6

What is the maximum dimension of a vector space of $\mathcal{M}_n(\mathbb{R})$ containing only nilpotent matrices ? ($\mathcal{M}_n(\mathbb{R})$ : matrices $n\times n$ with coefficients in $\mathbb{R}$)

I don't really know how to solve this problem.There must be a way to give some good upper bound to the dimension of the vector space, which would seem to be $(n^2-n)/2$, but I can't manage to get a good result ...

Hippalectryon
  • 7,830
  • @PedroTamaroff I don't know about Jordan forms. – Hippalectryon Nov 17 '14 at 01:00
  • 2
    An example for such a vector space is the space of strictly upper triangular matrices, which has dimension $(n^2-n)/2$. I wouldn't be surprised if this is the maximum, but I don't immediately see a proof. – Robert Israel Nov 17 '14 at 01:04
  • @RobertIsrael You're correct, and in fact any such vector space of maximal dimension is conjugate to the upper triangular matrices. This is a result of Gerstenhaber from 1958. I'm trying to write down a nicer proof in this special case but have yet to succeed. –  Nov 17 '14 at 06:07
  • @MikeMiller Fyk, this question was asked at an oral exam (30 mins) in 2011. I guess there must be a not too complex way to solve it. – Hippalectryon Nov 17 '14 at 11:42

1 Answers1

4

The dimension of the vector space of symmetric matrices $\mathcal{S}_n(\mathbb{R})$ is $\frac{n(n+1)}2$, and all nilpotent matrices are non symmetric. As the set of strictly upper triangular matrices $T_n^{++}(\mathbb{R})$ is a vector space of nilpotent matrices with dimension, $\frac{n(n-1)}2$ that gives us the maximum dimension.

Hippalectryon
  • 7,830
  • @loupblanc Uh ? Why is that ? It seems to me that all the useful information is here. Is is straightforward that $T_n^{++}$ is a vector space, and the dimensions are straightforward too. – Hippalectryon Nov 18 '14 at 21:45
  • @Hippalectryon The set of nilpotent matrices is not a vector subspace of $\mathcal{M}_n(\mathbb{R})$. If you let $N_n(\mathbb{R})$ denote a vector space of nilpotent real $n\times n$-matrices, then your result follows modulo the minor omission that you didn't say why no symmetric matrix other than $0$ is nilpotent. – Daniel Fischer Nov 18 '14 at 22:31
  • @loupblanc I unfortunately still do not see my error. Let $N_n(\mathbb{R})$ be the a subspace of nilpotent matrices of size $n$ with real coefficients, $$n^2=\dim_n(\mathbb{R})≥\dim(\mathcal{S}_n(\mathbb{R})+N_n(\mathbb{R}))$$ $$=\dim\mathcal{S}_n(\mathbb{R})+\dim N_n(\mathbb{R})−\dim(\mathcal{S}_n(\mathbb{R})\cap N_n(\mathbb{R}))=n(n+1)/2+\dim N_n(\mathbb{R})$$ hence $\dim N_n(\mathbb{R}))≤n(n−1)/2$, which is reached for $T^{++}_n$ (The result I was searching for is only on real matrices, not on any field as in Gerstenhaber's paper !) – Hippalectryon Nov 18 '14 at 22:35
  • Hippa, you are right and I am wrong. It is a pretty proof over $\mathbb{R}$ and it is unnecessary to prove the simultaneous triangularization. Note that the general case is a not obvious theorem from Gerstenhaber (cf. the Mike post). A simplified proof is in "B. Mathes, M. Omladiˇc, H. Radjavi, Linear spaces of nilpotent matrices,Linear Algebra Appl.149(1991) 215-225". –  Nov 18 '14 at 22:43