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Let $A$ and $B$ be two finite-dimensional Hilbert spaces, and let $M: A \rightarrow B$ be a $|B| \times|A|$ complex matrix. Show that the trace norm of $M$ can be expressed as $$ \|M\|_{1}=\max _{V: B \rightarrow A} \operatorname{Tr}[V M] $$ where the maximum is over all partial isometries $V: B \rightarrow A$.

$\mathbf{NOTE:}$ $V$ is a partial isometry if $V^*V=I^B$ or $VV^*=I^A$.

$\mathbf{ATTEMPT:}$ I have two ideas, one of them is that we have to show that $\max _{V: B \rightarrow A} \operatorname{Tr}[V M] \geqslant \|M\|_{1}$ and $\max _{V: B \rightarrow A} \operatorname{Tr}[V M] \leqslant \|M\|_{1}$. So then we can conclude $\|M\|_{1}=\max _{V: B \rightarrow A} \operatorname{Tr}[V M]$.

"$\leqslant$":The singular values of $M$ are $\mu_1 \geqslant \mu_2 \geqslant \cdots \geqslant \mu_n$ and the singular values of $V$ are $\nu_1 \geqslant \nu_2 \geqslant \cdots \geqslant \nu_n$. Now, using von Neumann's trace inequality: $$\operatorname{Tr}[V M] \leqslant \sum_{i=1}^{n} \mu_i \nu_i \leqslant \max _{i \in \{1,\cdots,n\}} \{\nu_i\}\sum_{i=1}^{n} \mu_i \leqslant \nu_{max} \|M\|_{1} \leqslant \|M\|_{1}$$

But for the other direction I have no idea and I'm not sure that the above proof is true. I should mention that I'm also struggling to understand the von Neumann's trace inequality.

Could you please help me to complete the proof or give me an alternative proof? Thanks

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1 Answer 1

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I think this answer is true, but not sure.

We can consider our partial isometry to be $V=\sum_{x=1}^{k} \left|u_x\right\rangle^A\left\langle v_x\right|^B $ where k is the minimum dimension between $|A|$ and $|B|$, $\{\left|u_x\right\rangle\}$ is an orthonormal set in $A$ and $\{\left|v_x\right\rangle\}$ is an orthonormal set in $B$. and we can write $M=\sum_{y=1}^{n} \left|s_y\right\rangle^B\left\langle r_y\right|^A$, where $\{\left|r_y\right\rangle\}$ and $\{\left|s_y\right\rangle\}$ are another orthonormal set in $A$ and $B$ respectively. So then $\operatorname{Tr}[VM]=\cdots=\sum_{x,y}^{} \mu_y \left\langle v_x \middle| s_y \right\rangle \left\langle r_y \middle| u_x \right\rangle$. So then, since $\left\langle v_x \middle| s_y \right\rangle , \left\langle r_y \middle| u_x \right\rangle \leqslant 1$ , then the maximum of the $\operatorname{Tr}[VM]$ occurs, if $\left\langle v_x \middle| s_y \right\rangle = \left\langle r_y \middle| u_x \right\rangle = \delta_{xy}$ and it depends on choosing the partial isometry. It means that $\max _{V: B \rightarrow A} \operatorname{Tr}[V M]=\max _{V: B \rightarrow A} \sum_{x,y}^{} \mu_y \left\langle v_x \middle| s_y \right\rangle \left\langle r_y \middle| u_x \right\rangle=\sum_{x,y}^{} \mu_y=\|M\|_1$.

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