Given a matrix $X \in \mathbb{R}^{m \times n},$ then the spectral norm is defined by

$$\left \| X\right\| = \max\limits_{i \in \{1, \dots, \min\{m,n\}\} }\sigma_i (X)$$

whereas the nuclear norm is defined by

$$\left \| X \right \|_* = \sum\limits_{i=1}^ {\min\{m,n\}} \sigma_i (X)$$

It is a well known fact that the dual norm of the spectral norm is the nuclear norm

https://math.stackexchange.com/questions/1158798/show-that-the-dual-norm-of-the-spectral-norm-is-the-nuclear-norm

This implies that $$\|M\| = \sup_{\|X\|_*\leq 1} \langle M, X \rangle.$$

My question is the following. Given a matrix $M \in \mathbb{R}^{n \times n},$ how to find a matrix $X^*$ such that

$$ X^* \in arg\sup_{\|X\|_*\leq 1} \langle M, X \rangle.$$

PD: The inner product here is $\langle A, B \rangle := \mathop{\textrm{Tr}}(A^TB)$

Given a matrix $X \in \mathbb{R}^{m \times n},$ then the spectral norm is defined by

$$\left \| X\right\| := \max\limits_{i \in \{1, \dots, \min\{m,n\}\} }\sigma_i (X)$$

whereas the nuclear norm is defined by

$$\left \| X \right \|_* := \sum\limits_{i=1}^ {\min\{m,n\}} \sigma_i (X)$$

It is a well-known fact that the dual norm of the spectral norm is the nuclear norm. This implies that

$$\|M\| = \sup_{\|X\|_*\leq 1} \langle M, X \rangle$$

where the inner product is defined by $\langle A, B \rangle := \mathop{\textrm{Tr}}(A^TB)$. Given a matrix $M \in \mathbb{R}^{n \times n},$ how to find a matrix $X^*$ such that the following holds?

$$ X^* \in \arg\sup_{\|X\|_*\leq 1} \langle M, X \rangle$$