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

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  • $\begingroup$ 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. $\endgroup$
    – Julien
    Mar 6, 2013 at 16:04

4 Answers 4

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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 $$P^{-1}e^TP=e^A\qquad \Rightarrow\qquad \det (e^T)=\det (P^{-1}e^TP)=\det(e^A).$$

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  • $\begingroup$ Nice answer (+1) $\endgroup$
    – Thomas
    Mar 6, 2013 at 16:09
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    $\begingroup$ @1015 Is it possible to generalize this to $det(f(A))=f(trA)$. Where $f(A)$ is some continuous differentiable function. i.e. use $P^{-1}T^kP=T^k$ in connection with the Taylor expansion of the function $f$. $\endgroup$ Nov 18, 2017 at 15:21
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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*}

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  • $\begingroup$ Can i have another hint. its a hard question $\endgroup$
    – John
    Mar 6, 2013 at 15:29
  • $\begingroup$ posted antoher hint $\endgroup$ Mar 6, 2013 at 15:33
  • $\begingroup$ is ta tthe jordan normal form $\endgroup$
    – John
    Mar 6, 2013 at 15:33
  • $\begingroup$ $A$ is the normal matrix and $D$ is the jordan normal form of $A$ $\endgroup$ Mar 6, 2013 at 15:33
  • $\begingroup$ Posted a more explizit proof $\endgroup$ Mar 6, 2013 at 15:40
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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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  • $\begingroup$ Ah! Finally an elementary answer...+1. I think you want $com(e^{sA})^te^{sA}=\det(e^{sA})I_n$. $\endgroup$
    – Julien
    Mar 6, 2013 at 16:18
  • $\begingroup$ @julien: Thank you, I edited my answer. $\endgroup$
    – Seirios
    Mar 6, 2013 at 16:52
  • $\begingroup$ Why $D \det(e^{tA}) \cdot Ae^{tA}=\text{tr} \left(^t \text{com}(e^{tA})Ae^{tA} \right)$? $\endgroup$
    – math.n00b
    Aug 16, 2014 at 14:44
  • $\begingroup$ The result is known as Jacobi's formula: en.wikipedia.org/wiki/Jacobi's_formula $\endgroup$
    – Seirios
    Aug 16, 2014 at 15:17
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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 $e^A = e^De^N.$

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  • $\begingroup$ so is this basically all i need to write out $\endgroup$
    – John
    Mar 6, 2013 at 15:38
  • $\begingroup$ @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. $\endgroup$
    – Thomas
    Mar 6, 2013 at 15:39
  • $\begingroup$ can you help me a bit more please regarding those details $\endgroup$
    – John
    Mar 6, 2013 at 15:40
  • $\begingroup$ @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. $\endgroup$
    – Thomas
    Mar 6, 2013 at 15:42
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    $\begingroup$ @julien: I like the idea of step (3). I guess it seems more clear to me... $\endgroup$
    – Thomas
    Mar 6, 2013 at 16:05

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