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

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

• All the answers so far use a triangularized form at some point. If you know that every complex square matrix is triangularizable, it brings the problem back to triangular matrices. Mar 6, 2013 at 16:04

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

• Nice answer (+1) Mar 6, 2013 at 16:09
• @1015 Is it possible to generalize this to $det(f(A))=f(trA)$. Where $f(A)$ is some continuous differentiable function. i.e. use $P^{-1}T^kP=T^k$ in connection with the Taylor expansion of the function $f$. Nov 18, 2017 at 15:21
• @AlexanderCska You can show, for any smooth path of complex square matrices $B(t)$ that passes through the identity, $$\left.\frac{d}{dt}\right|_{t=0}\det\left(B(t)\right)=\text{Tr}\left(B'(0)\right).$$ The equation essentially expresses the more general theorem that every Lie group homomorphism induces a Lie algebra homomorphism, so I doubt it holds outside of the exponential, which maps a Lie algebra into its Lie group. See Stillwell (2008) for more details. Sep 6, 2022 at 12:15

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*}

• Can i have another hint. its a hard question
– John
Mar 6, 2013 at 15:29
• posted antoher hint Mar 6, 2013 at 15:33
• is ta tthe jordan normal form
– John
Mar 6, 2013 at 15:33
• $A$ is the normal matrix and $D$ is the jordan normal form of $A$ Mar 6, 2013 at 15:33
• Posted a more explizit proof Mar 6, 2013 at 15:40

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

• Ah! Finally an elementary answer...+1. I think you want $com(e^{sA})^te^{sA}=\det(e^{sA})I_n$. Mar 6, 2013 at 16:18
• @julien: Thank you, I edited my answer. Mar 6, 2013 at 16:52
• Why $D \det(e^{tA}) \cdot Ae^{tA}=\text{tr} \left(^t \text{com}(e^{tA})Ae^{tA} \right)$? Aug 16, 2014 at 14:44
• The result is known as Jacobi's formula: en.wikipedia.org/wiki/Jacobi's_formula Aug 16, 2014 at 15:17

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

• so is this basically all i need to write out
– John
Mar 6, 2013 at 15:38
• @John: Yes. But you still have to write down the details of step (1) and (2) and there was some claims that I assumed you know. Mar 6, 2013 at 15:39
• can you help me a bit more please regarding those details
– John
Mar 6, 2013 at 15:40
• @John: What specific details? (Left is really just to write things down. For example, for step (1) try and write down a diagonal matrix $D$ and the figure out what $e^D$ is using that definition of the exponential map. Mar 6, 2013 at 15:42
• @julien: I like the idea of step (3). I guess it seems more clear to me... Mar 6, 2013 at 16:05

The other answers seem to overcomplicate things, and there's no need for any discussion of diagonalization or triangulation. Let $$A$$ have eigenvalues $$\lambda_j$$ and eigenvectors $$v_j$$. Either set can be degenerate, it makes no difference. Using the Taylor series $$e^A \equiv I + A + \frac{1}{2!} A^2 + \ldots$$, you can quickly see that $$e^A v_j = e^{\lambda_j} v_j$$. Thus $$e^M$$ has eigenvalues $$e^{\lambda_j}$$ and eigenvectors $$v_j$$.

Now, the determinant of any matrix is the product of its eigenvalues (1), and its trace is the sum of its eigenvalues (2). Thus $$\mathrm{det} \left(e^A \right) \, \overset{\scriptscriptstyle{(1)}}{=} \, e^{\lambda_1}e^{\lambda_2}e^{\lambda_3}\ldots = e^{\lambda_1 + \lambda_2 + \lambda_3 + \, \ldots} \, \overset{\scriptscriptstyle{(2)}}{=} \, e^{\mathrm{tr}(A)},$$ which is your desired result.

If you want to also prove (1) and (2) to be completely happy, we can do that as follows [this part is a modified version of Ted Schifrin's answer at https://math.stackexchange.com/a/546167/1096883]. The characteristic polynomial of an $$n \times n$$ matrix $$A$$ is $$p(t) = \det(A-tI) = (-1)^n \big(t^n - (\text{tr} A) \,t^{n-1} + \ldots + (-1)^n \det A\big)\, ,$$ which you can see from the standard tedious way ("Laplace expansion") of calculating a determinant. But from the fundamental theorem of algebra we also have $$p(t) = (-1)^n(t-\lambda_1) \ldots (t-\lambda_n)$$, where the $$\lambda_j$$ are the eigenvalues of $$A$$. We then equate the coefficients of $$t^{n-1}$$ and $$t^0$$ (i.e. the constant term) between the two forms of $$p(t)$$, which yields $$\det A = \lambda_1 \lambda_2 \ldots \lambda_n$$ and $$\text{tr}A = \lambda_1 + \lambda_2 + \ldots + \lambda_n$$.

Here is another alternative, set as a series of exercises in Naive Lie Theory by John Stillwell.

Lemma 1: For any complex square matrix $$B$$ $$\det\left(B\right)=\displaystyle\sum_{j=1}^{n} (-1)^{j+1}b_{1j}\det\left(B_{1j}\right),$$ where $$B_{1j}$$ denotes the matrix obtained by removing the first row and $$j^{\text{th}}$$ column from $$B$$.

