Taking derivative of $\log \det[x(I - y H)^{-1} + z I] \text{ w.r.t. }x, y, z$

Question

Let $C=x(I - y H)^{-1} + z I$ be an $n\times n$ symmetric positive definite matrix, where $H$ is an $n\times n$ symmetric matrix. Let $$ f(x, y, z) = \log \det[x(I - y H)^{-1} + z I]. $$ Is there any explicit formula for $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}$, and $\frac{\partial f}{\partial z}$?

If making change of variable $\rho = z/x$, then $$ g(x, y, \rho) = f(x, y, x\rho) = \log \det[x(I - y H)^{-1} + x \rho I], $$ and I could get explicit formula for $\frac{\partial g}{\partial x}$.

I am wondering whether we can find explicit formulas for $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}$, and $\frac{\partial f}{\partial z}$, respectively, or at least two of them.

Thanks.

Answer 1

Let $$\eqalign{ P&=(I-yH)^{-1}\cr M&=xP+zI \cr }$$ and find the differential of the function
$$\eqalign{ f &= {\rm log}({\rm det}(M))\cr &= {\rm tr}({\rm log}(M)) \cr\cr df &= M^{-T}:dM \cr }$$ Now we need to express $dM$ in terms of {$dx,dy,dz$} $$\eqalign{ dM &= P\,dx + x\,dP + I\,dz \cr &= P\,dx + xPHP\,dy + I\,dz \cr }$$ To find $\frac{\partial f}{\partial z}$ for example, substitute the $dz$ portion of $dM$ to obtain $$\eqalign{ df &= M^{-T}:I\,dz \cr &= (xP+zI)^{-T}:I\,\,dz \cr &= (xP+zI)^{-1}:I\,\,dz \cr &= {\rm tr}((xP+zI)^{-1}) \,dz \cr\cr \frac{\partial f}{\partial z} &= {\rm tr}\big((xP+zI)^{-1}\big) \cr &= {\rm tr}\Big((x(I-yH)^{-1}+zI)^{-1}\Big) \cr }$$

Answer 2

We assume that $U=x(I-y H)^{-1} + zI$ is invertible and let $f(U(x,y,z))=\log(|\det(U)|)$. Then $\dfrac{\partial f}{\partial t}=trace (\dfrac{\partial U}{\partial t}U^{-1})$. Thus it suffices to calculate $\dfrac{\partial U}{\partial t}$ when

$t=x$: $\dfrac{\partial U}{\partial x}=(I-yH)^{-1}$

$t=y$: $\dfrac{\partial U}{\partial y}=x(I-yH)^{-1}H(I-yH)^{-1}$.

$t=z$: $\dfrac{\partial U}{\partial z}=I$.