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

Question

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

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