In terms of matrix theory, the question I'm led to is the following: Start with an $n$-dimensional vector space over an algebraically closed field of characteristic 0 such as $\mathbb{C}$, which has a non-degenerate symmetric bilinear form. Consider all $n \times n$ skew-adjoint matrices $A$ relative to this form.

Given $n\geq 5$, what is the least power $k$ for which all

nilpotentmatrices of this type satisfy $A^k=0$?

For any nilpotent $n \times n$ matrix $A$, it follows from the Cayley-Hamilton Theorem that $A^n=0$. But in the special case here it seems plausible to expect a slightly smaller minimum: namely, $n-1$ if $n$ is odd and $n-2$ if $n$ is even. I also wonder what is written down in the literature along this line.

There is of course a hidden agenda, relative to simple Lie algebras attached to special orthogonal groups over a field like $\mathbb{C}$. In the classification of simple types $A_\ell-D_\ell$ of rank $\ell$, the respective Coxeter numbers are $\ell+1, 2\ell, 2\ell, 2(\ell-1)$. (Types $B, C$ share the same Weyl group.) Types $A, C$ are realized naturally as $n\times n$ matrices with $n=h$, but the other Lie algebras of orthogonal type yield $n=2\ell+1$$ and $n=2\ell$. So my question for the latter types means: does every nilpotent element $e$ of this natural matrix Lie algebra satisfy $e^h =0$ as in types $A,C$?

Behind this question is a related prime characteristic question for restricted Lie algebras, motivated in part by Kostant's classical 1959 paper in *Amer. J. Math.* (Corollary 5.4). In the general setting of simple Lie algebras he showed that *regular* (= principal) "nilpotent" elements $e$ are characterized by a condition on their adjoint operators: $(\mathrm{ad} \:e)^{2q}\neq 0$ where $q$ is the sum of coefficients of the highest root expressed relative to simple roots. Moreover, the next power annihilates all *regular* nilpotents. Earlier he showed that $q+1 = h$ is the Coxeter number of the Weyl group. (But there is a misprint in that corollary.)

ADDED: As Victor points out, except for a small decrease in type $D$ the four classical families of simple Lie algebras have index of nilpotence in their natural representations given by the Cayley-Hamilton approach in type $A$. My question arose from passing to characteristic $p>0$ via a Chevalley basis over $\mathbb{Z}$, then extending scalars. For $p \geq h$, results from the mid-1980s on cohomology of restricted Lie algebras and support varieties of modules (Jantzen, Friedlander-Parshall, ... ) reveal that for the built-in $[p]$ operation on such a Lie algebra one has $\text{ad}\: e^{[p]} = 0 = (\text{ad} \:e)^p$ for all nilpotents $e$. But in natural matrix representations like those for types $A-D$, the $[p]$ operation is the usual matrix power. Here the slightly modified Cayley-Hamilton power needed for vanishing agrees.

At the extreme of $E_8$ there is more divergence: Here the "natural" smallest faithful representation is given by the adjoint module with $n=248$, whereas $h=30$. For $p = 31$ this power of each $\text{ad} \:e$ vanishes, contrasting with Kostant's characteristic 0 result which requires a power at least $59$ when $e$ is regular.