IMC 2025 · Day 1 · P3

Let $n = 2025$. First, we give a charaterisation of matrices in $S$.

Suppose that $A = (a_{ij})_{i,j=1}^{n} \in S$. Since $\operatorname{rk} A = 1$, for every $1 < i, j \le n$, we have $$\det \begin{pmatrix} a_{11} & a_{1j} \\ a_{i1} & a_{ij} \end{pmatrix} = a_{11} a_{ij} - a_{i1} a_{1j} = a_{11} a_{ij} - a_{i1} a_{j1} = 0.$$ If $a_{11} = 1$ then this means $a_{ij} = a_{i1} a_{j1}$. In this case, let $u = \begin{pmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{n1} \end{pmatrix} = \begin{pmatrix} 1 \\ a_{21} \\ \vdots \\ a_{n1} \end{pmatrix}$; then we have $A = \bigl( a_{i1} a_{j1} \bigr) = u u^\top$. Otherwise, if $a_{11} = -1$, we have $a_{ij} = -a_{i1} a_{j1}$. In that case, let $u = \begin{pmatrix} -a_{11} \\ -a_{21} \\ \vdots \\ -a_{n1} \end{pmatrix} = \begin{pmatrix} 1 \\ -a_{21} \\ \vdots \\ -a_{n1} \end{pmatrix}$; then $A = -\bigl( a_{i1} a_{j1} \bigr) = -u u^\top$.

Hence, all matrices in $S$ can uniquely be written as $\pm u u^\top$ with a vector $u = \begin{pmatrix} 1 \\ u_2 \\ \vdots \\ u_n \end{pmatrix}$ such that $u_2, \dots, u_n \in \{\pm 1\}$. (Note that $\operatorname{rk}(u u^\top) = 1$ is satisfied.) In particular, we have $|S| = 2^n$, because the sign and the coordinates $u_2, \dots, u_n$ can be chosen independently.

Now, if $A = \pm u u^T$ and $B = \pm v v^T$ are elements of $S$, then $$AB = \pm (u u^\top)(v v^\top) = \pm u (u^\top v) v^\top = \pm u \cdot \langle u, v \rangle \cdot v^\top = \pm \langle u, v \rangle \cdot \bigl( u_i v_j \bigr)_{i,j=1}^{n}$$ and similarly $$BA = \pm \langle u, v \rangle \cdot \bigl( v_i u_j \bigr)_{i,j=1}^{n}.$$ Since $n = 2025$ is odd, it follows that $\langle u, v \rangle \ne 0$. The first columns of the matrices $u v^\top = \bigl( u_i v_j \bigr)_{i,j=1}^{n}$ and $v u^\top = \bigl( v_i u_j \bigr)_{i,j=1}^{n}$ are $u$ and $v$, respectively. Hence, $AB = BA$ if and only if $u = v$; in other words, if $A = \pm B$.

For each $A \in S$, there are precisely two suitable matrices $B \in S$, so the probability that $A, B$ commute is $\dfrac{2}{|S|} = \dfrac{1}{2^{n-1}}$.