# Derivative of transpose of inverse of matrix with respect to matrix

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.

• 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 – Amit Singh Mar 13 '15 at 10:20

@ 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 ?

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