Can an operator have Exp(z) as its characteristic "polynomial"? Let $\mathcal{H}$ be a Hilbert space, and let $T: \mathcal{H} \rightarrow \mathcal{H}$ be a trace-class operator.  Define
$$ f_T(z) = \sum_{i=0}^\infty \mbox{Tr}(\wedge^k T) \cdot z^k, $$
the ordinary generating function for traces of exterior powers of $T$.  Expressed another way,
$$ f_T(z) = \mbox{Det}(I + zT) $$
where $\mbox{Det}$ is the Fredholm determinant.  This function is entire, and can be considered a generalization of the characteristic polynomial.
I am wondering if 
$$f_T(z) = e^z $$
for some natural choice of $T$ on some nice incarnation of Hilbert space.
The motivation for this problem comes from representation theory, where finite minors of such a $T$ provide a formula for dimensions of irreps of $S_n$, the symmetric group on $n$ letters.
--edit--
In order to provide slightly more background on the sort of application I have in mind, here is an attractive identity which might pique your interest:
If $\beta : \mathbb{C} \rightarrow \mathbb{C}$ is any function, define $$\gamma(z) = \frac{1}{\Gamma(z+1)}, \hspace{.4in} \delta(z) = \frac{\beta(z)}{\Gamma(z+1)}$$
Now
$$
\left|
\begin{pmatrix} 
\gamma(3) & \gamma(4) & \gamma(5) \\\\
\gamma(-1) & \gamma(0) & \gamma(1) \\\\
\gamma(-2) & \gamma(-1) & \gamma(0)
\end{pmatrix}
\begin{pmatrix} 
\delta(3) & \delta(4) & \delta(5)\\\\
\delta(-1) & \delta(0) & \delta(1)\\\\
\delta(-2) & \delta(-1) & \delta(0)
\end{pmatrix}
\right|
+ $$
$$
\left|
\begin{pmatrix} 
\gamma(2) & \gamma(3) & \gamma(4) \\\\
\gamma(0) & \gamma(1) & \gamma(2) \\\\
\gamma(-2) & \gamma(-1) & \gamma(0)
\end{pmatrix}
\begin{pmatrix} 
\delta(2) & \delta(3) & \delta(4)\\\\
\delta(0) & \delta(1) & \delta(2)\\\\
\delta(-2) & \delta(-1) & \delta(0)
\end{pmatrix}
\right|
+ $$
$$
\left|
\begin{pmatrix} 
\gamma(1) & \gamma(2) & \gamma(3) \\\\
\gamma(0) & \gamma(1) & \gamma(2) \\\\
\gamma(-1) & \gamma(0) & \gamma(1)
\end{pmatrix}
\begin{pmatrix} 
\delta(1) & \delta(2) & \delta(3)\\\\
\delta(0) & \delta(1) & \delta(2)\\\\
\delta(-1) & \delta(0) & \delta(1)
\end{pmatrix}
\right| = \frac{\delta(1)^3}{3!}
$$
 A: This is not an answer, but just a sketch of a possible answer; I think the answer is no.
For $T$ as desired, consider $Det(I-z^2T^2)=Det(I-zT)\cdot Det(I+zT)=e^{-z}\cdot e^z\equiv 1$. Considering the Taylor expansion of this function, one can conclude that $Trace(T^{2n})=0$ for any natural $n$. I think this yields $T^2=0$, and then the claim follows.
Actually, I am not sure if such determinant may have positive exponential type - this could be another approach.
A: Here's a proof that $\exp(z)$ is not a characteristic function using the product expansion for the determinant, which is essentially equivalent to Lidskii's theorem stating that the trace of a trace class operator is the sum of its eigenvalues.

If $T$ is a trace class operator on a Hilbert space $\mathcal{H}$ with eigenvalues $\lambda_n$ (counted up to algebraic multiplicity), then $\sum_n\vert\lambda_n\vert < \infty$ and
  $$
{\rm det}(1+zT)\equiv\sum_{k=0}^\infty{\rm Tr}(\Lambda^kT)z^k=\prod_n(1+z\lambda_n)
$$
  for all $z\in\mathbb{C}$.

See Trace Ideals and Their Applications, Theorem 3.7. Actually, they use this result to prove Lidskii's theorem, that ${\rm Tr}(T)=\sum_n\lambda_n$.
Then, ${\rm det}(1+zT)$ cannot be equal to $\exp(z)$ everywhere. This is because $\exp$ has no zeros, so $\lambda_n$ would have to be all zero, giving ${\rm det}(1+zT)=1$.
