I have a question about a line in the textbook "Gaussian Processes for Machine Learning" by Rasmussen and Williams (available here). On page 125 (equation 5.23), they write the following:

$-\frac{1}{2} tr(B^{-1}K\frac{\partial W}{\partial \hat{f_i}}) = -\frac{1}{2}[(K^{-1} + W)^{-1}]_{ii}\frac{\partial^{3}}{\partial f_i^{3}} \textrm{log} p(y|\hat{f})$.


$W = -\nabla \nabla \textrm{log} p(y | \hat{f})$ (which is diagonal in this case, with all-positive entries).

$B = I + W^{\frac{1}{2}}KW^{\frac{1}{2}}$

$K$ is symmetric and positive definite.

By using the trace and the fact that the only element of $\frac{\partial W}{\partial \hat{f_i}}$ that is going to be non-zero is the $ii$ element, I think I managed to arrive at:

$-\frac{1}{2} tr(B^{-1}K\frac{\partial W}{\partial \hat{f_i}}) = -\frac{1}{2}(B^{-1}K)_{ii}(\frac{\partial W}{\partial \hat{f_i}})_{ii}$

But I don't understand how to get from $(B^{-1}K)_{ii}$ to $[(K^{-1} + W)^{-1}]_{ii}$, and I am not convinced I am on the right track. Any help would be appreciated!


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