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:



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



Die hard nilpotent spaces
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.


