$\det (e^A) = e^{\text{Tr}(A)}$ without Jordan canonical form, Schur decomposition?

Question

Let $V$ be a finite-dimensional vector space over $\mathbb{C}$. How do I show that $$\det\,e^A = e^{\text{Tr}\,A}$$for any $A \in \text{End}\,V$ without invoking Jordan canonical form or Schur decomposition?

Answer 1

Since so many proofs have been supplied that do not invoke the Schur decomposition nor Jordan canonical form, I see no harm in including two proofs that invoke them.

Proof 1: Schur decomposition.

The Schur decomposition states that if $A$ is a square matrix over an algebraically complete field (or if the characteristic polynomial for $A$ splits over the field we are in), then $A$ is similar to an upper triangular matrix. That is, there exists invertible $P$ such that $P^{-1}AP = T$, and $T$ is upper triangular.

Specifically for the complex numbers, we can say even more, in that we can choose $P$ to be unitary, i.e. $P^* P = I$. But we do not need that part here.

So we assume we have $P^{-1}AP = T$, which we can also write as $A = PTP^{-1}$. Two things to note:

1. $\text{Tr}(A) = \text{Tr}(T)$. Hence $e^{\text{Tr}(A)} = e^{\text{Tr}(T)}$.
2. For every nonnegative integer $k$, we have $A^n = PT^kP^{-1}$. Put all of these powers into the direction of the exponential, then factor $P$ out of one side and $P^{-1}$ out of the other, and we obtain $e^A = Pe^T P^{-1}$. It then follows that$$\det(e^A) = \det(P)\det(E^T)\det(P^{-1}) = \det(E^T).$$

So now, all we need to do is to prove the theorem for matrices that are already upper triangular.

Let$$T = \begin{pmatrix}\lambda_1&&&\\ &\lambda_2&*&\\ &0&\ddots&\\ &&&\lambda_n\end{pmatrix}.$$Then$$\text{Tr}(T)=\lambda_1+\dots+\lambda_n,\text{ }\det(T)=\lambda_1\lambda_2\dots\lambda_n.$$As always, the asterisk means "there are entries here that could be anything, which we are not tracking carefully because it does not really matter." By direct computation,$$T^k=\begin{pmatrix}\lambda_1^k&&&\\ &\lambda_2^k&*&\\ &0&\ddots&\\ &&&\lambda_n^k\end{pmatrix}.$$We compute that$$e^T=\begin{pmatrix}e^{\lambda_1}&&&\\ &e^{\lambda_2}&*&\\ &0&\ddots&\\ &&&e^{\lambda_n}\end{pmatrix}.$$From that,$$\det(e^T)=e^{\lambda_1}e^{\lambda_2}\dots e^{\lambda_n}=e^{\lambda_1+\cdots+\lambda_n}=e^{\mathrm{Tr}(T)},$$as desired.

Proof 2: Jordan canonical form.

The Jordan canonical form of $A$ is$$A = \begin{pmatrix} J_1 &&& \\ &J_2&& \\ &&\ddots& \\ &&&J_k\end{pmatrix},$$where we have Jordan blocks$$J_i = \begin{pmatrix} \lambda_i & 1 & & 0 \\ &\ddots & \ddots & \\ &&\ddots&1 \\ 0 &&& \lambda_i\end{pmatrix}.$$We have$$J_i = \lambda_i I_{n_i \times n} + \begin{pmatrix} 0 & 1 && 0 \\ &\ddots&\ddots& \\ &&\ddots&1 \\ 0 & & & 0 \end{pmatrix} = A_i + B_i \implies A_iB_i = B_iA_i.$$Set$$D = \begin{pmatrix} A_1 &&& \\ & \ddots && \\ &&\ddots& \\ &&&A_k\end{pmatrix}, \text{ }\, B = \begin{pmatrix} B_1 &&& \\ & \ddots && \\ &&\ddots& \\ &&&B_k\end{pmatrix}.$$Then $$A = D + B,\text{ }DB = BD \implies e^A = e^{D+B} = e^De^B \implies \det(E^A) = \det(E^D)\det(E^B).$$We have$$e^D = \begin{pmatrix} e^{A_1} &&& \\ & \ddots && \\ &&\ddots& \\ &&&e^{A_k}\end{pmatrix}, \text{ }\,\det(e^B) = 1.$$So$$\det(e^D) = e^{n_1\lambda_1 + \dots + n_k\lambda_k} \implies \det(e^A) = \det(e^D) = e^{n_1\lambda_1 + \dots + n_k\lambda_k} = e^{\text{Tr}(A)},$$as desired.

Answer 2

Let $f : \mathbb{R} \rightarrow \mathbb{C}$ be defined by $f(t) = \det e^{tA}$. This satisfies the ODE $\dot{f} = f \ \mathrm{Tr} A$, with the initial value $1$. Solving this and setting $t=1$ gives the RHS.

Added in response to comment: The map $\det : \mathrm{End}(V) \rightarrow \mathbb{C}$ is differentiable at the identity with derivative $\mathrm{Tr}$; this is a straightforward computation. We then have

$$\dot{f}(t_0) = \frac{\mathrm{d}}{\mathrm{d}t}\bigg|_{t_0} \det e^{(t-t_0)A} \det e^{t_0A}=\mathrm{Tr}\bigg(\frac{\mathrm{d}}{\mathrm{d}t}\bigg|_{t_0}(t-t_0)A\bigg) \det e^{t_0A}=f(t_0) \mathrm{Tr} A$$

for all $t_0$, using the chain rule.

