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Let $A_1,\dots,A_n$ be mutually commuting $m\times m$ matrices such that $A_i^2=0$ for all $1\le i \le n$. If $m<2^n$, prove that $A_1 A_2\cdots A_n=0$

Since $A_i^2=0$ So $\operatorname{Im}(A)\subset \ker(A) \implies \dim(\ker(A))\ge \frac m 2$ I think we have to use it to prove ..but don't know how go through ...

RAM_3R
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    I guess you mean $\dim(\ker(A_i)) \geq m/2$!? Probably it is useful to get a bound for $\dim(\ker(A_{n-1}A_n))$ and iterate – Stockfish Nov 23 '18 at 14:39
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    yes right ...... – RAM_3R Nov 23 '18 at 15:05
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    @Stockfish probably yes but i don't understand how relate this $m<2^n$?? – RAM_3R Nov 23 '18 at 15:41
  • Not a new question. That follows is a post with $m=n$. cf. https://math.stackexchange.com/questions/880429/pairwise-commuting-nilpotent-matrices-alternative-solution-needed?rq=1 –  Nov 23 '18 at 16:27
  • But the answer there cannot be applied directly here to the case where we have, e.g. m = 15, n = 4. We really should use that the $A_i$ of the current question satisfy $A^2 = 0$ which is much stronger than being 'merely' nilpotent – Vincent Nov 23 '18 at 16:55
  • Still we could steal some ideas from the answer to that other question... – Vincent Nov 23 '18 at 16:56

2 Answers2

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Let $x\in K^m$.

Then $A_1x\in Im(A_1)$, vector space of dimension $\leq m/2$.

In the sequel, $\tilde{U}$ denotes a restriction of $U$.

$\tilde{A_2}:Im(A_1)\rightarrow Im(A_1)$ is $2$-nilpotent. Then $A_2A_1x\in Im(\tilde{A_2})$,

vector space of dimension $\leq dim(Im(A_1))/2\leq m/2^2$, and so on $\tilde{A_3}:Im(\tilde{A_2})\rightarrow Im(\tilde{A_2})$ is $2$-nilpotent,....

$\tilde{A_n}:Im(\tilde{A_{n-1}})\rightarrow Im(\tilde{A_{n-1}})$ is $2$-nilpotent. Then $A_n\cdots A_1x\in Im(\tilde{A_n})$, vector space of dimension $\leq dim(Im(\tilde{A_{n-1}}))/2\leq m/2^n<1$.

Finally $Im(\tilde{A_n})=\{0\}$ and we are done.

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Edit. The answer below seems to be only the partial answer for the particular form of nilpotent matrices.

Mutually commuting means that all matrices can be represented simultaneously as, for example, upper triangular.

Of course they are all nilpotent and (here assumption for the form of matrices) the first non-zero "mini-diagonal" of this matrix is shifted from the main diagonal at least rounded $m/2$.
(do yourself analysis for dimension $m$ even and odd)

For example matrix

$\begin{bmatrix} 0\color{red}\rightarrow & 0\color{red}\rightarrow & 0\color{red}\rightarrow & \color{red}1 & 3 \\ 0 & 0\color{red}\rightarrow & 0\color{red}\rightarrow & 0\color{red}\rightarrow & \color{red}2 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix}$

has first minidiagonal (red) shifted by $3$ positions from the main diagonal.

In multiplication values of shifts for such matrices are added.
I mean that if $A=A_1A_2$ then shift($A$)=shift($A_1$)+shift($A_2$).

In this case $A^2$ should have to have shift $6>5$ what means that $A^2=0$.

(If shift were by $2$ with $A^2$ result shift would be $ 4$ what is too little to make matrix zero in general case.)

Multiplication $A_1A_2 \dots A_n$ means that the result matrix has shift at least $nm/2$ what is fully fulfilled if $m<2^n$.

Look at some examples. For $n=2$ we have limit for $m$ equal 3. Result the least shift $3$.

For $n=3$ we have limit for $m$ equal 7. Result shift $12$

For $n=4$ we have limit for $m$ equal 15. Result shift $30$,

Always result shift is bigger then $m$ what means that we get zero matrix.

Widawensen
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    When you simultaneously triangularize nilpotent matrices , they are strictly upper triangular, but each of them fills the upper part. You do not have the hand on the shape of these matrices. –  Nov 23 '18 at 16:36
  • @loupblanc So would it be only a partial answer for particular form of nilpotent matrices? – Widawensen Nov 23 '18 at 16:45
  • I think that it is difficult to show that we can put our matrices in the form you consider. –  Nov 23 '18 at 17:32
  • @loupblanc yes, you are right, it is only the answer for some subset of nilpotent matrices, yours is much more precise I hope the OP will transfer the green mark to you.. – Widawensen Nov 23 '18 at 17:34