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Consider first general matrices in $\mathbb{C}^{n \times m}$. Using norm duality, the nuclear norm (sum of singular values) can be expressed as

$$\|A\|_* = \max \{ | \langle A, B \rangle | : \|B\|_2 \leq 1 \}$$

where $\|\|_2$ denotes the spectral norm (largest singular value) and $\langle A,B \rangle = \mathrm{Tr} AB^*$.

How can I show that, if $A$ is self-adjoint, it is sufficient to consider self-adjoint $B$ in the above?

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  • $\begingroup$ Do you mean $\langle A, B \rangle = \text{Tr}(AB^*)$? $\endgroup$
    – angryavian
    Sep 3 '19 at 16:01
  • $\begingroup$ @angryavian Yes of course! Corrected. $\endgroup$
    – YLee
    Sep 3 '19 at 16:21
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When $A$ is self-adjoint, it is diagonalizable, and the nuclear norm becomes the sum of the absolute values of the eigenvalues, say, $|\lambda_1| + \cdots + |\lambda_n|$ where $A = UDU^*$ and the diagonal entries of $D$ are $\lambda_1, \ldots, \lambda_n$.

Let $\tilde{D}$ be diagonal with diagonal entries $\lambda_1 / |\lambda_1|, \ldots, \lambda_n / |\lambda_n|$ and let $B = U\tilde{D} U^*$. You can check that $\|A\|_* = |\langle A, B \rangle|$ and that $\|B\|_2 \le 1$.

Thus, if $A$ is self-adjoint, there exists a self-adjoint $B$ with $\|B\|_2 \le 1$ such that $\|A\|_* = |\langle A, B \rangle|$, so the maximization can be restricted to self-adjoint $B$.

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