Infinite dimensional trace vs finite dimensional trace

For a Hilbert space $H$, and a linear operator $T:H\to H$, I've seen the trace of $T$ defined as $\sum_k<Te_k,e_k>$ (where $\{e_k\}$ is any orthonormal basis) provided this is finite.

How does this generalize the finite dimensional case? I see how this would generalize for self-adjoint operators (or normal on complex hilbert spaces) since these operators are unitarily diagonalizable. But what about a general linear operator on a finite dimensional space?

• Can you give the definition of the trace of a general linear operator on a finite dimensional space you want to use? – Dirk May 10 '17 at 9:37
• You might be interested in the following: en.wikipedia.org/wiki/Hilbert%E2%80%93Schmidt_operator Also I think your question is how to generalize it to the infinite-dimensional setting, it says finite for the moment. – Mathematician 42 May 10 '17 at 9:37
• No need for diagonalisability. Consider the matrix of $T$ with respect to the basis $\langle e_1, e_2, \dotsc, e_{\dim H}\rangle$. – Daniel Fischer May 10 '17 at 9:38
• @Bemte sum of complex eigenvalues – Tom Chalmer May 10 '17 at 9:38
• @DanielFischer Ah, ok I understand. Thank you. – Tom Chalmer May 10 '17 at 9:40