A sketch of the proof is given in an addendum. $$\qquad\square$$

Lemma 2: For any smooth (i.e. differentiable) path of complex square matrices $$B(t)$$, with $$B(0)=\mathbf{1}$$, $$\left.\frac{d}{dt}\right|_{t=0}\det\left(B(t)\right)=\text{Tr}\left(B'(0)\right)$$

Proof: Consider $$\left.\frac{d}{dt}\right|_{t=0}b_{1j}(t)\det\left(B_{1j}(t)\right)=b'_{1j}(0)\det\left(B_{1j}(0)\right)+b_{1j}(0)\left.\frac{d}{dt}\right|_{t=0}\det\left(B_{1j}(t)\right).$$ Since $$B(0)=\mathbf{1}$$ both $$b_{1j}(0)$$ and $$\det(B_{1j}(0))$$ are equal to $$0$$ when $$j\neq 1$$ and $$1$$ when $$j=1$$. Hence by Lemma 1 $$\left.\frac{d}{dt}\right|_{t=0}\det\left(B(t)\right)=b'_{11}(0)+\left.\frac{d}{dt}\right|_{t=0}\det\left(B_{11}(t)\right)$$ and, by induction on $$\det\left(B_{11}(t)\right)$$, $$\left.\frac{d}{dt}\right|_{t=0}\det\left(B(t)\right)=b'_{11}(0)+b'_{22}(0)+\cdots +b'_{nn}(0)=\text{Tr}\left(B'(0)\right).$$ $$\square$$

Lemma 3: Let $$B(t)=e^{tA}$$, for which $$B'(0)=A$$, then for the smooth function $$f(t)=\det(e^{tA})$$, with $$f(0)=1$$, $$f'(t)=f(t)\text{Tr}(A).$$

Proof: $$$$\label{eq1} \begin{split} f'(t) & = \lim_{h\rightarrow 0}\frac{1}{h}\left[\det\left(e^{(t+h)A}\right)-\det\left(e^{tA}\right)\right] \\ & = \lim_{h\rightarrow 0}\frac{1}{h}\left[\det\left(e^{tA}e^{hA}\right)-\det\left(e^{tA}\right)\right] \\ & = \det\left(e^{tA}\right)\lim_{h\rightarrow 0}\frac{1}{h}\left[\det\left(e^{hA}\right)-1\right] \\ & =\det\left(e^{tA}\right)\left.\frac{d}{dt}\right|_{t=0}\det\left(B(t)\right) \\ & =f(t)\text{Tr}(A).\end{split}$$$$ The second and third equalities follow from the commutative and product laws of the exponential and determinant map, respectively. The fourth equality follows from Lemma 2. $$\qquad\square$$

Lemma 4: $$\det\left(e^A\right)=e^{\text{Tr}(A)}.$$

Proof: If $$h(t)=e^{t\cdot\text{Tr}(A)}$$ then $$h'(t)=\text{Tr}(A)h(t).$$ Hence $$f(t)=f'(t)\frac{1}{\text{Tr}(A)}=f'(t)\frac{1}{h'(t)}e^{t\cdot\text{Tr}(A)}=g(t)e^{t\cdot\text{Tr}(A)}.$$ Evaluating the function $$g$$ shows $$g'(0)=\mathbf{0}\implies g(t)=\text{constant}$$ and since $$f(0)=1$$, this constant is also $$1$$ and we conclude the result by evaluating $$f$$ when $$t=1$$. $$\qquad\square$$

Addendum: Here is a sketch of the proof for Lemma 1. The determinant of a complex square matrix $$A=(a_{ij})$$ is a map $$A\mapsto \displaystyle\sum_{\sigma\in S_n}\text{sgn}(\sigma)a_{\small{1\sigma(1)}}\cdots a_{\small{n\sigma(n)}},\tag{*}$$ which uniquely satisfies the three axioms of multilinearity, anti-symmetry, and normalisation. We claim the map $$B\mapsto \displaystyle\sum_{j=1}^{n} (-1)^{j+1}b_{1j}\det\left(B_{1j}\right)$$ satisfies these three axioms and thus is equal to the determinant. Normalisation follows almost immediately from the definition and we skip the proof. Roughly, to prove anti-symmetry, show that switching any two columns of $$B$$ is equivalent to multiplying each permutation $$\sigma$$ by a $$2$$-cycle. Since this changes the sign of a permutation and the even and odd permutations divide $$S_n$$ evenly, the sign of every product flips and so the map is anti-symmetric. To prove multilinearity, write $$\overline{b_i}=a\overline{u}+b\overline{v}$$ for the $$i$$-th column, let $$U=\left(\overline{b_1},\cdots,\overline{b_{i-1}},\overline{u},,\overline{b_{i+1}},\cdots, \overline{b_n}\right)$$ and similarly for the matrix $$V$$. Then for all $$j\neq i$$ $$ab_{1j}\det(U_{1j})+bb_{1j}\det(V_{1j})= b_{1j}\det(B_{1j}),$$ relying mostly on $$(*)$$ and commutativity of products in $$\mathbb{C}$$. The case $$j=i$$ is even simpler to check and multilinearity follows.