# Operator from 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 ?

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What is $s$ in your notation? –  Willie Wong Jun 13 '12 at 16:11

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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what is $I^{-1}(g)$ ?? is the inverse function of $g(s)$ thanks –  Jose Garcia Jun 13 '12 at 16:02
$I$ is the mapping defined (as in the displayed math line) by associating to the compact operator $T$ the function $g(s)$. So by definition $I^{-1}$ would map $g(s)$ to the operator $T$. –  Willie Wong Jun 13 '12 at 16:06
Willie Wong, I'm not sure abut I remember that read it somwhere. Is it true there exist such a pairing between this classes? –  Norbert Jun 13 '12 at 16:09
@Norbert: it is true (if I remember right) if $s$ is interpreted as an element of $S^q(H)$. I am however not sure whether that is the OP's intent. –  Willie Wong Jun 13 '12 at 16:11
's' here is a real number $s >0$ –  Jose Garcia Jun 13 '12 at 18:43