Derivative of Log Determinant of a Matrix w.r.t a parameter

  1. Hi,
    I'm trying to see why the following theorem is true. It concerns the derivative of the log of the determinant of a symmetric matrix.

    Here's the theorem as stated:

    For a symmetric matrix A:
    [tex]\frac{d}{dx} ln |A| = Tr[A^{-1} \frac{dA}{dx}][/tex]

    Here's what I have so far, I'm almost at the answer, except I can't get rid of the second term at the end:

    [tex]A = \sum_{i} \lambda_{i} u_{i} u_{i}^{T}[/tex]
    [tex]A^{-1} = \sum_{i} \frac{1}{\lambda_{i}} u_{i} u_{i}^{T}[/tex]

    So
    [tex]A^{-1} \frac{dA}{dx} = \sum_{i} \frac{1}{\lambda_{i}} u_{i} u_{i}^{T} \frac{d}{dx}(\sum_{j}\lambda_{j} u_{j} u_{j}^{T})
    =\sum_{i}\sum_{j}\frac{1}{\lambda_{i}}\frac{d\lambda_{j}}{dx}u_{i} u_{i}^{T}u_{j} u_{j}^{T} + \sum_{i}\sum_{j}\frac{\lambda_{j}}{\lambda_{i}}u_{i} u_{i}^{T}\frac{d}{dx}u_{j} u_{j}^{T}
    =\sum_{i}\frac{1}{\lambda_{i}}\frac{d\lambda_{j}}{dx}u_{i} u_{i}^{T} + \sum_{i}\sum_{j}\frac{\lambda_{j}}{\lambda_{i}}u_{i} u_{i}^{T}\frac{d}{dx}u_{j} u_{j}^{T}[/tex]

    And this would be just perfect if the second term was equal to zero. But I can't see how that could be made to happen.

    Thanks a lot for your help
    -Patrick
     
  2. jcsd
  3. This theorem is true indeed, and doesn't even need A to be symmetric.

    Using :
    [tex] \frac{\partial}{\partial x} ln det A = \sum_{i,j} \frac{\partial a_{ij}}{x} \frac{\partial}{\partial a_{i,j}}[\tex]

    with :
    [tex]\frac{\partial }{\partial c_{ij}} ln det A = (A^{-1})_{ji}[\tex]

    you get :
    [tex] \frac{\partial}{\partial x} ln det A = Tr(A^{-1}\frac{\partial A}{\partial x}) = Tr(\frac{\partial A}{\partial x}A^{-1})[\tex]

    I hope that will help...

    Canag
     
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