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On page 16 of Prof. Jones' online von Neumann algebras notes(you can find the book here: von Neumann algebras notes

There is an exercise 3.3.9(iv) (labeled as a harder one),

Suppose $H$ is a separable Hilbert space, show that there is no nonzero linear map $tr: B(H)\rightarrow \mathbb{C}$ satisfying $~tr(ab)=tr(ba)$.

I can only figure out that if such $tr$ exists, it must be zero on any finite rank operator by using the right shift operator $S$, but how to process from finite dimensional operators to all operators in $B(H)$ by just using the property $tr(ab)=tr(ba)$, can any one give some hint?

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    for part ii, just use $Tp_k\rightarrow T$, but $tr(Tp_k)=0$, where $p_k$ is the orthognormal projection on the n-dimensional subspace; while for iii, my method only works if we assume tr(x^*x)>0 for all nonzero operators.2012-10-31

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The same idea that you use to show that $tr$ is zero on finite-rank operators, can be used to see that it is zero on any projection.

Consider a projection $q$ with infinite dimension and codimension. Consider an orthonormal basis $\{\xi_n\}$ such that $\{\xi_{2n}\}$ spans the range of $1-q$. Let $\{e_{kj}\}$ be the matrix units associated with our orthonormal basis, and let $ x=\sum_je_{j,2j}. $ Then $ x^*x=\sum_je_{2j,2j}=1-q,\ \ xx^*=\sum_je_{j,j}=1. $ So $tr(1)=tr(xx^*)=tr(x^*x)=tr(1-q)$. Then $tr(q)=0$. As $q$ was any projection with infinite dimension and co-dimension, we also have $tr(1-q)=0$. Thus $ tr(1)=tr(q+(1-q))=tr(q)+tr(1-q)=0+0=0. $ Now if $p$ is a projection with finite co-dimension, then $tr(1-p)=0$ (as $1-p$ is finite-rank), but then $tr(p)=0$ since $tr(1)=0$.

We have shown that $tr(p)=0$ for all projections in $B(H)$. It is well-known (see here for references) that every operator in $B(H)$ is a finite linear combination of projections. So $tr=0$.

(in part iii one can avoid this last argument: the fact that $tr$ is positive allows one to use Cauchy-Schwarz, and this together with $tr(1)=0$ can be used to see that $tr=0$: indeed, in that case for any $x$ $ |tr(x)|=|tr(1x)|\leq tr(1)^{1/2}tr(x^*x)^{1/2}=0. $ so $tr=0$)

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    No problem. I edited the answer at the end to clarify how to solve iii without using the result about the linear combinations of projections).2012-10-31