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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

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

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