17
$\begingroup$

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.

$\endgroup$

1 Answer 1

10
+25
$\begingroup$

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: $\|BA\|_1 = \sup_{\|C\|\le 1}|\mathrm{tr}(CBA)| \le \sup_{\|C\|\le 1}|\|B\|\mathrm{tr}(CA)| = |\|B\| \|A\|_1$
$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – Guldam
    Mar 7, 2016 at 2:44
  • $\begingroup$ @Guldam Remember that the trace of a squared matrix $A$ is the sum of its eigenvalues. $\endgroup$
    – EA304GT
    Sep 27, 2018 at 1:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .