- #1
CuppoJava
- 24
- 0
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
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