Let $W$ be a real and symmetric matrix $m \times m$ from the set $M_{m,m}$, and $T:M_{m,m} \rightarrow \mathbb{R}$ a function defined by $T(W) = \operatorname{trace}(W^3)$.
We are interested to find the direction $V$ in $M_{m,m}$ such that, the directional derivative of $W$
\begin{equation} \nabla T(W)_V = \lim_{h \rightarrow 0} \frac {T(W+hV)-T(W)}{h} \end{equation}
attains the maximum.
Intuitively, the direction $V$ must be to the 1-rank matrix $vv^T$, where $v$ is the eigenvector associated with the highest eigenvalue of $W$.
Is this intuition correct ? Even in the Hilbert spaces?
Since $W\in S_m$, $DT_W:V\in S_m\rightarrow 3tr(W^2V)$. We may assume that $W^2=diag((\lambda_i)_i)$ where $\lambda_i\geq 0$. We consider the Frobenius inner product (cf. joriki post) $<A,B>=tr(A^TB)$. We seek $\sup_{||V||=1,V\in S_m}tr(W^2V)$. One has $tr(W^2V)=<W^2,V>=\sum_i\lambda_i v_{i,i}=\leq ||W^2||||V||=||W^2||=\sqrt{\sum_i\lambda_i^2}$. We put $V=\dfrac{W^2}{||W^2||}$; then $||V||=1$ and $<W^2,V>=\sqrt{\sum_i\lambda_i^2}$, that is the required maximum.