How to prove $\det(e^A) = e^{\operatorname{tr}(A)}$?

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

Solution 1:

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

Solution 2:

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*}

Solution 3:

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

Solution 4:

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 $e^A = e^De^N.$