Show that \[ \det(\exp M)=e^{\operatorname{tr}M} \] for every complex \(n\times n\) matrix \(M\).
Solution
1. There is an invertible matrix \(S\) such that\[ T\coloneqq S^{-1}MS \]is upper triangular.
Over \(\CC\), the characteristic polynomial of \(M\) splits completely. Hence the standard triangularization theorem gives a basis in which \(M\) is upper triangular.
2. If the diagonal entries of \(T\) are \(\lambda_1,\ldots,\lambda_n\), then \(\exp T\) is upper triangular with diagonal entries\[ e^{\lambda_1},\ldots,e^{\lambda_n}. \]
Every power \(T^k\) is upper triangular, and its \(i\)th diagonal entry is \(\lambda_i^k\). Therefore the power series \[ \exp T = \sum_{k=0}^{\infty}\frac{T^k}{k!} \] is upper triangular, and its \(i\)th diagonal entry is \[ \sum_{k=0}^{\infty}\frac{\lambda_i^k}{k!} = e^{\lambda_i}. \]
3. One has\[ \boxed{\det(\exp M)=e^{\operatorname{tr}M}}. \]
The exponential power series commutes with similarity, so step 1 gives \[ \exp T=S^{-1}(\exp M)S. \] Hence \[ \det(\exp M) = \det(\exp T) = \prod_{i=1}^n e^{\lambda_i} = e^{\lambda_1+\cdots+\lambda_n}. \] Since \(T\) is upper triangular, \[ \lambda_1+\cdots+\lambda_n = \operatorname{tr}T = \operatorname{tr}M, \] where the last equality uses invariance of trace under similarity. This gives the displayed identity.
4. Q.E.D.
Step 3 is the required formula.