# Absolute convergence of trace of trace class operator

Wikipedia states that if a bounded operator $A$ on a separable Hilbert space satisfies $\operatorname{Tr}|A| = \sum \langle |A|e_i,e_i\rangle<\infty$, then also $\operatorname{Tr}(A)$ converges absolutely. I wonder why.

To show this, I wanted to use $|\langle Ax,x\rangle| \leq \langle |A|x,x\rangle$ but this is not always true: Is $|\langle Ax,x\rangle| \leq \langle |A|x,x\rangle$ in a complex Hilbert space?

I know that: the trace of a positive operator does not depend on the basis. Every bounded operator is the difference of positive operators. The product of two Hilbert-Schmidt operators has absolutely convergent trace.

Option 1 (Spectral theorem) Since $\vert A\vert$ is trace-class, it is compact and hence also $A=U\vert A\vert$ (for some unitary operator $U$) is compact. Assume for now that $A$ is also self-adjoint, then $H$ admits a basis consisting of eigenvectors $e_1,e_2,\dots$ of $A$ with corresponding eigenvalues $\lambda_1,\lambda_2,\dots\in\mathbb{R}$. Then $\vert A\vert$ is diagonalizable with eigenvalues $\vert \lambda_1\vert,\vert\lambda_2\vert,\dots$ and the statement that it has finite trace is equivalent to $$\sum_i \vert \lambda_i\vert <\infty$$ Consequently $\sum_i\vert\langle A e_i,e_i\rangle\vert=\sum_i \vert\lambda_i\vert<\infty$ and the series in the definition of $\mathrm{tr}(A)$ converges absolutely. For arbitrary $A$, let $A_0=(A+A^*)/2$ and $A_1=(A-A^*)/2i$, then $A=A_0+iA_1$ and $A_0$ and $A_1$ are self-adjoint and you can use the previous result.
Option 2 Look into any textbook that deals with traceclass operators, for example Pedersen, Analysis Now. There they prove a result of the form $\vert\mathrm{tr}(ST)\vert\le \Vert S\Vert \mathrm{tr}{\vert T\vert}$ (Lemma 3.4.10 in Pedersens book) which can be applies to $A=U\vert A\vert=:ST$ and the proof should also reveal that the series converges absolutely.
This follows from polar decomposition: There exists $U$ with $A=U\cdot|A|$. If $|A|$ has finite trace then $\sqrt{|A|}$ is Hilbert-Schmidt (by definition of HS), so $A = U\sqrt{|A|} \cdot \sqrt{|A|}$ is the product of two Hilbert-Schmidt operators.