If one were being super rigorous about the ODE step, one could define $g:\mathbb{R} \rightarrow \mathbb{C}$ by $g(t)=e^{t\mathrm{Tr}A}$ and consider $(f/g)^\dot{}$.

Answer 3

It is a consequence of Jacobi formula. You can see the proof at: https://en.wikipedia.org/wiki/Jacobi%27s_formula.

Answer 4

We know that

1. $Tr(A) = \sum_{\forall k} \lambda_k(A)$
2. $\lambda_k(e^A) = e^{\lambda_k(A)}, \forall k$
3. $\det(A) = \prod_{\forall k} \lambda_k(A)$

These together with "logarithm laws": $e^{\sum a} = \prod e^{a}$ can show what we want.

Answer 5

You can derive this from the Lie product formula. Write $A = B + C + D$ where $B$ and $C$ are is the strictly upper triangular and strictly lower triangular parts respectively of $A$, and $D$ is the diagonal. Thus $\text{Tr}(A) = \text{Tr}(D)$. For scalars $t$, $e^{tB}$ and $e^{tC}$ are upper and lower triangular with $1$'s on the diagonal, so $\det e^{tB} = \det e^{tC} = 1$, while $$\det e^{tD} = \prod_i e^{tD_{ii}} = \exp\left(\sum_i tD_{ii}\right) = e^{\text{Tr}(tD) }$$

Now using the Lie product formula $$\eqalign{\det \exp(B+D) &= \lim_{N \to \infty} \left((\det \exp(B/N)) (\det \exp(D/N))\right)^N \cr &= \lim_{N \to \infty} (1 \cdot e^{\text{Tr}(D)/N})^N = e^{\text{Tr}(D)} }$$ and similarly $$ \det \exp(A) = \det \exp(B+D+C) = \lim_{N \to \infty} \left((\det \exp(B+D/N)) (\det \exp(C/N))\right)^N = e^{\text{Tr}(D)} $$

Answer 6

We can show that the function $f(A)=det(e^A)-e^{tr(A)}$ doesn't depend on any of the variables of $A$. We can do this as follows. We first show that for each matrix $A$, $f(A+\lambda E_{ij})=f(A)$ for $i\neq j$. Then $f(A+\lambda E_{ij})=det(e^{A+\lambda E_{ij}})-e^{tr(A+\lambda E_{ij})}=det(e^{A+\lambda E_{ij}})-e^{tr(A)}$. We compute $det(e^{A+\lambda E_{ij}})=det(e^Ae^{\lambda E_{ij}})=det(e^A)det(1+\lambda E_{ij})=det(e^A)$, where the first equality metrits explanation. We have $e^{A+\lambda E_{ij}}=e^{A}e^{\lambda E_{ij}}\prod_i e^{p_i([A, E_{ij}])}$ where $p_i$ involves taking taking commutators with $\lambda E_{ij}$ and $A$. We only need to show thus that $det(e^{\prod_i p_i([A, E_{ij}]})=1$. But $tr([A, E_{ij}])=0$, which is the condition for its exponential to be in $SL(n)$, by basic lie-theory, and $e^{p_i([A, E_{ij}])}$ consists of only sums of commutators, we have the desired result. Thus $f(A)$ only depends on the diagonal elements. Thus we only need to show that $det(e^A)=e^{tr(A)}$ for diagonal $A$. This trival though, since $e^A$ just the exponential of the diagonal entries, and so we are reduced to the sum formula.

Answer 7

We may assume that $A\in M_n(\mathbb C)$. Recall that if $\lambda_1,\lambda_2,\ldots,\lambda_n$ are the eigenvalues of $A$ and $p$ is a polynomial, then $\det(p(A))=\prod_ip(\lambda_i)$. (This is a direct consequence of the fact that determinant is multiplicative. No diagonalisation or triangulation is involved. See theorem 1 in this answer for instance.)

By Cayley-Hamilton theorem, for each $m$ the partial sum $\sum_{k=0}^m\frac{1}{k!}A^k$ lies inside the matrix subspace $W=\operatorname{span}\{I,A,A^2,\ldots,A^n\}$. Hence $e^A$ also lies inside $W$ because $W$ is closed. That is, $e^A=p(A)$ for some polynomial $p$. However, for each (possibly non-semisimple) eigenvalue $\lambda$ of $A$, if $v$ is a corresponding eigenvector, then $p(\lambda)v=p(A)v=e^Av=e^\lambda v$. Therefore $p(\lambda)$ is necessarily equal to $e^\lambda$. Hence $$ \det(e^A)=\det(p(A))=\prod_ip(\lambda_i)=\prod_ie^{\lambda_i}=e^{\sum_i\lambda_i}=e^{\operatorname{tr}(A)}. $$

Answer 8

So many good answers, I'll take my shot as well. You can use the already mentioned fact the the gradient of the determinant at the identity matrix is the trace, together with $ e^{A}=\lim_{n\to \infty}(I+\frac{1}{n}A)^n $. Then, from standard calculus, $$ \det e^A = \lim \det (I+\frac{1}{n}A)^n = \lim (1 +\frac{1}{n}Tr\; A + o(\frac{1}{n}))^n = e^{Tr\; A}. $$