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Assume A(t) is a time dependent n by n Hermition matrix. What is time-derivative of log(A(t)). I remember seeing this somewhere, but now I have a hard time finding it again.

fred
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  • $\frac{d\log(A(t))}{dt}=A^{-1}(t)\frac{dA(t)}{dt}$. –  Feb 06 '14 at 01:32
  • Are you sure? I was wondering why not $\frac{dA}{dt}A^{-1}$ ? – fred Feb 06 '14 at 01:35
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    Note that the chain rule is given by $\mathbf D(f\circ g) = \mathbf Df(g), \mathbf Dg$, in that order. The derivative of the outer function comes first. – Ben Grossmann Feb 06 '14 at 01:50
  • Is there any reference which I could cite for my question, any book for example which contains this particular example. – fred Feb 06 '14 at 01:56
  • @fred You're right. I was mistaken. I'd forgotten that matrix multiplication isn't commutative. –  Feb 06 '14 at 04:18
  • Thanks for your responses. I am confused now, which one is correct? your answer or mine? Is there any references? – fred Feb 06 '14 at 04:24

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Both answers are false ! In fact, the correct formula is complicated. To see that, let $A=I+B$ where $||B||<1$. Then $\log(A)=B-1/2B^2+1/3B^3+\cdots$. Thus $(log(A))'=B'-1/2(BB'+B'B)+1/3(B^2B'+BB'B+B'B^2)+\cdots$. In particular, $B'$ is not a factor of the result except if $BB'=B'B$.