2
$\begingroup$

I want to calculate
$$ \frac{\mathrm{d}\left(\mathbf{C}^{-1}\right)^T}{\mathrm{d}\mathbf{C}} = \quad? $$

From The Matrix Cookbook I know that $$ \frac{\mathrm{d}\left(\mathbf{X}^{-1}\right)_{kl}}{\mathrm{d}X_{ij}}=-\left(\mathbf{X}^{-1}\right)_{ki}\left(\mathbf{X}^{-1}\right)_{jl} $$ I was thinking that I can do
$$ \frac{\mathrm{d}\left(\mathbf{C}^{-1}\right)^T}{\mathrm{d}\mathbf{C}} =\frac{\mathrm{d}\left(\mathbf{C}^{-1}\right)^T}{\mathrm{d}\mathbf{C}^{-1}}\frac{\mathrm{d}\left(\mathbf{C}^{-1}\right)}{\mathrm{d}\mathbf{C}} $$ In index notation $[\mathbf{A}] = A_{ij}$ and its transpose is $[\mathbf{A}^T] = A_{ji}$. Derivative of transpose of a matrix with itself, then, should be $$ \frac{\mathrm{d}A_{ji}}{\mathrm{d}A_{kl}} = \delta_{jk}\delta_{il} $$ But I am not sure whether my application of chain rule above is correct and also how do I proceed further. What would my answer look like in index notation or matrix notation? I doubt that matrix notation would work because I expect the answer to be a four dimensional array.

P.S.:

If anyone needs context of the problem then I am trying to find the second derivative of compressible neo-Hookean strain energy.
$$ w(\mathbf{C}) = \frac{\lambda}{2}\ln^2{J} - \mu\ln{J} + \frac{mu}{2}(I_1 - 3) $$ where $\mathbf{C} = \mathbf{F}^T\mathbf{F}$ and $J = \lvert \mathbf{F}\lvert$ and also $I_1 = \mathrm{tr}(\mathbf{C})$. I am able to find the first derivative $\frac{\mathrm{d}w}{\mathrm{d}\mathbf{C}}$ but the second derivative has terms involving the derivative of inverse transpose of C with C. This is where I am stuck.

$\endgroup$
  • $\begingroup$ I realized the importance of giving context to my problem. It turns out the matrix $\mathbf{C}$ is symmetric. So $\mathbf{C}^{-T} = \mathbf{C}^{-1}$. So I don't need the derivative of transpose anymore. Also I figured out how to implement derivative of the inverse in Matlab using the information from this website $\endgroup$ – Amit Singh Mar 13 '15 at 10:20
2
$\begingroup$

@ Amit Singh , if you use Matrix Cookbook, then I am sure that you do not understand one word about the calculation of derivatives.

Since $I_1$ is linear, its second derivative is $0$. Then the function to be considered is $f:C\rightarrow (\lambda/2)\log|C|-\mu\sqrt{\log(|C|]}$.

$Df_C:H\rightarrow tr(HC^{-1})(\lambda/2-(\mu/2)\log^{-1/2}(|C|))$.

$D^2f_C:(H,K)\rightarrow -tr(HC^{-1}KC^{-1})(\lambda/2-(\mu/2)\log^{-1/2}(|C|))+tr(HC^{-1})(\mu/4)\log^{-3/2}(|C|)tr(KC^{-1})$.

Why do you want the second derivative ?

$\endgroup$
0
$\begingroup$

If you use the notation from Magnus, Neudecker, "Matrix Differential Calculus with Applications to Simple, Hadamard, and Kronecker Products", 1985, then you have $$ \frac{\partial X^{-1}}{\partial X} = \frac{\partial X^{-1}}{\partial vec(X)} = -X^{-T}\otimes X^{-1} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.