Finding gradient of $\det(\mathbf{\Sigma})^{\frac{-1}{2}}$ with respect to $\mathbf{\Sigma}$ (a Covariance matrix)

Question

In the book Mathematics for Machine Learning (chapter on Gaussian Mixture Models), I got stuck trying to figure out how they came up with

$$\frac{\partial \det(\mathbf{\Sigma})^{\frac{-1}{2}}}{\partial \mathbf{\Sigma}} = \frac{-1}{2}\det(\mathbf{\Sigma})^\frac{-1}{2} \mathbf{\Sigma}^{-1}$$

where $\mathbf{\Sigma}$ is the covariance matrix. The book mentions that the following identity was used: $$\frac{\partial \det(f(\mathbf{X}))}{\partial \mathbf{X}} = \det(f(\mathbf{X}))Trace\left( f(\mathbf{X})^{-1} \frac{\partial f(\mathbf{X})}{\partial \mathbf{X}} \right)$$

But the question is, how did they use it to arrive at the first equation ? I can't seem to figure it out. Been stuck for quite a while.

Thanks!

Answer 1

It's probably easier to work with the logarithmic form of the Jacobi formula $$\eqalign{ \frac{\partial \log\det X}{\partial X} &= X^{-T} &\quad\big({\rm assuming\;}\det X > 0\big) \\ }$$ and to use $S=\Sigma\;$ for ease of typing. $$\eqalign{ c &= (\det S)^{-1/2} &\qquad\qquad&\big({\rm Cost\;function}\big) \\ \log(c) &= -\tfrac 12\log(\det S)\\ \frac{1}{c}\frac{\partial c}{\partial S} &= -\tfrac 12S^{-T} &\qquad&\big({\rm Logarthmic\;derivative=Jacobi}\big) \\ \frac{\partial c}{\partial S} &= -\tfrac 12cS^{-1} &\qquad&\big(S\,{\rm \,is\,symmetric}\big) \\ }$$

Answer 2

Your question has two parts: an apparent matrix/scalar mismatch in the general result, and how to choose and substitute a suitable $f$. To address the first part, I'll assume your $\operatorname{Tr}$ shouldn't be in the general result; after all, you've already edited it out of the special case of interest. So onto the second part: if we take $f(y)=y^{-1/2}$ so $f^\prime(y)=-\tfrac12y^{-3/2}$, the putative identity gives$$\partial_\Sigma(\det \Sigma)^{-1/2}=\det(\Sigma^{-1/2})\operatorname{Tr}(\Sigma^{1/2}\cdot-\tfrac12\Sigma^{-3/2})=-\tfrac12(\det\Sigma)^{-1/2}\Sigma^{-1}.$$