Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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))
share|improve this question
prove it for diagonal matrices and use denseness of diagonalizable matrices. notice that the trace of $A$ is the sum of the eigenvalues –  lyj Mar 1 '13 at 20:34

4 Answers 4

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.

share|improve this answer

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

share|improve this answer

You just use the Jordan normal form, $A$ is you matrix, $D$ is the Joran Form of $A$ and $S$ is the transfomration matrix, for the second equal just remember of the definition of the Matrixexponential, 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.

share|improve this answer

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 $t=1$, $\det(e^{A})= e^{\text{tr}(A)}$.

share|improve this answer
What kind of commutator is $^t\text{com}(e^{tA})$? Never seen this notation. –  Love Learning Jan 29 '14 at 10:13
$\mathrm{com}(A)$ is the matrix of cofactors associated to $A$. Possibly, it is a French notation, where $\mathrm{com}$ stands for comatrice. –  Seirios Jan 29 '14 at 11:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.