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How to calculate the gradient with respect to $X$ of: $$ \log \mathrm{det}\, X^{-1} $$ here $X$ is a positive definite matrix, and det is the determinant of a matrix.

How to calculate this? Or what's the result? Thanks!

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Note that $\log\det\mathbf X^{-1}=\log\frac1{\det\mathbf X}=-\log\det\mathbf X$... – J. M. May 12 '11 at 14:48
And note that $\log \det X =\text{tr} \log X$... – Fabian May 12 '11 at 14:50
Somehow I wonder if what you actually need is the Gâteaux or the Fréchet derivative... where did you encounter this, and what are you actually doing? – J. M. May 12 '11 at 15:00
I encounter this when deriving a lower bound of D-optimal experimental design using dual theory (an exercise of Convex Optimization). I want to find the optimal of a function which involves $\log\mathrm{det}\,(X^{-1})$. – pluskid May 12 '11 at 15:25
up vote 15 down vote accepted

I assume that you are asking for the derivative with respect to the elements of the matrix. In this cases first notice that

$$\log \det X^{-1} = \log (\det X)^{-1} = -\log \det X$$

and thus

$$\frac{\partial}{\partial X_{ij}} \log \det X^{-1} = -\frac{\partial}{\partial X_{ij}} \log \det X = - \frac{1}{\det X} \frac{\partial \det X}{\partial X_{ij}} = - \frac{1}{\det X} \mathrm{adj}(X)_{ji} = - (X^{-1})_{ji}$$

since $\mathrm{adj}(X) = \det(X) X^{-1}$ for invertible matrices (where $\mathrm{adj}(X)$ is the adjugate of $X$, see

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Thank you very much! This solved my problem! – pluskid May 12 '11 at 15:26

Or you can check section A.4.1 of the book (link: for an alternative solution, where they compute the gradient without using the adjugate.

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