International Mathematics Competition
for University Students
2021

IMC2021: Day 2, Problem 5

Problem 5. Let \(\displaystyle A\) be a real \(\displaystyle n\times n\) matrix and suppose that for every positive integer \(\displaystyle m\) there exists a real symmetric matrix \(\displaystyle B\) such that

\(\displaystyle 2021B=A^m+B^2.\)

Prove that \(\displaystyle |\det{A}|\le 1\).

Rafael Filipe dos Santos, Instituto Militar de Engenharia, Rio de Janeiro

Solution. Let \(\displaystyle B_m\) be the corresponding matrix \(\displaystyle B\) depending on \(\displaystyle m\):

\(\displaystyle 2021B_m=A^m+B_m^2. \)

For \(\displaystyle m=1\), we obtain \(\displaystyle A=2021B_1-B_1^2\). Since \(\displaystyle B_1\) is real and symmetric, so is \(\displaystyle A\). Thus \(\displaystyle A\) is diagonalizable and all eigenvalues of \(\displaystyle A\) are real.

Now fix a positive integer \(\displaystyle m\) and let \(\displaystyle \lambda\) be any real eigenvalue of \(\displaystyle A\). Considering the diagonal form of both \(\displaystyle A\) and \(\displaystyle B_m\), we know that there exists a real eigenvalue \(\displaystyle \mu\) of \(\displaystyle B_m\) such that

\(\displaystyle 2021\mu = \lambda^m+\mu^2 \Rightarrow \mu^2-2021\mu+\lambda^m=0. \)

The last equation is a second degree equation with a real root. Therefore, the discriminant is non-negative:

\(\displaystyle 2021^2-4\lambda^m \ge 0 \Rightarrow \lambda^m \le \frac{2021^2}{4}. \)

If \(\displaystyle |\lambda|>1\), letting \(\displaystyle m\) even sufficiently large we reach a contradiction. Thus \(\displaystyle |\lambda|\le1.\)

Finally, since \(\displaystyle \det{A}\) is the product of the eigenvalues of \(\displaystyle A\) and each of them has absolute value less then or equal to \(\displaystyle 1\), we get \(\displaystyle |\det{A}|\le 1\) as desired.

Solution. Different solution can be found in paper s2002