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Derivative of Log Determinant of a Matrix w.r.t a parameter 
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#1
Nov1409, 06:28 PM

P: 24

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
Apr310, 07:19 AM

P: 1

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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