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I've seen on Wikipedia that for a complex matrix $X$, $\det(e^X)=e^{\operatorname{tr}(X)}$.

It is clearly true for a diagonal matrix. What about other matrices ?

The series-based definition of exp is useless here.

Harald Hanche-Olsen
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Gabriel Romon
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2 Answers2

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A alternative to doing this by normal forms which perhaps assumes more but is much more natural to me is (as suggested in the comment on $\det(\exp X)=e^{\mathrm{Tr}\, X}$ for 2 dimensional matrices) to note that it clearly holds for diagonalizable matrices (see the duplicate How to prove $\det(e^A) = e^{\operatorname{tr}(A)}$? ...), and by

  1. the continuity of $\det, \mathrm{tr}$ and $\exp$
  2. the density of diagonalizable matrices in the space of all complex matrices (Diagonalizable matrices with complex values are dense in set of $n\times n$ complex matrices.)

we have the result more generally for all matrices.

Community
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not all wrong
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  • I've thought of this proof too. Are you sure it is valid ? You're dealing with diagonalizable matrices, not diagonal ones. – Gabriel Romon Jan 15 '14 at 17:12
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    Yes, it is. It's enough for $X$ to be diagonalisable for the argument to work (because the trace and determinant are the same for similar matrices). So, equality holds on a dense set of matrices. Everything in the equality is continuous, so we can pass from a dense set to the full space. – Jakub Konieczny Jan 15 '14 at 17:15
  • OK. @Sharkos: you should add that the trace and determinant invariance in your answer to make it perfect. – Gabriel Romon Jan 15 '14 at 17:22
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If $X$ is upper triangular, then this is clear.

If the claim holds for a matrix $Y$, then it holds for any $X$ similar to $Y$.

By Jordan decomposition, each $X$ is similar to an upper triangular matrix $Y$ (of a special form, but never mind).

Thus, the claim holds for all $X$.

Jakub Konieczny
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  • Well the Jordan decomposition is not a cakewalk to prove, is it ? Anyway, I'm glad it does the trick. – Gabriel Romon Jan 15 '14 at 17:13
  • @GabrielR. Noone is saying it is. But once you have it, why not use it? The argument only uses that a matrix can be brought into an upper triangular form - this may be more elementary. – Jakub Konieczny Jan 15 '14 at 17:17
  • It definitely is more elementary. You need the existence of an eigenvector, choose a complementary subspace, do induction and you are done. – Carsten S Jan 15 '14 at 18:47