IMC 2026 · Day 1 · P2

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

$$\begin{gathered} 0 = \operatorname{tr} \left( \mathrm{A}^{\top} \mathrm{A} + \mathrm{B} \mathrm{B}^{\top} - \mathrm{A} \mathrm{B} - \mathrm{B} \mathrm{A} \right) = \operatorname{tr} \left( \mathrm{A}^{\top} \mathrm{A} \right) + \operatorname{tr} \left( \mathrm{B} \mathrm{B}^{\top} \right) - \operatorname{tr} (\mathrm{A} \mathrm{B}) - \operatorname{tr} (\mathrm{B} \mathrm{A}) \\ = \sum_{i, j = 1}^{n} a_{i j}^{2} + \sum_{i, j = 1}^{n} b_{j i}^{2} - 2 \sum_{i, j = 1}^{n} a_{i j} b_{j i} = \sum_{i, j = 1}^{n} \left( a_{i j} - b_{j i} \right)^{2}. \end{gathered}$$ Hence, $a_{i j} = b_{j i}$ for every $i, j$, so $B = A^{\top}$.

By substituting $B = A^{\top}$ back into the condition, we get $$\begin{gathered} 2 B A = A^{\top} A + B B^{\top} = A B + B A, \\ B A = A B, \end{gathered}$$ so the answer is YES.

Remark. If $B = A^{\top}$ and $A B = B A$ then $A^{\top} A = A^{\top} A$,so $A$ must be a normal matrix.

Then the condition is satisfied as $$A^{\top} A + B B^{\top} = A B + B A = 2 A B.$$

Observe that the desired equality can be written as $$\left( A - B^{\top} \right)^{\top} \left( A - B^{\top} \right) = A B - A^{\top} B^{\top}$$ where the left-hand side is a symmetric and non-negative definite matrix over $\mathbb{R}$, meaning that it has only real non-negative eigenvalues. On the other hand, for the right-hand side we can compute $$\operatorname{tr} \left( \mathrm{A} \mathrm{B} - \mathrm{A}^{\top} \mathrm{B}^{\top} \right) = \operatorname{tr} (\mathrm{A} \mathrm{B}) - \operatorname{tr} \left( \mathrm{A}^{\top} \mathrm{B}^{\top} \right) = \operatorname{tr} (\mathrm{A} \mathrm{B}) - \operatorname{tr} (\mathrm{B} \mathrm{A})^{\top} = 0.$$ Hence, the matrix $\left( A - B^{\top} \right)^{\top} \left( A - B^{\top} \right) = A B - A^{\top} B^{\top}$ must have all its non-negative eigenvalues equal to 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 $A - B^{\top} = 0$ so that $A = B^{\top}$. Then, the solution can be completed like in the first solution.