On maximizing $\operatorname{Trace}(AX)$ subject to $\|X\|=1$ Let $\Bbb S_n (\Bbb R)$ denote the set of $n \times n$ real symmetric matrices. Given $A \in \Bbb S_n (\Bbb R)$, let
$$ X^* := \arg\max_{X \in\Bbb S_n (\Bbb R)} \operatorname{Trace}(AX) \quad\mbox{s.t.}\quad \|X\| = 1 $$
For what set of matrix norms (e.g., Frobenius, spectral, $p$-Schatten, etc) in the problem above, would the maximizer have the following form?
$$ X^* = \frac{1}{\|A\|} A $$
 A: Sometimes the simplest thing to do is draw a picture

Note: This shows the entry-wise norms $L_{p,p}$, i.e. treating $X$ as a $n^2$ vector.
A: Let $A=\|a_{ij}\|\ne 0$ and $X=\|x_{ij}\|$. Then $\operatorname{Trace} AX=\sum_{i,j=1}^n a_{ij}x_{ji}$. If the Frobenius norm $\|X\|_F=\sqrt{\sum_{i,j=1}^n x_{ij}^2}$ of the matrix $X$ equals $1$ then by the Cauchy-Schwarz inequality
a set $\mathcal X$ of $\arg\max_X\operatorname{Trace}(AX)$ consists of a unique element  $X^*=\|A\|_F^{-1}A^t$.
For other matrix norms this conclusion can fail.
For instance, for $L_{1,1}$ norm of $X$, we have if $\|X\|=1$ then $X\in\mathcal X$ iff for each $1\le i,j\le n$ we have $a_{ij}x_{ji}\ge 0$, if $(i,j)$ is $\arg\max_{i’,j’} |a_{i’,j’}|$ and $x_{ji}=0$, otherwise. Since for each $p\ge 1$, $L_{1,1}$ norm coincides with $L_{p,1}$ norm for diagonal matrices, $X^*$ can fail to belong to $\mathcal X$ for these norms either.
For max norm of $X$, we have $X\in\mathcal X$ iff for each $1\le i,j\le n$ such that $a_{ij}\ne 0$ we have $x_{ji}=\operatorname{sgn} a_{ij}$. Since max norm coincides with $\|\cdot \|_1$ and $\|\cdot \|_\infty$ norms for diagonal matrices, $X^*$ can fail to belong to $\mathcal X$ for these norms either.
