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If $A\colon H\to S$ is a bounded operator on a Hilbert space $H$, and $S\subset H$. It is known that $\operatorname{trace}(A)=\sum_{n} \langle Af_n,f_n\rangle$ for any orthonormal basis $\{f_{n}\}$. Is there a relation between $\operatorname{trace}(A)$, $\operatorname{rank}(A)$, and dimension of $\operatorname{range}(S)$?

Edit: What if $A$ is a composition of two orthogonal projections $A_{1}:H\to S_{1}$,$A_{2}:H\to S_{2}$, such that $A=A_{1}oA_{2}$, for $S_{1},S_{2}\subset H$. I need to show that $\operatorname{trace}(A)\leq \operatorname{rank}(A)\leq \dim(S_{2})$

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What is $\operatorname{range}(S)$? Why do you expect such a relation? –  Davide Giraudo Apr 30 '12 at 15:51
    
Also please clarify what you mean by rank; one possible dimension is that the rank is the dimension of the range, which makes your question seem redundant. –  Nate Eldredge Apr 30 '12 at 17:15
    
Of course, for most $A$ that trace series diverges... The so-called "trace class" is where it makes sense. –  GEdgar Apr 30 '12 at 17:17
    
Question: If $A=A_{1}oA_{2}$ as above, is it a projection in this case!! –  Nichole Apr 30 '12 at 21:06
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2 Answers

Note that the trace is sensitive to scalar multiplication, while the range isn't (i.e. $A$ and $2A$ have the same range but different trace). So there cannot be relation in general.

Now, if for example $A$ is a projection, then it is indeed true that the trace of $A$ agrees with the dimension of the range.

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One thing you can say is that if $A$ is a trace class operator it is a compact operator, so that $\text{Ran}(A)$ is $\sigma$-compact (i.e. the union of countably many compact sets).

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