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Let $||A||_1=\operatorname{trace}(\sqrt{A^* A})$. I already proved that for arbitrary unitary matrices $U$ and $V$, $||UAV^*||_1=||A||_1$ and $||A||_1=\sigma_1+\dots+\sigma_k$. Now I would like to prove that $||A||_1$ defines a matrix norm, $A\in M_{m\times n}\mathbb (C)$.

1) $||A||_1=0\Leftrightarrow A=0$. I already proved that.

2) $||\lambda A||_1=|\lambda|||A||_1$.This also.

3) $|\operatorname{trace}(A)|\leqslant ||A||_1|$. I am not sure, my idea is to use $A=U\Sigma V^*$.

4) $||BA||_1\leqslant ||B||||A||_1$ for $B\in M_{l\times m}\mathbb (C)$ and $||B||=\sup\frac{||Bx||}{||x||}=\max\{\sigma_1,\dots,\sigma_k\}$. My idea is again using singular value decomposition for $A$ and a polar decomposition for $BA$.

5)$||A||_1=\sup_{||B||\leqslant 1}|\operatorname{trace}(BA)|$ with $B\in M_{n\times m}\mathbb (C)$ and $A\in M_{m\times n}\mathbb (C)$ Here I have no idea.

6) $||A+A'||_1\leqslant||A||_1+||A'||_1$ with $A,A'\in M_{m\times n}\mathbb (C)$ This can be followed from 5).

If you could help me with 3)-5) I would really appreciate it.

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up vote 5 down vote accepted

Here are some (edit: more) ideas: First, it seems useful to restrict oneself to square matrices by "squaring" A as in this reference (p. 2 bottom of - just add zeros to $A$ to make it square which does not affect the SVD except some diagonal ones to $U$ or $V$ and some zeros to $\Sigma$).

3) (I think this should anyway only hold if $A \in M_{n\times n}(\mathbb{C})$ is a square matrix.) In that case, I believe you can prove this by considering $$\mathrm{tr}(A) = \mathrm{tr}(U\Sigma V^*) = \mathrm{tr}(\Sigma V^*U) = \sum_i(\Sigma e_i)\cdot(V^*U e_i) \le \sum_i \| \Sigma e_i \| \| V^*U e_i \|\\ \le \sum_i \sigma_i.$$ (Note $U,V,\Sigma$ are all square now.)

5) First of all, let us assume von Neumann's trace inequality ( ). This inequality in particular implies $|\mathrm{tr}(BA)| \le \|B\| |\mathrm{tr}(A)|$. Therefore, together with 3), we obtain $|\mathrm{tr}(BA)| \le \|B\| \|A\|_1$, i.e.\ $\sup_{\|B\|\le 1} |\mathrm{tr}(BA)| \le \|B\| \|A\|_1$. The other direction follows with the choice $B=VU^*$, where $U,V$ unitary are such that $A=U\Sigma V^*$, i.e. $\sqrt{A^*A} = V \Sigma V^*$, because then $\mathrm{tr}(BA) = \mathrm{tr}(VU^* U\Sigma V^*) = \mathrm{tr}(V\Sigma V^*) = \|A\|_1$.

4) Now we deduce 4) from 5) and von Neumann's trace inequality: $\|BA\|_1 = \sup_{\|C\|\le 1}|\mathrm{tr}(CBA)| \le \sup_{\|C\|\le 1}|\|B\|\mathrm{tr}(CA)| = |\|B\| \|A\|_1$

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von Neumann's trace inequality implies $|\mathrm{tr}(BA)| \le \|B\| \|A\|_1$ , but not $|\mathrm{tr}(BA)| \le \|B\| |\mathrm{tr}(A)|$, doesn't it? – Guldam Mar 7 at 2:44

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