---
tag: W69VC0G18A
kind: proof
name: Proof that the determinant of a matrix exponential is the exponential of the trace
msc: [15A16, 15A15]
proves: 0A1QQN333A
uses: [0K665H5V1S, 0C3CFHDN6R, 07PTPSZPME, 0363JMQ42Y, 0MKABWDRNC, 054VXHMCA3, 0K1KKMFHYH, 04BD23PSK5, 0Y4NF1DRKD, 04CPJ6AF00, 0Z4BWARC38, 07XBM3A9J5]
sources:
  - url: https://faculty.tcu.edu/richardson/Seminars/TrDet.pdf
    locator: Theorem det(exp A) = e^{tr A}
---

## Proof

We triangularise $A$ over $\C$ and compute both sides for a triangular matrix.

1. If $\mathbb{K} = \R$, we view $A$ as a complex matrix. Its exponential is given by the same series, and its determinant and trace do not change. So we may assume that $A$ is a complex matrix.
2. By [[triangularisation|0363JMQ42Y]] over $\C$, there is an invertible complex matrix $P$ such that $U = P^{-1} A P$ is [[upper triangular|0MKABWDRNC]]. Write $U = (u_{ij})$.
3. Let $S_N(X) = \sum_{k=0}^{N} X^k / k!$. By induction on $k$, $U^k = P^{-1} A^k P$. So $S_N(U) = P^{-1} S_N(A) P$.
4. Each entry of $P^{-1} S_N(A) P$ is a fixed linear combination of the entries of $S_N(A)$. By the [[definition of $e^A$|0K665H5V1S]] and the [[algebra of limits|07XBM3A9J5]], letting $N \to \infty$ in step 3 gives $e^U = P^{-1} e^A P$.
5. Let $X, Y$ be upper triangular. For $i > j$, each term $x_{il} y_{lj}$ of $(X Y)_{ij}$ needs $i \le l \le j$, which is impossible. So $X Y$ is upper triangular. The only nonzero term of $(X Y)_{ii}$ has $l = i$, so $(X Y)_{ii} = x_{ii} y_{ii}$.
6. By step 5 and induction on $k$, $U^k$ is upper triangular with diagonal entries $u_{ii}^k$. This also holds for $k = 0$, with $U^0 = I_n$.
7. So $S_N(U)$ is upper triangular, and its entry $(i, i)$ is $\sum_{k=0}^{N} u_{ii}^k / k!$.
8. Letting $N \to \infty$, $e^U$ is upper triangular. Its entry $(i, i)$ is the [[complex exponential|054VXHMCA3]] $e^{u_{ii}}$.
9. By the [[determinant of a triangular matrix|0K1KKMFHYH]], $\det e^U = e^{u_{11}} e^{u_{22}} \cdots e^{u_{nn}}$.
10. For complex numbers $z, w$, the $1 \times 1$ matrices $(z)$ and $(w)$ commute, and their exponentials are $(e^z)$ and $(e^w)$. By the [[exponential of a sum of commuting matrices|04BD23PSK5]], $e^{z + w} = e^z e^w$.
11. By step 10 and induction on $n$, $e^{u_{11}} \cdots e^{u_{nn}} = e^{u_{11} + \cdots + u_{nn}} = e^{\tr U}$, where $\tr$ is the [[trace|07PTPSZPME]].
12. Since the [[trace of a product is symmetric|0Y4NF1DRKD]], $\tr U = \tr\big(P^{-1} (A P)\big) = \tr\big((A P) P^{-1}\big) = \tr A$.
13. By step 4, $e^A = P e^U P^{-1}$. The [[determinant is multiplicative|04CPJ6AF00]], and complex numbers commute. So $\det e^A = \det e^U \cdot \det P \det P^{-1} = \det e^U \det(P P^{-1})$.
14. Since $P P^{-1} = I_n$ and [[$\det I_n = 1$|0Z4BWARC38]], step 13 gives $\det e^A = \det e^U$.
15. By steps 9, 11, 12 and 14, $\det e^A = e^{\tr U} = e^{\tr A}$.

## Idea

Both sides are unchanged when $A$ is replaced by a similar matrix. Over $\C$, every matrix is similar to a triangular one. For a triangular matrix $U$, the exponential is triangular with diagonal entries $e^{u_{ii}}$, so its determinant is $\prod_i e^{u_{ii}} = e^{\tr U}$.