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I need help finding the Jacobian of the matrix logarithm function, i.e. $\log{M} = R$ defined by $e^R = M = V\begin{bmatrix}e^{\lambda_1} & & \\ & \ddots & \\ & & e^{\lambda_n} \end{bmatrix}V^{-1}$ (where $V\begin{bmatrix} \lambda_1 & & \\ & \ddots & \\ & & \lambda_n \end{bmatrix}V^{-1}$ is the eigendecomposition of $M$).

Anyway, I've found some presentations like this one but found them totally impenetrable.

If it helps, I don't need the derivative in a completely general situation. In my case $M$ is a $3 \times 3$ rotation matrix, and immediately after taking the logarithm I take the Frobenius norm, so I can shortcut those two operations with $||\log{M}||_F = 2\sqrt{\theta}$ as detailed on Wikipedia. However, I'm not sure if I can extend the shortcut to say $\frac{\partial}{\partial M}||\log{M}||_F = 2\theta^{-\frac{1}{2}}$. And anyway, the calculation of $\theta$ depends on arccosine which has nasty discontinuities so that is not an ideal method.

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Derivative with respect to what? You talk about rotation matrices meaning that you might be interested in a single parameter group of matrices. In a more general setting the derivative becomes a linear functional, and it doesn't sound like you are looking for something like that! –  Jyrki Lahtonen Feb 25 '13 at 20:42
I'm not sure that its a straightforward calculation... –  copper.hat Feb 25 '13 at 20:42
If you are interested in rotations about a fixed axis only, then see this recent answer of mine. There I am looking at the derivative of the exponential, but the general recipe is: If $A(\vec{n})$ is the matrix of the linear mapping from $R^3$ to itself gotten by taking the crossproduct with the axis $\vec{n}$, then the rotation matrices are $R(x)=e^{xA}$, where the length of $\vec{n}$ multiplied by $x$ gives the angle of the rotation. For such rotation matrices a logarithm (it's multiply valued) is $xA$, and differentiating w.r.t. $x$ is easy. –  Jyrki Lahtonen Feb 25 '13 at 20:48
Scratch that, may be? I am no longer sure that my idea of a matrix logarithm matches with yours... –  Jyrki Lahtonen Feb 25 '13 at 20:51
Sorry, I'll edit the question as well. To clarify, I mean the Jacobian, so the derivative of the elements of $\log{M}$ with respect to the elements of $M$. –  durka42 Feb 25 '13 at 21:34

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