International Mathematics Competition
for University Students
2024

IMC2024: Day 2, Problem 7

Problem 7. Let \(\displaystyle n\) be a positive integer. Suppose that \(\displaystyle A\) and \(\displaystyle B\) are invertible \(\displaystyle n\times n\) matrices with complex entries such that \(\displaystyle A+B=I\) (where \(\displaystyle I\) is the identity matrix) and

\(\displaystyle (A^2 + B^2)(A^4 + B^4) = A^5 + B^5. \)

Find all possible values of \(\displaystyle \det(AB)\) for the given \(\displaystyle n\).

Sergey Bondarev, Sergey Chernov, Belarusian State University, Minsk

Solution 1. Notice first that \(\displaystyle AB=A(I-A)=A-A^2=(I-A)A=BA\), so \(\displaystyle A\) and \(\displaystyle B\) commute.

Let \(\displaystyle C=AB=BA\); then

$$\begin{align*} A^2+B^2 &= (A+B)^2-2AB = I-2C, \\ A^4+B^4 &= (A+B)^4-4AB(A+B)^2+2A^2B^2 = I-4C+2C^2, \\ A^5+B^5 &= (A+B)^5-5AB(A+B)^3+5A^2B^2(A+B) = I-5C+5C^2, \end{align*}$$

so

$$\begin{align*} 0 &= (A^5+B^5)-(A^2+B^2)(A^4+B^4) = (I-5C+5C^2)-(I-2C)(I-4C+2C^2) \\ &= 4C^3-5C^2+C = 4C(C-I)(C-\tfrac14I); \end{align*}$$

since \(\displaystyle C\) is invertible, we have

\(\displaystyle (C-I)(C-\tfrac14I) = 0. \)

Hence, the polynomial \(\displaystyle p(x)=(x-1)(x-\tfrac14)\) annihilates the matrix \(\displaystyle C=AB\) and therefore all eigenvalues of \(\displaystyle C\) are roots of \(\displaystyle p(x)\), so the possible eigenvalues are \(\displaystyle 1\) and \(\displaystyle \tfrac14\). The determinant is the product of the \(\displaystyle n\) eigenvalues, so

\(\displaystyle \det(AB) = \det C\in \Big\{1,\tfrac14,\tfrac1{4^2},\ldots,\tfrac1{4^n} \Big\}. \)

Now show that these values are indeed possible.

If

\(\displaystyle A = \mathrm{diag}\Big( \underbrace{\tfrac12,\ldots,\tfrac12}_{k}, \underbrace{e^{i\pi/3},\ldots,e^{i\pi/3}}_{n-k}\Big) \quad\text{and}\quad B = \mathrm{diag}\Big( \underbrace{\tfrac12,\ldots,\tfrac12}_{k}, \underbrace{e^{-i\pi/3},\ldots,e^{-i\pi/3}}_{n-k}\Big), \)

then \(\displaystyle A+B=I\), \(\displaystyle AB= \mathrm{diag}\Big( \underbrace{\tfrac14,\ldots,\tfrac14}_{k}, \underbrace{1,\ldots,1}_{n-k}\Big)\) and \(\displaystyle \det(AB)=\tfrac1{4^k}\).