Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For each positive integer $k$, find the smallest number $n_k$ for which there exist real $n_k$ by $ n_k$ matrices $A_1; A_2; ....; A_k$ such that all of the following conditions hold:

$$ \text{ 1. } A_1^2 = A_2^2 = ... = A_k^2 = 0 $$

$$ \text{ 2. } A_jA_i = A_iA_j \forall1 \leq i \text{ and } j \leq k$$

$$ \text{ 3. } A_1A_2 .... A_k \neq 0 $$

share|cite|improve this question
Where does this problem come from? A very similar one was asked and answered at MO a few months ago… – m_t_ Dec 3 '12 at 8:44
This seems to be Problem 5 from the 2007 International Mathematics Competition for University Students (IMC2007); see ref. [spoiler alert]. – Douglas S. Stones Dec 4 '12 at 3:03
up vote 1 down vote accepted

(I find this question very interesting. I couldn't come up with a solution, but below are some ideas that maybe someone can use/improve)

Fix $k$. From 2, we know that the matrices $A_1,\ldots,A_k$ are mutually commuting. So they can be triangularized simultaneously. This means that we can assume right off the bat that $A_1,\ldots,A_k$ are upper triangular.

Claim 1: the matrices $\prod_{j=n}^mA_nA_{n+1}\cdots A_m$, $1\leq n\leq m\leq k$, are linearly independent.

Claim 2: the set of matrices in Claim 1 has cardinality $2^k-1$.

These two claims together imply that we have $2^k-1$ mutually commuting linearly independent upper triangular matrices with null square. The dimension of the space of such $n_k\times n_k$ matrices is certainly less than $n_k-1$ (not that I have a proof, so this is a missing step too). So we would have $n_k-1\geq 2^k-1$, i.e. $n_k\geq 2^k$.

The challenge would be, assuming the estimate above is sharp, to construct $A_1,\ldots,A_k$ of size $2^k\times 2^k$ satisfying the required conditions.

Proof of Claim 2: we can write our products as $\prod_{j=1}^kA_1^{s_1}\cdots A_k^{s_k}$, where $s_1,\ldots,s_k\in\{0,1\}$. We have $2^k-1$ choices for $s_1,\ldots,s_k$ (as we don't allow them to be all zero). Note that, since $A_1\cdots A_k\ne0$, any partial product cannot be zero either.

Proof of Claim 1: suppose $$ \sum_{s\in\{0,1\}^n}\lambda_s\prod_{j=1}^kA_1^{s_1}\cdots A_k^{s_k}=0. $$ If we multiply by $A_2\cdots A_k$, all the terms containing any of $A_2,\ldots,A_k$ vanish. This shows that the coefficient of $A_1$ is zero. Similarly, all coefficients of terms of degree $1$ vanish. Now we multiply by $A_3\cdots A_k$ and the only term remaining will be $\lambda_{1,1,0,\ldots,0}A_1A_2=0$; so this coefficient is zero and similarly we can show that all coefficients corresponding to terms of degree two are zero. Going on with this process we can deduce that all $\lambda_s$ are zero, and we get linear independence.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.