Trace norm of adjoint operator For a positive operator $A$ on a Hilbert space $H$, the trace of $A$ is defined by
$$Tr (A)=\sum_i\langle Ae_i,e_i\rangle$$
where $\{e_i\}$ is any orthonormal basis of $H$.
For a general bounded operator $A$ on $H$, $A$ is said to be of trace class if the trace of $|A|:=\sqrt{A^*A}$ is finite. We can also define the trace norm of a trace class oeprator to be $||A||_1:=Tr(|A|)$.
I am trying to prove (or disprove) that $||A^*||_1 = ||A||_1$. I am stuck because there seems to be no easy way to relate the operators $|A|$ and $|A^*|$. Any help will be appreciated.
It is so much easier to prove that the Frobenius norm $||A^*||_2 = ||A||_2$. The inner product form plus the absolute value makes the trace norm quite nasty!
 A: As per my comment, being trace class implies that $A$ is compact. Then $|A|$ is compact, so by the spectral theorem $|A|$ has a decreasing sequence of positive eigenvalues $s_n(A)$ tending to $0$. If we let $\{e_i\}$ be an orthonormal basis of eigenvectors for $|A|$, then $|A| = \sum_i s_i(A) e_i\otimes e_i^\ast$  and also
$$\|A\|_1 = \sum_i \langle|A|e_i,e_i\rangle = \sum_i s_n(A)$$
Moreover by the polar decomposition we can write
$$A = U|A|$$
where $U$ is a partial isometry. If we let $f_n = Ue_n$ then we have
$$A = \sum_is_n(A)f_i \otimes e_i^\ast$$
The following is a lemma 1.4 from Ken Davidson's book "Nest Algebras" (which is free to download on his website).
Claim: Fix $n \ge 1$. Then
$$s_n(A)= \inf\{\|A - F\|: rank F \le n-1\}$$
Proof: Let $F_n = \sum_{k=1}^{n-1} s_k f_k \otimes e_k^\ast$ and then $A - F_n = \sum_{k=n}^\infty s_k f_k \otimes e_k^\ast$. Thus
$s_n(A)  = \|A - F_n\|$ and the $\ge$ inequality follows.
Conversely if $rank F \le n-1$ then choose a unit vector $x \in ker F \cap span \{e_1,\dots, e_n\}$ and then
$$\|A - F\| \ge \|(A - F)x\| = \|Ax\| 
    = \left\|\sum_{k=1}^n s_k(A)(x, e_k)f_k\right\| 
     \ge s_n(A) \sum_{k=1}^n |(x,e_k)|^2 
      = s_n(A)$$
Which proves the claim.
Finally, for any finite rank $F$ we have that $dim(Range(F)) = dim(Range(F^\ast))$ and
$$\|A - F\| = \|(A-  F)^\ast\| = \|A^\ast - F^\ast\|$$
And hence it follows that
$$s_n(A) = \inf\{\|A - F\|: rank F \le n-1\}=\inf\{\|A^\ast - F^\ast\|: rank F \le n-1\} = s_n(A^\ast)$$
And so
$$\|A\|_1 = \sum_i s_i(A) = \sum_i s_i(A^\ast) = \|A^\ast\|_1$$
