Is $\operatorname{tr}|A|=\operatorname{tr}|A^\dagger|$? Is it true that $\operatorname{tr}|A|=\operatorname{tr}|A^\dagger|$ for any operator in a Hilbert space? I can prove this statement for normal operators such that $[A,A^\dagger]=0$. I want to know is there any proof or counter example for general case?
 A: Note: Any references to theorems etc. I will make in the following are with respect to the book "Methods of Modern Mathematical Physics. I: Functional Analysis" by Reed & Simon (1980).
If your Hilbert space $\mathcal H$ is separable and $A\in\mathcal B(\mathcal H)$, i.e. $A$ is a bounded operator on $\mathcal H$, your assertion is indeed true for the following reason. As you might know, on separable Hilbert spaces one defines the trace class to be
$$
\mathcal B^1(\mathcal H):=\lbrace A\in\mathcal B(\mathcal H)\,|\,\operatorname{tr}(\sqrt{A^\dagger A})<\infty\rbrace\,.
$$
Here one uses that on positive semi-definite operators, the trace is well-defined (independent of the chosen orthonormal basis), linear and takes values in $[0,\infty]$. Now if $A\in\mathcal B^1(\mathcal H)$ then $A$ is compact (Theorem VI.21) which is equivalent to admitting a singular value decomposition
$$
A=\sum_{n\in\mathbb N}\sigma_n(A)\langle \psi_n,\cdot\rangle\phi_n\tag{1}
$$
with unique non-negative numbers $(\sigma_n(A))_{n\in\mathbb N}$ and orthonormal systems $(\psi_n)_{n\in\mathbb N}$, $(\phi_n)_{n\in\mathbb N}$ in $\mathcal H$ where the sum converges in the operator norm (Theorem VI.17). Due to the above singular value decompoition, it is evident that every compact operator $A$ on $\mathcal H$ satisfies
$$
\operatorname{tr}(\sqrt{A^\dagger A})=\sum_{n\in\mathbb N}\sigma_n(A)\,,\tag{2}
$$
where the right-hand side (and thus both sides) might take the value $\infty$. This is enough preparation to prove the statement in question.

Proposition. Let $A\in\mathcal B(\mathcal H)$. Then $\operatorname{tr}(\sqrt{A^\dagger A})=\operatorname{tr}(\sqrt{A A^\dagger})$, where one side (and thus both sides) might take the value $\infty$.

Proof. We may assume that at either $\operatorname{tr}(\sqrt{A^\dagger A})$ or $\operatorname{tr}(\sqrt{A A^\dagger})$ (or both) are finite - otherwise we would obviously be done due to $\operatorname{tr}(\sqrt{A^\dagger A})=\infty=\operatorname{tr}(\sqrt{A A^\dagger})$. W.l.o.g. $\operatorname{tr}(\sqrt{A^\dagger A})<\infty$ so $A\in\mathcal B^1(\mathcal H)$ and $\operatorname{tr}(\sqrt{A^\dagger A})=\sum_{n\in\mathbb N}\sigma_n(A)$. By (1), obviously $A^\dagger$ is compact as well with $\sigma_n(A^\dagger)=(\sigma_n(A))^*=\sigma_n(A)$ and thus
$$
\operatorname{tr}(\sqrt{A^\dagger A})=\sum_{n\in\mathbb N}\sigma_n(A)=\sum_{n\in\mathbb N}\sigma_n(A^\dagger)=\operatorname{tr}(\sqrt{A A^\dagger})<\infty
$$
which concludes the proof. $\quad\square$
