I don't understand how to define the trace of a matrix with values in operators. This occurred in the following situation: Suppose that $H$ is an Hilbert space and $K$ is the algebra of compact operators on $H$. We proved in the last lecture that $$K_0(K(H))=\mathbb{Z}$$
In the proof we have seen that every $x\in K_0(K(H))$ has the form $x=[e]-[1_n]$, where $e\in M_N(K)$ is an idempotent with $\epsilon_K(e) \sim 1_n $ and such that $e$ is of finite rank. The map $\epsilon_K$ is defined to be the induced mapping from the mapping $\epsilon$, given by : $ 0\to K \to \widetilde{K} \xrightarrow{\epsilon} \mathbb{Z}$. Now we defined the mapping: $$tr: K_0(K(H)) \to \mathbb{Z}$$ $$ x\mapsto tr(e-1_n)$$
as the simple trace function.
What is the trace of an operator matrix? The whole subject of $K$-Theory is new to me, so maybe I got it totally wrong...
Any explanations are welcome. Thanks.
I'm not in position to explain any $K$-theory. But it is not hard to see that $M_n(K(H))\simeq K(\bigoplus_1^nH)$ canonically. So you can see matrices of compact operators as compact operators on a bigger Hilbert space, and you can calculate its trace if it is trace-class.
More concretely, if you have a matrix of operators $T=(T_{kj})_{k,j}\in M_n(B(H))$, its canonical trace (provided that $T$ is trace-class) will be $$ \mbox{Tr}(T)=\sum_{j=1}^n\mbox{Tr}(T_{jj}), $$ where the trace on the left is the one in $B(\oplus_1^n H)$ and the one on the right is that of $B(H)$.