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Let $X$ be a finite dimensional complex vector space. For $A \in L_\mathbb{C}(X)$ (the set of all complex linear maps from $X$ to itself), we define $$ e^A = \sum_{k=0}^\infty \frac{A^k}{k!}. $$ Show that if $B \in L_\mathbb{C}(X)$ with $\det B \neq 0$, then for some $A \in L_\mathbb{C}(X)$, $e^A=B$.

My original attempt was since $B$ has some Jordan decomposition, i.e., $B = PJP^{-1}$, I can apply the log function to show that $A = P\log(J)P^{-1}$. However, I see that I can run into cases where the diagonal entries of the Jordan form are negative. How should I go about doing this?

Rócherz
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    Some posts on this site about this question: post 1, post 2, post 3. Perhaps one of these suffice for your purposes – Ben Grossmann Mar 06 '25 at 22:50
  • The issue is you talk about "the log function" but there is no such thing in complex analysis. In fact you are looking for a choice of logarithm. Your matrix has finitely many eigenvalues so choose a branch of the logarithm that has all of them in its domain. – user8675309 Mar 07 '25 at 00:45

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