I'm trying take the derivative of a symmetrix matrix $\mathbf{C}$ with respect to itself. $$ \begin{equation} \frac{\partial \mathbf{C}^{-1}}{\partial \mathbf{C}} \end{equation} $$

Using the indicial notation, above equation can be rewritten as follows $$ \begin{equation} \frac{\partial C_{ij}^{-1}}{\partial C_{kl}} \end{equation} $$

At first I've used the following formula, $$ \begin{equation} \frac{\partial C_{ij}^{-1}}{\partial C_{kl}} = -C^{-1}_{ik}C^{-1}_{lj} \end{equation} $$

But I quickly realized that we've lost the symmetry of the problem now.

I read The Matrix Cookbook and the other posts about the same problem but unfortunately, I couldn't understand the things they've done.

For example in this article, at Eq.(100) authors have used the property below when taking the derivative of Eq.(99) $$ \begin{equation} \frac{\partial \mathbf{C}^{-1}}{\partial \mathbf{C}} = -\mathbf{C}^{-1} \boxtimes \mathbf{C}^{-T} \mathbf{I}_s \end{equation} $$ Where $\boxtimes$ is the square product, $\mathbf{I}_s$ is the symmetric fourth-order identity tensor and they are defined as follows $$ \begin{align} (\mathbf{A} \boxtimes \mathbf{B})_{ijkl} &= \mathbf{A}_{ik}\mathbf{B}_{jl} \\ (\mathbf{I}_s)_{ijkl} &= \frac{1}{2}(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}) \end{align} $$

I couldn't understand how did they achieve this result and how can I derive it myself.

  • $\begingroup$ You need a $4$-dimensional matrix. $\endgroup$ May 19 at 11:13

2 Answers 2


$ \def\p{\partial}\def\o{{\tt1}} \def\E{{\cal E}}\def\F{{\cal F}}\def\G{{\cal G}} \def\C{C^{-1}}\def\Ct{C^{-T}} \def\LR#1{\left(#1\right)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\c#1{\color{red}{#1}} $Once you learn the technique, the problem can be solved very briefly $$\eqalign{ d\C &= -{\C\,dC\,\C} \\ &= -\LR{\C\E\Ct}:dC \\ \grad{\C}{C} &= -{\C\E\Ct} \\ }$$ The details are as follows...

Introduce a fourth-order tensor $\E$ with components $$\eqalign{ \E_{ijk\ell} = \delta_{ik}\,\delta_{j\ell} = \begin{cases} \o\quad{\rm if}\; i=k\;\;{\rm and}\;\;j=\ell \\ 0\quad{\rm otherwise} \end{cases} \\ }$$ The most useful property of this tensor is its ability to rearrange matrix products $$\eqalign{ ABC &= \LR{A\E C^T}:B \;=\; \F:B \\ }$$ where juxtaposition implies a single-dot product and a colon $(:)$ denotes the double-dot product $$\eqalign{ &\F_{ijk\ell} = \sum_{p=1}^n\sum_{r=1}^n A_{i\c{p}}\E_{\c{p}jk\c{r}}\,C_{\c{r}\ell}^T \;=\; A_{ik}C_{\ell j} \\ &\LR{\F:B}_{ij} = \sum_{k=1}^n\sum_{\ell=1}^n \F_{ij\c{k\ell}}\,B_{\c{k\ell}} = \sum_{k=1}^n\sum_{\ell=1}^n A_{i\c{k}}B_{\c{k\ell}}C_{\c{\ell}j} \\ }$$ Start with the differential of the matrix inverse identity $$\eqalign{ &I = \C C \\ &dI = \c{d\C}C + \C dC \;\doteq\; 0 \\ &\c{d\C} = -\C\,dC\,\C \\ }$$ Then use $\E$ to rearrange the terms and recover the gradient $$\eqalign{ d\C &= -{\C\E\Ct}:dC \\ \grad{\C}{C} &= -{\C\E\Ct} \\ }$$ Or in component notation $$\eqalign{ \grad{\C_{ij}}{C_{k\ell}} &= -\sum_{p=1}^n\sum_{r=1}^n \C_{i\c{p}}\E_{\c{p}jk\c{r}}\C_{\ell\c{r}} \;=\; -\C_{ik}\C_{\ell j} \\ }$$


The comments have become a rehash of the old "symmetric gradient" debate.

