IMC 2024 · Day 2 · P7

Hint: Find a polynomial $p(x)$ such that $p(AB) = 0$.

Notice first that $AB = A(I - A) = A - A^2 = (I - A)A = BA$, so $A$ and $B$ commute. Let $C = AB = BA$; 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 $$\begin{align*} 0 &= (A^5 + B^5) - (A^2 + B^2)(A^4 + B^4) = (I - 5C + 5C^2) - (I - 2C)(I - 4C + 2C^2) \\ &= 4C^3 - 5C^2 + C = 4C (C - I) \bigl( C - \tfrac14 I \bigr); \end{align*}$$ since $C$ is invertible, we have $$(C - I) \bigl( C - \tfrac14 I \bigr) = 0.$$ Hence, the polynomial $p(x) = (x - 1) \bigl( x - \frac14 \bigr)$ annihilates the matrix $C = AB$ and therefore all eigenvalues of $C$ are roots of $p(x)$, so the possible eigenvalues are 1 and $\frac14$. The determinant is the product of the $n$ eigenvalues, so $$\det(AB) = \det C \in \left\{ 1, \frac14, \frac{1}{4^2}, \dots, \frac{1}{4^n} \right\}.$$ Now show that these values are indeed possible.

If $$A = \operatorname{diag} \bigl( \underbrace{\tfrac12, \dots, \tfrac12}_{k}, \underbrace{e^{i\pi/3}, \dots, e^{i\pi/3}}_{n-k} \bigr) \quad \text{and} \quad B = \operatorname{diag} \bigl( \underbrace{\tfrac12, \dots, \tfrac12}_{k}, \underbrace{e^{-i\pi/3}, \dots, e^{-i\pi/3}}_{n-k} \bigr),$$ then $A + B = I$, $AB = \operatorname{diag} \bigl( \underbrace{\tfrac14, \dots, \tfrac14}_{k}, \underbrace{1, \dots, 1}_{n-k} \bigr)$ and $\det(AB) = \dfrac{1}{4^k}$.