# Nuclear norm of self-adjoint matrix

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?

• Do you mean $\langle A, B \rangle = \text{Tr}(AB^*)$? Sep 3 '19 at 16:01
• @angryavian Yes of course! Corrected.
– YLee
Sep 3 '19 at 16:21

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 $$B$$.