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Show that finite rank operators are $\|\cdot \|_1$ dense in $F_1$

Here $F_1$ denote the trace class of Hilbert space $H$ . $A \in F_1$ iff tr$(|A|)\lt +\infty$ . $\| A \|_1:= $tr$(|A|)$.

We know that finite rank operator are dense in $F_1$ in operator norm . To prove the density with trace class norm , we need to find $\{A_n \}$ such that $tr|A-A_n| \to 0$ . In Michael Reed's functional analysis Page $209$, it said this is a corollary of the following theorem .

Every $A\in F_1$ is compact . A compact operator $A$ is in $F_1$ if and only if $\sum \lambda_n \lt \infty$ where $\{\lambda_n\}_{n=1}^{\infty}$ are the singular values of $A$ .

In the proof here I see that $$A_N =\sum_{n=1}^N \lambda_n(A)\langle \cdot,e_n\rangle f_n$$

Norm converges to $A$ . If $|A-A_N|$ are positive , then $$\sum_1^{\infty} \langle(|A-A_N|)e_n,e_n \rangle=\sum_{N+1}^{\infty}\lambda_n(A) \langle f_n, e_n \rangle \to 0$$ However , If $A-A_N$ not positive , we can not have $$tr(A-A_N)=\sum \langle (A-A_N)e_n,e_n \rangle$$ I'm stuck in proving this , any hint or solution would be very appreciate .

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  • With $A_n = \sum_{n=1}^N \langle .,e_n\rangle Ae_n$ and $e_n$ the largest eigenvector of $(A^A)^{1/2}$ then $((A-A_N)^(A-A_N))^{1/2}$ is $(A^A)^{1/2}$ with its largest eigenvalues set to $0$. Try with a matrix $A = U S V $, replacing $A$ by $A-<.,v_1>Av_1$ is the same as setting the first entry of $S$ to $0$, and it is compatible with $A\to A^ A\to (A^A)^{1/2}$. Equivalently $v_1$ maximizes $\frac{|Ax|}{|x|}=\frac{|(A^A)^{1/2}x|}{|x|}$ – reuns Jan 15 '20 at 07:16

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