International Mathematics Competition
for University Students
2026

IMC2026: Day 1, Problem 2

Problem 2. Let \(\displaystyle n\) be a positive integer. Suppose that \(\displaystyle A\) and \(\displaystyle B\) are \(\displaystyle n\times n\) matrices with real entries such that

\(\displaystyle A^\top A + BB^\top = AB + BA,\)

where \(\displaystyle X^\top\) denotes the transpose of matrix \(\displaystyle X\).

Does this imply that \(\displaystyle AB = BA\)?

Nikolaos Kolliopoulos, University of Cyprus

Solution 1. Let \(\displaystyle A=(a_{ij})_{i,j=1}^n\) and \(\displaystyle B=(b_{ij})_{i,j=1}^n\). By taking traces of both sides,

\(\displaystyle 0 = \tr(A^\top A+BB^\top-AB-BA) = \tr(A^\top A)+\tr(BB^\top)-\tr(AB)-\tr(BA) \)

\(\displaystyle = \sum_{i,j=1}^n a_{ij}^2 +\sum_{i,j=1}^n b_{ji}^2-2\sum_{i,j=1}^n a_{ij}b_{ji} =\sum_{i,j=1}^n (a_{ij}-b_{ji})^2. \)

Hence, \(\displaystyle a_{ij}=b_{ji}\) for every \(\displaystyle i,j\), so \(\displaystyle B=A^\top\).

By substituting \(\displaystyle B=A^\top\) back into the condition, we get

\(\displaystyle 2BA = A^\top A+BB^\top = AB + BA, \)

\(\displaystyle BA=AB, \)

so the answer is YES.

If \(\displaystyle B=A^\top\) and \(\displaystyle AB=BA\) then \(\displaystyle A^\top A=A^\top A\),so \(\displaystyle A\) must be a normal matrix. Then the condition is satisfied as

\(\displaystyle A^\top A + BB^\top = AB + BA = 2AB. \)

Solution 2. Observe that the desired equality can be written as

\(\displaystyle (A - B^\top)^\top(A - B^\top) = AB - A^\top B^\top\)

where the left-hand side is a symmetric and non-negative definite matrix over \(\displaystyle \mathbb{R}\), meaning that it has only real non-negative eigenvalues. On the other hand, for the right-hand side we can compute

\(\displaystyle \tr (AB - A^\top B^\top) = \tr (AB) - \tr (A^\top B^\top) = \tr (AB) - \tr (BA)^\top = 0. \)

Hence, the matrix \(\displaystyle (A - B^\top)^\top(A - B^\top) = AB - A^\top B^\top\) must have all its non-negative eigenvalues equal to \(\displaystyle 0\), and because it is a symmetric matrix with real entries and thus a diagonalizable matrix, it has to be the zero matrix. This means that \(\displaystyle A - B^\top = 0\) so that \(\displaystyle A = B^\top\). Then, the solution can be completed like in the first solution.