Let $H$ be a complex, separable Hilbert space and let $\mathcal{B}(H)$ denote the algebra of linear bounded operators on $H$. If $K \in \mathcal{B}(H)$ is a compact operator, there exist two orthonormal sequences $(e_n)$ and $(f_n)$ in $H$ and a $c_0$ sequence $(s_n)$ such that $$ T = \sum_n s_n e_n\otimes f_n.$$ We say that $T$ is a trace class operator if $(s_n) \in \ell_1$ and that $T$ is a Hilbert-Schmidt operator if $(s_n) \in \ell_2.$ Moreover, if an operator is Hilbert-Schmidt we have that $$ \sum_n |s_n|^2 = ||mat(T)||_{\ell_2(\mathbb{N}\times\mathbb{N})},$$ where, given an orthonormal basis $(g_n)$ $$mat(T) = (\langle Tg_i,g_j\rangle)_{i,j \geq 0},$$ is the matrix of $T$ with respect to this basis. This quantity is independent of the chosen basis.
My question is: Does this matrix norm characterization also work for trace class operators? Id est, an operator $T$ is trace class if and only if its matrix $mat(T)$ belongs to $\ell_1(\mathbb{N}\times\mathbb{N})$? I think it is false, and probably we have that every operator $T$ such that $mat(T)$ belongs to $\ell_1(\mathbb{N}\times\mathbb{N})$ is trace class, but I think the contrary does not hold. Anyone can help me?
Thank you very much!
Best Answer
In the space $\ell^2(\mathbb N\backslash\{0\})$ consider the operator $T$ defined by the symmetric matrix $A = (a_{ij})$, where $a_{ii} = 1/i^2$, $a_{ij}=0$ for $|i-j|>1$ and $a_{i+1,i} = 1/i$. This operator is trace class but $mat(T)\notin\ell^1(\mathbb N^2)$.