On the other hand, if a small set of scalar parameters are used to construct a tensor quantity, then the derivative of the tensor components with respect to one of those scalar parameters can exhibit any number of interesting symmetries. A large part of Continuum Mechanics is devoted to studying the implications of such symmetries.

But that's a different problem than calculating the derivative of one tensor component with respect to another tensor component. But many people (even professors and famous authors) often conflate these two problems.

  • $\begingroup$ Nice explanation but is this result still valid for symmetric C? It looks like you ended up with the same expression I've used in my post. $\endgroup$ May 18 at 18:30
  • 1
    $\begingroup$ When it comes to gradients, symmetric matrices are terribly misunderstood. If you're interested you should really study this paper. In short, a gradient expression which is valid for a general matrix is also valid for a symmetric matrix. When it comes to higher-order tensors, the concept of symmetry itself gets ambiguous, i.e. is it symmetric in its first 2 indexes? Its last 2? The first and the last indexes? It's not worth the headache to worry about such things if all you need is a valid expression for a gradient descent algorithm. $\endgroup$
    – greg
    May 18 at 20:09
  • 1
    $\begingroup$ BTW, you haven't "lost" any of the symmetry inherent in the problem. In the indexed expression, $(i,j)$ can be swapped on the RHS without affecting its validity. Likewise, and independently, $(k,\ell)$ could be swapped. $\endgroup$
    – greg
    May 18 at 20:48
  • $\begingroup$ For example if I swap the indices of the nominator I get $\frac{\partial C_{ji}^{-1}}{\partial C_{kl}} = -C^{-1}_{jk}C^{-1}_{li}$ but is this equal to $-C^{-1}_{ik}C^{-1}_{lj}$ $\endgroup$ May 19 at 8:05

In the single-variable case, we have that


This can obtained by differentiating the expression $C(t)C(t)^{-1}=I$ on both sides with some simple algebra. This directly generalizes to the multivariable case by expressing


Then, we have that


from which we get


When $C$ is symmetric, then it can be written as


$$\begin{array}{rcl} \dfrac{d}{dc_{kl}}C^{-1}&=&-C^{-1}\dfrac{dC}{dc_{kl}}C^{-1}=-C^{-1}(e_ke_l^T+e_le_k^T)C^{-1},\ \mathrm{for}\ k\ne l\\ \dfrac{d}{dc_{kk}}C^{-1}&=&-C^{-1}\dfrac{dC}{dc_{kk}}C^{-1}=-C^{-1}e_ke_k^TC^{-1} \end{array}$$

then we have that

$$\begin{array}{rcl} \dfrac{d}{dc_{kl}}C_{ij}^{-1}&=&-e_i^TC^{-1}(e_ke_l^T+e_le_k^T)C^{-1}e_j=-C_{ik}^{-1}C_{lj}^{-1}-C_{il}^{-1}C_{kj}^{-1},\ \mathrm{for}\ k\ne l\\ \dfrac{d}{dc_{kk}}C_{ij}^{-1}&=&-e_i^TC^{-1}e_ke_k^TC^{-1}e_j=-C_{ik}^{-1}C_{kj}^{-1}.\end{array}$$

  • $\begingroup$ What about the symmetry? $\endgroup$ May 18 at 15:52
  • $\begingroup$ @MuratGüven Edited the post. $\endgroup$
    – KBS
    May 18 at 16:59
  • $\begingroup$ Consider a diagonal (and therefore symmetric) matrix $$ C = {\rm Diag}\Big(a\;\;b\;\;c\;\;\ldots\Big) \quad\implies\quad C^{-1} = {\rm Diag}\Big(a^{-1}\;\;b^{-1}\;\;c^{-1}\;\;\ldots\Big) $$ According to your general formula the derivative of the first component is $$\frac{da^{-1}}{da} = -a^{-2}\qquad \Big(i=j=k=\ell=\tt1\Big)$$ which is correct, but according to your symmetric formula it should be doubled for some reason. That doesn't make any sense. $\endgroup$
    – greg
    May 19 at 11:05
  • $\begingroup$ Okay, you've fixed the diagonal terms, but the whole idea that symmetric matrices require special treatment is bogus. Please read this paper for a detailed explanation. $\endgroup$
    – greg
    May 19 at 11:10
  • 1
    $\begingroup$ @MuratGüven I've read the paper and updated my answer with an explanation. $\endgroup$
    – greg
    May 19 at 13:25

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.