We fix an integer $n\geq 2$. Let $S_n$ be the set of real symmetric matrices in $M_n(\mathbb{R})$. We consider the algebraic sets $Y_k=\{A\in M_n(\mathbb{R});A^k\in S_n\},k\geq 2$ and the sequence $d_k=dim(Y_k),k\geq 2$.

It is not too difficult to prove that the minimum of $(d_k)$ is $\dfrac {n(n+1)}{2}$ and is reached for $k=2$.

$\textbf{Conjecture}$. The maximum of $(d_k)$ is $n^2-n+1$ and is reached for $k=n$.

$\textbf{Remarks}$. i) Note that $Y_n\supset Z_n=\{A\in M_n(\mathbb{R}); A^n$ is a scalar matrix$\}$ and we can prove that $dim(Z_n)=n^2-n+1$.

ii) On the other hand, a formal calculation, using Grobner bases, shows that the conjecture is true for $n=2,3,4$ (at least with a great probability, because we cannot test all values of $k$).

$\textbf{Question}$. Is the above conjecture true for every $n$ ?

Thanks in advance.

  • 1
    $\begingroup$ Is your Gröbner basis computation really saying much about matrices over $\mathbb{R}$, or only over $\mathbb{C}$? (Not that I particularly cared about $\mathbb{R}$.) $\endgroup$ Sep 8, 2018 at 21:15
  • $\begingroup$ Indeed, we should keep in mind that $Y_k=\mathbb{Y}_k(\mathbf{R})$ (for some obviously defined subvariety of the spaces of matrices), and thus $\dim(Y_k)\le\dim(\mathbb{Y}_k)$ (unclear a priori if equality holds). $\endgroup$
    – YCor
    Sep 8, 2018 at 22:52
  • $\begingroup$ @darij grinberg , YCor in fact, I am not very comfortable when I am in front of a problem concerning real varieties; the softwares working about real Grobner bases are not very efficient, at least as far as I know; it is true that, in this case, one has to solve very difficult gradient problems. However, $\mathbb{C}$C-based software works well for not too complicated systems. As YCor wrote, the dimension wrt. $\mathbb{R}$ is $\leq$ than the dimension wrt. $\mathbb{C}$. To be continued. $\endgroup$
    – loup blanc
    Sep 11, 2018 at 15:33
  • $\begingroup$ @YCor , darij grinberg , In the instances I worked on, I showed (up to error) that the dimension on $\mathbb{R}$ is at least the one found on $\mathbb{C}$ by Grobner method. $\endgroup$
    – loup blanc
    Sep 11, 2018 at 15:34


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