Let $n$ be a positive integer and let $U$ be a finite subset of $M_{n\times n}(\mathbb{C})$ which is closed under multiplication of matrices. Show that there exists a matrix $A$ in $U$ satisfying $\text{trace}(A) \in \{1,...,n\}.$

I was thinking if we look at this problem by contradiction. Then we have that for all matrices $A$ $\text{trace}(A)>n$ or $\text{trace}(A)=0.$ In the first case, we get that $\det(A)<1$ and the second case we get that $\det{A}=0.$ I am not sure how to proceed further, so any hint would be much appreciated.

  • $\begingroup$ The $n\times n$ identity matrix has trace $n \in A$ $\endgroup$
    – amWhy
    Oct 18, 2018 at 16:51
  • $\begingroup$ @amWhy Set $U$ is given, it's not as easy $\endgroup$
    – Jakobian
    Oct 18, 2018 at 16:52
  • $\begingroup$ @Jakobian In fact it is easy and the above comment gives an excellent hint: a non-empty finite subset of a group closed under the group's operation is in fact a subgroup. $\endgroup$
    – DonAntonio
    Oct 18, 2018 at 16:54
  • 1
    $\begingroup$ @DonAntonio But it's not a group, it's a semigroup $\endgroup$
    – Jakobian
    Oct 18, 2018 at 16:55
  • 7
    $\begingroup$ I think the set $U=\{0_{n\times n}\}$ is a counter-example. $\endgroup$
    – awllower
    Oct 18, 2018 at 17:02

3 Answers 3


Let $A\in U$. Since $U$ is finite, there exist $a,b\in\mathbb N$ such that $A^a=A^b$ and $a < b$. So the minimal polynomial of $A$ divides $x^a-x^b$. This shows that any eigen-value of $A$ is either $0$ or a root of unity.

Consider the matrices $A^k$ for $k\in\mathbb N$.

By this, the eigen-values of $A^k$ are the $k$-th powers of eigen-values of $A$, so we can find $k$ such that the eigen-values of $A^k$ are either $0$ or $1$. Then the trace of $A^k$ is a sum of $n$ numbers in $\{0,1\}$, so is an integer between $0$ and $n$.

Hope this helps.


A hint: Consider an element $A$. Since $U$ is closed under multiplication, for every natural number $m$, $A^m \in U$. However, $U$ is finite--so that means $A^m$ takes on finitely many values. What does that imply about $A$'s eigenvalues? The trace is the sum of its eigenvalues--must there be a $m$ such that $\mathrm{Tr}(A^m)\in \mathbb N$?


This is wrong unless the set contains zero

Pick any $A \in U$ and consider the sequence $A,A^2,A^3,...$. Since $U$ is finite, there must exist $s<t$ such that $A^s = A^t$.

In particular, if $\lambda$ is an eigenvalue of $A$ we see that $\lambda^s = \lambda^t$ and hence either $\lambda=0$ or $\lambda^{t-s} = 1$. In other words, all eigenvalues of $A^{t-s}$ (which must be a member of $U$) satisfy $\lambda^{t-s} \in \{0,1\}$.

Consequently, $\operatorname{tr} A^{t-s} \in \{0,...,n\}$.


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