Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $V\subset\mathbb{C}^{n\times n}$ be a linear space consisting of $n\times n$ complex matrices. Say that $V$ is nilpotent if every matrix $v\in V$ is nilpotent; denote by $V^k$ the subspace spanned by all possible products $v_1\ldots v_k$ with $v_i\in V$. The conjecture is:

Assume that both $V^k$ and $V^{k+1}$ are nilpotent spaces. Then, $\dim V^k\geqslant \dim V^{k+1}$.

Some experiments with upper-triangular nilpotent spaces make me hope that this conjecture is true in general. I would be very happy if someone was able to help me with either a proof or a counterexample, even for $k=1$.

share|improve this question

1 Answer 1

up vote 12 down vote accepted

You conjecture is not true. Let $P$ be the $3\times 3$ matrix $$ \left( \begin{array}{ccc} 0&1&0\\ 0&0&1\\ 0&0&0\end{array} \right) $$ which is nilpotent with $P^2\neq 0$.

Consider the subspace of $9 \times 9$ uppertriangular matrices spanned by $$ A=\left( \begin{array}{ccc}P&0&0\\0&P&0\\0&0&0\end{array} \right),~~ B= \left( \begin{array}{cc}0&0&0\\0&P&0\\0&0&P\end{array} \right). $$ Then $A^2$, $AB$ and $B^2$ are linearly independent.

share|improve this answer
    
Thank you for the example! May I ask you a bit more, namely, do you see a way to generalize your example to matrices which do not have such a nice block-diagonal structure? For instance, under an additional assumption that matrices from $V$ do not possess a common invariant subspace? –  user56203 Aug 2 at 18:50
2  
First of all, my example shows that if you take two generic uppertriangular matrices $A$ and $B$ of size $9$, then $A^2,AB,B^2$ are linearly independent. I suspect that in fact $BA$ would also be independent from them in this case, but I did not check it. I also think that taking $V$ to be a generic dimension two subspace of uppertriangular size $4$ matrices, would give $V^2$ of dimension $3$ (elements of $V^2$ can only have three possible nonzero entries, but I see no reason for why the span of them would be any less than that). –  Lev Borisov Aug 2 at 20:43
    
I believe that every nilpotent subspace $V$ can be conjugated inside the space of uppertriangular matrices, regardless of your assumption on dimensions of $V^k$. So they will have a common invariant zero eigenspace. –  Lev Borisov Aug 2 at 21:01
2  
Thanks for your further explanations, considerations with generic spaces seem helpful. Concerning the second comment, unfortunately, there are nilpotent spaces $V$ not equivalent to triangular matrices; for instance, that spanned by $A=\left(\begin{smallmatrix}0&1&0\\0&0&-1\\0&0&0\end{smallmatrix}\right)$ and $B=\left(\begin{smallmatrix}0&0&0\\1&0&0\\0&1&0\end{smallmatrix}\right)$ holds this property because $AB$ has a non-zero eigenvalue. –  user56203 Aug 2 at 21:24
    
Thank you, this is a nice example. I got fooled by my commutative heritage. –  Lev Borisov Aug 2 at 23:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.