The determinant of a matrix exponential is the exponential of the trace

Let A be an n \times n matrix over \mathbb{C}. Prove that \mathsf{det}\ \mathsf{exp}(A) = e^{\mathsf{tr}\ A}.

Say P^{-1}AP = \bigoplus_i J_{k_i}(\lambda_i) is in Jordan canonical form, where J_k(\lambda) is the Jordan block of size k with eigenvalue \lambda. Now \mathsf{det}\ \mathsf{exp}(A) = \mathsf{det}\ P^{-1}\mathsf{exp}(A)P = \mathsf{det}\ \mathsf{exp}(P^{-1}AP) = \mathsf{det}\ \mathsf{exp}(\bigoplus_i J_{k_i}(\lambda_i)) = \mathsf{det}\ \bigoplus_i \mathsf{exp}(J_{k_i}(\lambda_i)). By this previous exercise, \mathsf{exp}(J_k(\lambda)) is upper triangular with diagonal entries e^\lambda. In particular, the determinant of this direct sum is just \prod_i \prod_{k_i} e^{\lambda_i} = e^{\mathsf{tr}\ A}, since the product is over all the eigenvalues of A (with multiplicity).

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