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Suppose $A$ is a $n \times n$ constant matrix. How can I prove $\det(e^A) = e^{\displaystyle \sum_{\lambda_i\in\sigma(A)} \lambda_i}$,

where $\sigma(A)$ is the multiset of eigenvalues of $A$?

The following matlab code shows this is true in $n = 3$ case:

A = rand(3)
detA = exp(sum(eig(A)))
detmA = det(expm(A))
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    $\begingroup$ prove it for diagonal matrices and use denseness of diagonalizable matrices. notice that the trace of $A$ is the sum of the eigenvalues $\endgroup$ – cats Mar 1 '13 at 20:34
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If you're afraid of the density of diagonalizable matrices, simply triangularize $A$. You get $$A=P^{-1}UP,$$ with $U$ upper triangular and the eigenvalues $\{\lambda_j\}$ of $A$ on the diagonal.

Then $$ \mbox{det}\;e^A=\mbox{det}(P^{-1}e^UP)=\mbox{det}\;e^U. $$

Now observe that $e^U$ is upper triangular with $\{e^{\lambda_j}\}$ on the diagonal.

So $$ \mbox{det} \;e^A=\mbox{det} \;e^U=e^{\lambda_1}\cdots e^{\lambda_n}=e^{\lambda_1+\ldots+\lambda_n}. $$

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    $\begingroup$ Anyone trying to understand this solution: The main idea is that if the eigenvalues of $A$ are $\lambda_i$, then the eigenvalues of $e^A$ are $e^{\lambda_i}$, which you can see by using the series definition of $e^A$ and multiplying nonzero eigenvector $v_i$ as in $e^A v_i = \lambda_i v_i$. Distribute the $v_i$ through the series. Finally, the determinant of a matrix is the product of the eigenvalues, and the trace of a matrix is the sum of the eigenvalues. This explains the second to last equality in the proof above. $\endgroup$ – wkschwartz Oct 25 '15 at 23:28
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You can reduce problem for diagonal matrix using the fact that every matrix can be approximated with any given precision and both functions: $e^X$ and $\det X$ are continuous. The problem with diagonal matrix is obvious.

P.S. It seems that lyj was faster than me.

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For an analytic method, using differential equations:

Let $f(t)= \det(e^{tA})$. Then $f'(t)=D \det(e^{tA}) \cdot Ae^{tA}=\text{tr} \left(^t \text{com}(e^{tA})Ae^{tA} \right)$. But $A$ and $e^{tA}$ commute, and $^t\text{com}(e^{tA})e^{tA}=\det(e^{tA}) \operatorname{I}_n$. Therefore, $f'(t)=\text{tr}(A)f(t)$ and $f(0)=1$, hence $f(t)=e^{\text{tr}(A)t}$. For $t=1$, $\det(e^{A})= e^{\text{tr}(A)}$.

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  • $\begingroup$ What kind of commutator is $^t\text{com}(e^{tA})$? Never seen this notation. $\endgroup$ – Your Majesty Jan 29 '14 at 10:13
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    $\begingroup$ $\mathrm{com}(A)$ is the matrix of cofactors associated to $A$. Possibly, it is a French notation, where $\mathrm{com}$ stands for comatrice. $\endgroup$ – Seirios Jan 29 '14 at 11:56
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    $\begingroup$ Hello from 2016! Could you please give a reference to this proof? $\endgroup$ – Rubi Shnol May 17 '16 at 6:35
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You just use the Jordan normal form, $A$ is your matrix, $D$ is the Jordan normal form of $A$ and $S$ is the transformation matrix, for the second equal just remember the definition of matrix exponential, and that $$(SAS^{-1})^n = S A S^{-1}S A S^{-1} S A \dots S A S^{-1}=S A^n S^{-1}$$ \begin{align*} \det(\exp(A))&=\det(\exp(S D S^{-1}))\\ &=\det(S \exp(D) S^{-1})\\ &=\det(S) \det(\exp(D)) \det (S^{-1})\\ &=\det(\exp (D))\\ &=\prod_{i=1}^n e^{d_{ii}}\\ &=e^{\sum_{i=1}^n{d_{ii}}}\\ &=e^{\text{tr}D} \end{align*} As the trace is invariant this works.

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