matrix such that P^tAP=D=\textrm{diag}\;(\lambda_1,\ldots,\lambda_n  ) and P^tBP=I being \lambda_j the generalized eigenvalues associated to A and B .
    Last edited by FernandoRevilla; November 20th 2011 at 09:22 PM.
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    Re: Generalized Rayleigh Quotient

    We aren't given that A is positive definite. Also, why does substituting that particular matrix into the equation prove there is a critical point?
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Generalized Rayleigh Quotient

    Quote Originally Posted by tubetess123 View Post
    We aren't given that A is positive definite.
    I meant symmetric.

    Also, why does substituting that particular matrix into the equation prove there is a critical point?
    Suppose \lambda_1\leq\lambda_2\leq\ldots\leq\lambda_n , then

    Q(x)=\dfrac{x^tAx}{x^tBx}=\frac{(Py)^tA(Py)}{(Py)^  tB(Py)}=\dfrac{y^tP^tAPy}{y^tP^tBPy}=\dfrac{y^tDy}  {y^tIy}=

    \frac{\lambda_1y_1^2+\ldots+\lambda_ny_n^2}{y_1^2+  \ldots+y_n^2} \geq \lambda_1\quad (\forall y=(y_1,\ldots,y_n)^t\in\mathbb{R}^n)

    If y_0=(1,0,\ldots,0)^t and x_0=Py_0 then, Q(x_0)=\lambda_1\geq Q(x) for all x\in\mathbb{R}^n . This implies that x_0 is a point of minimum for Q and as a consequence, a critical point of Q .
    Last edited by FernandoRevilla; November 20th 2011 at 09:23 PM.
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    Re: Generalized Rayleigh Quotient

    Thank you for that explanation. Also, how do you make the nice equations show up instead of what I had shown above?
    And is matrix Py a column vector? Because to do the dot product we need two column vectors. And why can we go from (Py)^tB(Py) to y^Py? Why can we change the order of y^t and P^t and why does the B change to A?
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    Re: Generalized Rayleigh Quotient

    The next thing I need to do is prove that x is a CP of Q if and only if Ax = (lambda)Bx for some real lambda. I see that you've already proved one portion of this, but I don't know how I'd go about proving the other part?

    Thanks!
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    Re: Generalized Rayleigh Quotient

    So in your hint above, do you mean to say that P^{t}BP = I and that \lambda _{j} are the eigenvalues of A? It doesn't make sense to me if this is not the case.

    Thanks!
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    Re: Generalized Rayleigh Quotient

    Quote Originally Posted by tubetess123 View Post
    Also, how do you make the nice equations show up instead of what I had shown above?
    (Ax)\cdot x=(Ax)^tx=x^tA^tx=x^tAx . The same for B .

    And is matrix Py a column vector? Because to do the dot product we need two column vectors.
    Yes, it is, y=(y_1,\ldots,y_n)^t is a column vector, so Py is also a column vector.

    And why can we go from (Py)^tB(Py) to y^Py?
    We can't. It is (Py)^tB(Py)=y^tP^tBPy=y^tIy=y^ty=y_1^2+y_2^2+...+y  _n^2


    Why can we change the order of y^t and P^t
    Well known property: (MN)^t=N^tM^t .

    and why does the B change to A?
    A typo on my part. I've just corrected it.
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    Re: Generalized Rayleigh Quotient

    Thanks! Those clarifications are great help!

    Last thing, could you help me prove that if all the above given information is true, and that at a critical point x it's true that Ax = \lambda Bx, that the determinant of A-\lambda B = 0. I don't know how I'd go about showing this. Not sure if I said this, but x can't be 0.
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  10. #10
    MHF Contributor FernandoRevilla's Avatar
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    Re: Generalized Rayleigh Quotient

    Quote Originally Posted by tubetess123 View Post
    Thanks! Those clarifications are great help!
    Last thing, could you help me prove that if all the above given information is true, and that at a critical point x it's true that Ax = \lambda Bx, that the determinant of A-\lambda B = 0. I don't know how I'd go about showing this. Not sure if I said this, but x can't be 0.
    \exists x\neq 0:Ax=\lambda Bx\Leftrightarrow \exists x\neq 0: (A-\lambda B)x=0

    \Leftrightarrow \textrm{rank}\;(A-\lambda B)<n \Leftrightarrow \det (A-\lambda B)=0
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