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

by CuppoJava
Tags: derivative, determinant, matrix, parameter
CuppoJava is offline
Nov14-09, 06:28 PM
P: 24
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]

[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
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canag is offline
Apr3-10, 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...


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