Is there any integral expression for $\log (X + Y) - \log (X)$ if $X$ and $Y$ are positive definite matrices?
Could anyone give some suggestion as to how to find such an integral expression if there is any?
Thanks a bunch.
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1$\begingroup$ can you give an example of the type of integral expression you are hoping for? I am not even sure what would be an "integral expression" for the matrix $X$ by itself... $\endgroup$– Carlo BeenakkerCommented Sep 5, 2021 at 6:25
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$\begingroup$ @CarloBeenakker: I suspect that the OP is looking for a matrix-valued generalization of $\log (x+y) - \log (x) = \int _0 ^1 \frac y {x + ty} \, \mathrm d t$. $\endgroup$– Alex M.Commented Sep 5, 2021 at 7:45
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$\begingroup$ @Daflo Beenakker$:$ Perhaps it is related to spectral theorem for self-adjoint operators. $\endgroup$– RKCCommented Sep 5, 2021 at 10:12
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$\begingroup$ @AlexM. -- thanks, I don't think a single integral of this type will work, I have written a double integral expression in the answer box. $\endgroup$– Carlo BeenakkerCommented Sep 5, 2021 at 21:18
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1 Answer
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The formula $$\frac{d}{ds}\log Z(s) = \int_0^1 [(1-t)I+tZ(s)]^{-1}Z'(s) [(1-t)I+tZ(s)]^{-1}\, dt,$$ with $Z(s)=X+sY$, gives upon integration of $$\int_0^1 \frac{d}{ds}\log Z(s)\,ds=\log Z(1)-\log Z(0)$$ an integral expression for $$\log(X+Y)-\log X=\int_0^1\int_0^1 [(1-t)I+t(X+sY)]^{-1} Y [(1-t)I+t(X+sY)]^{-1}\, dt\,ds.$$
answered Sep 5, 2021 at 15:33