let 's assume we have a self-adjoint operator whose trace is known
$ \mathrm{Tr}(sT)= g(s) $ for a known function $ g(s) $.
My quesiton is, can we recover the operator simply by knowing the trace ?
let 's assume we have a self-adjoint operator whose trace is known
$ \mathrm{Tr}(sT)= g(s) $ for a known function $ g(s) $.
My quesiton is, can we recover the operator simply by knowing the trace ?
No, but we can recover quite a bit of information if we know the eigenvalues.
Any operator $F\colon V\to V$ acting on a vector space $V$, if it is nonsingular, has $n=\mathrm{dim}(V)$ vectors $\vec{v}_{1},\dots,\vec{v}_{n}\in V$ which are "stretched" by $F$: $ F(\vec{v}_{i})=\lambda_{i}\vec{v}_{i} $ where $\lambda_{i}$ are scalars.
We have $\mathrm{tr}(F)=\lambda_{1}+\dots+\lambda_{n}$ and $\det(F)=\lambda_{1}(\dots)\lambda_{n}$.
How can we reconstruct the operator? Well, we need the eigenvectors and eigenvalues. Eigenvalues alone don't cut it. This is the spectral theorem for finite dimensional vector spaces.
The question will be much more meaningful if $s$ will denote an operator.
Trace of operator is well defined for compact operators. Hence I'll assume that $T\in S^p(H)$ for some $p\in[1,+\infty]$. By Schatten- Von-Neumann theorem there exist isometric isomorphism $ I:S^p(H)\to S^q(H)^*: T\mapsto (s\mapsto\mathrm{Tr}(sT)) $ Hence, your function $g$ is a an element of $S^q(H)^*$. The desired operator is $T=I^{-1}(g)$. Well this is not constructive proof, but it shows that such a method does exist.
If $s$ is just a positive constant, then you know very little about $T$, and your function $g$ is necessary linear. Indeed $ g(s)=\mathrm{Tr}(sT)=s\mathrm{Tr}(T). $ The only thing that you get from this function is its slope, i.e. $\mathrm{Tr}(T)$. Knowledge of trace of course is not sufficient for reconstructing $T$. See explanation in Alex Nelson's answer.