# Trace norm of a complex $|B| \times|A|$ matrix $M$.

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.

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 $$\max _{V: B \rightarrow A} \operatorname{Tr}[V M]=\max _{V: B \rightarrow A} \sum_{x,y}^{} \mu_y \left\langle v_x \middle| s_y \right\rangle \left\langle r_y \middle| u_x \right\rangle=\sum_{x,y}^{} \mu_y=\|M\|_1$$.