Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove $$\det(e^A) = e^{\operatorname{tr}(A)}$$ for all matrices $A \in \mathbb{C}_{n×n}$.

share|cite|improve this question
All the answers so far use a triangularized form at some point. If you know that every complex square matrix is triangularizable, it brings the problem back to triangular matrices. – 1015 Mar 6 '13 at 16:04

Both sides are continuous. A standard proof goes by showing this for diagonalizable matrices, and then using their density in $M_n(\mathbb{C})$.

But actually, it suffices to triangularize $$ A=P^{-1}TP $$ with $P$ invertible and $T$ upper-triangular. This is possible as soon as the characteristic polynomial splits, which is obviously the case in $\mathbb{C}$.

Let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $A$.

Observe that each $T^k$ is upper-triangular with $\lambda_1^k,\ldots,\lambda_n^k$ on the diagonal. It follows that $e^T$ is upper triangular with $e^{\lambda_1},\ldots,e^{\lambda_n}$ on the diagonal. So $$ \det e^T=e^{\lambda_1}\cdots e^{\lambda_n}=e^{\lambda_1+\ldots+\lambda_n}=e^{\mbox{tr}\;T} $$

Finally, observe that $\mbox{tr} \;A=\mbox{tr}\;T$, and that $P^{-1}T^kP=A^k$ for all $k$, so $$P^{-1}e^TP=e^A\qquad \Rightarrow\qquad \det e^T=\det (P^{-1}e^TP)=\det A.$$

share|cite|improve this answer
Nice answer (+1) – Thomas Mar 6 '13 at 16:09

Hint: Use that every complex matrix has a jordan normal form and that the determinant of a triangular matrix is the product of the diagonal.

use that $\exp(A)=\exp(S^{-1} J S ) = S^{-1} \exp(J) S $

And that the trace doesn't change under transformations.

\begin{align*} \det(\exp(A))&=\det(\exp(S J S^{-1}))\\ &=\det(S \exp(J) S^{-1})\\ &=\det(S) \det(\exp(J)) \det (S^{-1})\\ &=\det(\exp (J))\\ &=\prod_{i=1}^n \exp(j_{ii})\\ &=\exp(\sum_{i=1}^n{j_{ii}})\\ &=\exp(\text{tr}J) \end{align*}

share|cite|improve this answer
Can i have another hint. its a hard question – John Mar 6 '13 at 15:29
posted antoher hint – Dominic Michaelis Mar 6 '13 at 15:33
is ta tthe jordan normal form – John Mar 6 '13 at 15:33
$A$ is the normal matrix and $D$ is the jordan normal form of $A$ – Dominic Michaelis Mar 6 '13 at 15:33
Posted a more explizit proof – Dominic Michaelis Mar 6 '13 at 15:40

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|cite|improve this answer
Ah! Finally an elementary answer...+1. I think you want $com(e^{sA})^te^{sA}=\det(e^{sA})I_n$. – 1015 Mar 6 '13 at 16:18
@julien: Thank you, I edited my answer. – Seirios Mar 6 '13 at 16:52
Why $D \det(e^{tA}) \cdot Ae^{tA}=\text{tr} \left(^t \text{com}(e^{tA})Ae^{tA} \right)$? – math.n00b Aug 16 '14 at 14:44
The result is known as Jacobi's formula:'s_formula – Seirios Aug 16 '14 at 15:17

You can do it in these steps (still requires some work):

$\quad \bf (1)$ $A$ is diagonalizable

$\quad \bf (2)$ $A$ is nilpotent

$\quad \bf (3)$ $A$ is arbitrary

$\bf (1)$ This shouldn't be too hard. Start with assuming that $A = CDC^{-1}$ for $D$ a diagonal matrix.

$\bf (2)$ Use that every nilpotent matrix is similar to a upper triangular matrix $D$ with $0$s on the diagonal. So $A = CDC^{-1}$.

$\bf (3)$ Use that every matrix can be written as the sum $A = D + N$ of a nilpotent matrix $N$ and a diagonalizable matrix $D$ and $D$ and $N$ commute. So $$ \det(e^{A}) = \det(e^De^N) =\det(e^{D})\det(e^{N}) = e^{\text{Tr}(D)}e^{\text{Tr}(N)} = e^{\text{Tr}(D) + \text{Tr}(N)} = e^{\text{Tr}(A)}. $$ We have used here that $D$ and $N$ commute so that $e^A = e^De^N.$

share|cite|improve this answer
so is this basically all i need to write out – John Mar 6 '13 at 15:38
@John: Yes. But you still have to write down the details of step (1) and (2) and there was some claims that I assumed you know. – Thomas Mar 6 '13 at 15:39
can you help me a bit more please regarding those details – John Mar 6 '13 at 15:40
@John: What specific details? (Left is really just to write things down. For example, for step (1) try and write down a diagonal matrix $D$ and the figure out what $e^D$ is using that definition of the exponential map. – Thomas Mar 6 '13 at 15:42
You need to add that $D$ and $N$ commute. Also, since you triangularize $D$, why don't you triangularize $A$ directly (which is what I did)? – 1015 Mar 6 '13 at 15:50

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.