Trace formula on finite-dimensional Hilbert space Let $\mathscr{H}$ be a finite-dimensional complex Hilbert space and let $A$ be a linear operator on $\mathscr{H}$. Suppose I want to evaluate $\operatorname{Tr}e^{-tA}$, where $t$ is a nonnegative real parameter. The book I was reading states that, because $\mathscr{H}$ is finite-dimensional, $e^{-tA}$ is an entire function of $t$ so the following equation must hold:
$$\operatorname{Tr}e^{-tA} = \lim_{n \to \infty}\operatorname{Tr}\bigg{(}1-\frac{t A}{n}\bigg{)}^{n}$$
At first, I thought this formula was a consequence of Trotter's formula, but it seem it is not. Could someone help me understand why does it hold? I believe this is probably quite simple, but I could not deduce it by myself.
 A: Suppose $A$ is $m\times m$ and write $A=VTV^*$ the Schur-Decomposition (the Jordan Form also works here, the point is that you get a triangular matrix with the eigenvalues of $A$ in the diagonal), with $T$ triangular. Then
$$
e^{-tA}=\sum_{k=0}^\infty \frac{(-1)^kt^kA^k}{k!}=\sum_{k=0}^\infty \frac{(-1)^kt^kVT^kV^*}{k!}=V\,\sum_{k=0}^\infty \frac{(-1)^kt^kT^k}{k!}\,V^*.
$$
As $T$ is triangular, the diagonal entries of $T^k$ are $T_{11}^k,\ldots,T_{mm}^k$. It follows that
$$
\operatorname{Tr}(e^{-tA})=\operatorname{Tr}\Big(V\,\sum_{k=0}^\infty \frac{(-1)^kt^kT^k}{k!}\,V^*\Big)
=\operatorname{Tr}\Big(\sum_{k=0}^\infty \frac{(-1)^kt^kT^k}{k!}\Big)
=\sum_{k=0}^\infty \frac{(-1)^kt^k\operatorname{Tr}(T^k)}{k!},
$$
where the exchange between the trace and the series is due to the trace being continuous. Now,
\begin{align}
\operatorname{Tr}(e^{-tA})&=\sum_{k=0}^\infty \frac{(-1)^kt^k\sum_{j=1}^mT_{jj}^k}{k!}=\sum_{j=1}^m\sum_{k=0}^\infty \frac{(-1)^kt^kT_{jj}^k}{k!}=\sum_{j=1}^me^{-t\,T_{jj}}
=\sum_{j=1}^m\lim_{n\to\infty}\Big(1-\frac{t\,T_{jj}}n\Big)^n\\[0.3cm]
&=\lim_{n\to\infty}\sum_{j=1}^m\Big(1-\frac{t\,T_{jj}}n\Big)^n=\lim_{n\to\infty}\operatorname{Tr}\Big(1-\frac{t\,T}n\Big)^n=\lim_{n\to\infty}\operatorname{Tr}V\Big(1-\frac{t\,T}n\Big)^nV^*\\[0.3cm]
&=\lim_{n\to\infty}\operatorname{Tr}\Big(1-\frac{t\,A}n\Big)^n
\end{align}
A: Let $\lambda_1, \ldots, \lambda_n$ be the eigenvalues of $A$, with multiplicities.
According to the spectral mapping theorem (with multiplicities), the eigenvalues of $e^{-tA}$ are exactly
$$e^{-t\lambda_1}, \ldots, e^{-t\lambda_n}$$
so since the trace is sum of eigenvalues we get
$$\operatorname{Tr} e^{-tA} = e^{-t\lambda_1}+ \cdots+ e^{-t\lambda_n}.$$
On the other hand, for each $k\in\Bbb{N}$, the eigenvalues of $I - \frac{tA}k$ are exactly
$$1 - \frac{t\lambda_1}k, \ldots, 1 - \frac{t\lambda_n}k$$
so we get
$$\operatorname{Tr} \left(I - \frac{tA}k\right) = \left(1 - \frac{t\lambda_1}k\right)+ \cdots + \left(1 - \frac{t\lambda_n}k\right).$$
Therefore
$$\lim_{k\to\infty}\operatorname{Tr} \left(I - \frac{tA}k\right) = \lim_{k\to\infty}\left(1 - \frac{t\lambda_1}k\right)+ \cdots + \lim_{k\to\infty}\left(1 - \frac{t\lambda_n}k\right) = e^{-t\lambda_1}+ \cdots+ e^{-t\lambda_n} = \operatorname{Tr} e^{-tA}$$
by the formula in $\Bbb{C}$.
