I have a function: $$ \psi(t) = \frac{1}{2} \ln det\left(-\frac{\Omega(t)^{-1}}{2}\right) $$ where $\Omega(t)$ is a symmetric matrix semi-definite positive of size $k$. When I want to calculate the gradient, I have three different solutions, so I would like to know what mistakes I have made, what is the correct solution (for me case(1))

To calculate the gradient $[\ln det(X)]$ where $X$ is a matrix, I used [1], where $[\nabla \ln det(X)] = X^{-1}$

Case (1): Using the fact that $det(a\Sigma) = a^k det(\Sigma)$ where $a$ is a constant and $\Sigma$ a matrix of size $k \times k$, and [1]: \begin{align} \psi(t) &= \frac{1}{2} \left[ \ln \frac{1}{(-2)^k} - \ln det(\Omega(t)) \right]\\ \nabla_t \psi(t) &= - \frac{\Omega(t)^{-1}}{2}\\ \end{align}

Case (2): Using the fact that $det(A^{-1}) = det(A)^{-1}$, $\ln det(A)^{-1} = -\ln det(A) $ and [1]: \begin{align} \psi(t) &= -\frac{1}{2} \ln det(-2\Omega(t))\\ \nabla_t \psi(t) &= \frac{\Omega(t)^{-1}}{4}\\ \end{align}

Case(3): Using the fact that $\Xi(t) = (-2\Omega(t))^{-1}$ and [1]: \begin{align} \psi(t) &= \frac{1}{2} \ln det(\Xi(t))\\ \nabla_t \psi(t) &= \frac{1}{2} \Xi(t)^{-1} = -\Omega(t) \\ \end{align}

Thanks in advance for your help

[1] How to calculate the gradient of log det matrix inverse? or How to calculate the gradient of log det matrix inverse?

  • $\begingroup$ All of your cases attempt to calculate $\frac{\partial\psi}{\partial\Omega}$ but your question (and nabla notation) suggest that you actually wanted to calculate $$\nabla_t\phi=\frac{d\psi}{dt} = -\tfrac{1}{2}{\rm Tr}\big(\Omega^{-1}\tfrac{\partial\Omega}{\partial t}\big)$$ $\endgroup$ – greg Mar 11 at 19:13

Another way to solve this is by a change of variables.

You already know that for symmetric matrices

$$d \operatorname{logdet}(X) = X^{-1} : dX$$

Let $$ U = -\frac{1}{2}\Omega^{-1}; \quad U^{-1}=-2\Omega$$

then $$ dU = \frac{1}{2}{\Omega^{-1} d\Omega \Omega^{-1}} $$

We have that

$$ \eqalign{ \phi &= \frac{1}{2} \operatorname{logdet}U \\ d\phi &= \frac{1}{2}U^{-1} : dU \\ &=-\Omega : \frac{1}{2}\Omega^{-1}d\Omega\Omega^{-1} \\ &=-\frac{1}{2}\Omega^{-1} : d\Omega }$$

Thus $$\frac{\partial \phi}{\partial \Omega}= -\frac{1}{2}\Omega^{-1}$$


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