Trace norm of rank one operator $x\otimes y$ for $x,y\in H$

Question

Let $H$ be a Hilbert space. The trace norm on $B(H)$ is defined as $$\|u\|_{1}:=\operatorname{tr}(|u|):=\sum_{e\in E}\langle|u|(e),e\rangle,$$ where $|u|:=(u^{*}u)^{1/2}$ and $E$ is (any) orthonormal basis for $H$. This may be $+\infty$. It can be shown that this definition is independent of $E$. For $x,y\in H$ we have a rank one operator $x\otimes y\colon H\to H$ defined by $(x\otimes y)(h):=\langle h, y\rangle x$. I want to show that $\|x\otimes y\|_{1}=\|x\|\|y\|$. I have computed $$(x\otimes y)^{*}(x\otimes y)=\|x\|^{2}(y\otimes y).$$ So if $y\neq0$, then $$(x\otimes y)^{*}(x\otimes y)=\|x\|^{2}\|y\|^{2}(u\otimes u),\quad\text{where}\quad u:=y/\|y\|.$$ But how do I proceed? I think that I have to pick a specific orthonormal basis $E$, but I don't see how. Any help would be greatly appreciated! Thanks in advance!

Answer 1

Note that $|u|$ is defined to be the unique positive operator for which $|u|^2 = u^*u$. Thus, with your work so far, we can now deduce that $$ |x \otimes y| = \|x\|\,\|y\|\,u \otimes u. $$ Now, the easiest way to compute $\operatorname{tr}(|x\otimes y|)$ is to select an orthonormal basis $E$ whose first element is $u$. For the purposes of your proof/computation, it suffices to note that such an orthonormal basis necessarily exists.