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.