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Dec 26, 2023 at 16:16 comment added Robert Israel Why don't you think it shows that?
Dec 26, 2023 at 6:31 comment added Kanghun Kim It does not seem to show that all logarithms of A(z) are non-entire...
Dec 25, 2023 at 18:54 comment added Christian Remling This is a nice simple argument. I had a slightly more complicated one, which however shows that there almost never will be an entire matrix logarithm: The eigenvalues $\lambda_j$ of $A$ will typically have Puiseux type branch points and hence so will those of $B$ (call them $\mu_j=\log\lambda_j$), but then $\prod\mu_j=\det B$ would have to be entire, which is unlikely.
Dec 25, 2023 at 18:24 history edited Robert Israel CC BY-SA 4.0
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Dec 25, 2023 at 18:24 comment added Robert Israel Yes, it is a counterexample.
Dec 25, 2023 at 18:23 comment added Kanghun Kim Looks like "proof by counterexample" to me...
Dec 25, 2023 at 18:21 history answered Robert Israel CC BY-SA 4.0