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Let $A$ be an $n \times n$ matrix over a field $\mathbb{K}$ where $\mathbb{K} = \mathbb{C}$ or $\mathbb{K} = \mathbb{R}$. Does $\det(e^A) = e^{\mathrm{tr}(A)}$ always hold?

If the field is $\mathbb{C}$, this can be easily proven using the Jordan canonical form. Now, if the field is $\mathbb{R}$, the Jordan canonical form may not exist for $A$. So does $\det(e^A) = e^{\mathrm{tr}(A)}$ hold for any square matrix in $\mathbb{R}$ or only when $A$ meets certain conditions (e.g. $A$ has a Jordan canonical form)?

mathjacks
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    I don't know. What is $e^A$ if we are working over an arbitrary field? – André Nicolas Apr 23 '16 at 04:27
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    For $\Bbb K = \Bbb R$, you can always think of the matrix as over $\Bbb C$, determinant and trace don't change because of that. – Seven Apr 23 '16 at 04:37
  • I think if $\mathbb{K}$ can be viewed as a subfield of $\mathbb{C}$ then it should be fine, because you just view $A$ as a complex matrix, see that the result is true there, and recover the result in the original matrix ring. @AndréNicolas makes a good point though - the matrix exponential is not necessarily defined for arbitrary fields, so you need to narrow the definition somewhat. – πr8 Apr 23 '16 at 04:44
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    Even in the case $\mathbb{K}=\mathbb{R}$, why don't you view $A$ as a complex matrix, as the conclusion is independent of the containing field. – Quang Hoang Apr 23 '16 at 06:50
  • The identity is conveniently proved by going to complex numbers, but is holds for all matrices. Even for those with rational entries :) – Hans Engler May 13 '16 at 22:04

3 Answers3

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For an alternative argument, you can compute $f'(t)$, with $f(t)=\det e^{tA}$. Since $f(t_0+s)=f(t_0)f(s)$, it suffices to do this at $t=0$. Then $e^{tA}=1+tA+O(t^2)$, so $$ \det e^{tA}=1+t\,\textrm{tr}\, A +O(t^2) , $$ because the only way to obtain a contribution linear in $t$ is to multiply one of the $ta_{jj}$ on the diagonal with the $1$'s in the other diagonal entries. This says that $f'(0)=\textrm{tr}\, A$.

So $f$ solves the IVP $f'=(\textrm{tr}\, A)f$, $f(0)=1$. Thus $f(t)=e^{t\,\textrm{tr} A}$, as claimed.

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An Argument from Lie Theory

Using the Jordan-Chevalley Decomposition Theorem (for the Lie algebra $\mathfrak{sl}_n(\mathbb{C})$, which consists of traceless $n$-by-$n$ matrices), $A=t\,I+S+N$ where $t:=\frac{\text{trace}(A)}{n}$, $S$ is semisimple (i.e., diagonalizable over $\mathbb{C}$) of trace $0$ , and $N$ is nilpotent such that $SN=NS$. Then, because $t\,I$, $S$, and $N$ commute, $\exp(A)=\exp(t\,I)\,\exp(S)\,\exp(N)$. That is, $$\det\big(\exp(A)\big)=\det\big(\exp(t\,I)\big)\cdot\det\big(\exp(S)\big)\cdot\det\big(\exp(N)\big)\,.$$ Obviously, the first factor is $\det\big(\exp(t\,I)\big)=\big(\exp(t)\big)^n=\exp(nt)=\exp\big(\text{trace}(A)\big)$. The last factor is $1$ because $\exp(N)=I+M$ for some nilpotent matrix $M$ (whence the eigenvalues of $M$ are all $0$). The middle factor is $1$ because $S$ is diagonalizable as $TDT^{-1}$ for some invertible matrix $T$ and for some diagonal traceless matrix $D$. Then, $$\det\big(\exp(S)\big)=\det\big(T\exp(D)T^{-1}\big)=\det\big(\exp(D)\big)=\exp\big(\text{trace}(D)\big)=1\,.$$ Consequently, $\det\big(\exp(A)\big)=\exp\big(\text{trace}(A)\big)$.

Batominovski
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Consider the field $\mathbb{C}$, from which $\mathbb{R}$ is a special case.

Both sides of your equality are continuous functions on $L(\mathbb{C}^n, \mathbb{C}^n)$, which clearly coincide in diagonizable operators. Since those are dense when we consider $\mathbb{C}$, we have that they must coincide in the whole space.

Aloizio Macedo
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