# Convention for absolute value of operator in the definition of trace-class operators

Let $$H$$ be a separable complex Hilbert space and $$T\in\mathcal{B}(H)$$. Let $$\{v_i\}_{i\in\mathbb{N}}$$ be an orthonormal basis for $$H$$. Then $$T$$ is of trace class if $$\sum_{i\in\mathbb{N}}\langle |T|v_i,v_i\rangle_H<\infty,$$ where $$|T|$$ is defined to be $$\sqrt{T^*T}$$.

I have a couple of very basic questions relating to this definition.

Question 1: Would this definition change if we instead defined $$|T|$$ to be $$\sqrt{TT^*}$$?

For a trace class operator, its trace is defined to be $$\text{tr}(T):=\sum_{i\in\mathbb{N}}\langle Tv_i,v_i\rangle_H.$$

Question 2: What is the motivation for defining the notion of trace class using the absolute value, instead of just saying that $$T$$ is trace class if its trace is finite, i.e. if $$\sum_{i\in\mathbb{N}}\langle Tv_i,v_i\rangle_H<\infty?$$

Q1: I'll assume you already know that the definition does not depend on the choice of the ONB. Let $$T=U|T|$$ be the polar decomposition of $$T$$, where $$U$$ is a partial isometry with initial space $$\overline{\mathrm{ran}}(T^\ast)$$ and range $$\overline{\mathrm{ran}}(T)$$. Then $$TT^\ast=U|T|^2U^\ast$$, hence $$|T^\ast|=U|T|U^\ast$$.
Now let $$(v_j)$$ be an ONB of $$\overline{\mathrm{ran}}(T)$$ and $$(w_k)$$ an ONB of $$\overline{\mathrm{ran}}(T)^\perp$$. Then $$(U^\ast v_j)$$ is an ONB of $$\overline{\mathrm{ran}}(T^\ast)=(\ker T)^\perp=(\ker |T|)^\perp$$ and $$U^\ast w_k=0$$. Thus $$\mathrm{tr}(|T^\ast|)=\sum_j \langle |T^\ast|v_j,v_j\rangle+\sum_k \langle |T^\ast|w_k,w_k\rangle=\sum_j \langle |T|U^\ast v_j,U^\ast v_j\rangle+\sum_k\langle |T|U^\ast w_k,U^\ast w_k\rangle=\mathrm{tr}(|T|).$$ In particular, the left side is finite if and only if the right side is. More generally, the same argument shows that $$\mathrm{tr}(f(T^\ast T))=\mathrm{tr}(f(TT^\ast))$$ for any postive bounded Borel function $$f$$.