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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 ?

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    What is $s$ in your notation?2012-06-13

2 Answers 2

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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.

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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.

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    's' here is a real number s >0 2012-06-13