# Trace Norm properties

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.

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 http://www.drhea.net/wp-content/uploads/2011/11/vonNeumann.pdf - 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 $$\Sigma$$).
1. (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.)
2. First of all, let us assume von Neumann's trace inequality ( http://en.wikipedia.org/wiki/Von_Neumann%27s_trace_inequality ). This inequality implies $$|\mathrm{tr}(BA)| \le \|B\| \|A\|_1$$, 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 $$\mathrm{tr}(BA) = \mathrm{tr}(VU^* U\Sigma V^*) = \mathrm{tr}(V\Sigma V^*) = \|A\|_1$$.
The statement $$|\mathrm{tr}(AB)|\le ||B||\cdot |\mathrm{tr}(A)|$$ is untrue in general, consider the matrix $$A=B=\begin{bmatrix} 0 & 1\\ 1 & 0\\ \end{bmatrix}$$. $$|\mathrm{tr}(AB)|=2$$ while $$|\mathrm{tr}(A)| = 0$$.
1. Now we deduce 4) from 5) and von Neumann's trace inequality: $$\|BA\|_1 = \sup_{\|C\|\le 1}|\mathrm{tr}(CBA)| \le \sup_{\|C\|\le 1}|\|B\|\mathrm{tr}(CA)| = |\|B\| \|A\|_1$$
• 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? Mar 7, 2016 at 2:44
• @Guldam Remember that the trace of a squared matrix $A$ is the sum of its eigenvalues. Sep 27, 2018 at 1:56