International Mathematics Competition
for University Students
2020

IMC2020: Day 1, Problem 2

Problem 2. Let \(\displaystyle A\) and \(\displaystyle B\) be \(\displaystyle n\times n\) real matrices such that

\(\displaystyle \textrm{rk}(AB-BA+I)=1 \)

where \(\displaystyle I\) is the \(\displaystyle n\times n\) identity matrix.

Prove that

\(\displaystyle \tr(ABAB)-\tr(A^2B^2)=\frac12 n(n-1). \)

(\(\displaystyle \textrm{rk}(M)\) denotes the rank of matrix \(\displaystyle M\), i.e., the maximum number of linearly independent columns in \(\displaystyle M\). \(\displaystyle \tr(M)\) denotes the trace of \(\displaystyle M\), that is the sum of diagonal elements in \(\displaystyle M\).)

Rustam Turdibaev, V. I. Romanovskiy Institute of Mathematics

Solution. Let \(\displaystyle X=AB-BA\). The first important observation is that

\(\displaystyle \tr(X^2)=\tr(ABAB-ABBA-BAAB+BABA)=2\tr(ABAB)-2\tr(A^2B^2)\)

using that the trace is cyclic. So we need to prove that \(\displaystyle \tr(X^2)=n(n-1)\).

By assumption, \(\displaystyle X+I\) has rank one, so we can write \(\displaystyle X+I=v^tw\) for two vectors \(\displaystyle v,w\). So

\(\displaystyle X^2=(v^tw-I)^2=I-2v^tw+v^twv^tw=I+(wv^t-2)v^tw.\)

Now by definition of \(\displaystyle X\) we have \(\displaystyle \tr(X)=0\) and hence \(\displaystyle wv^t=\tr(wv^t)=\tr(v^tw)=n\) so that indeed

\(\displaystyle \tr(X^2)=n+(n-2)n=n(n-1).\)

An alternative way to use the rank one condition is via eigenvalues: Since \(\displaystyle X+I\) has rank one, it has eigenvalue \(\displaystyle 0\) with multiplicity \(\displaystyle n-1\). So \(\displaystyle X\) has eigenvalue \(\displaystyle -1\) with multiplicity \(\displaystyle n-1\). Since \(\displaystyle \tr(X)=0\) the remaining eigenvalue of \(\displaystyle X\) must be \(\displaystyle n-1\). Hence

\(\displaystyle \tr(X^2)=(n-1)^2+(n-1) \cdot 1^2=n(n-1).\)