Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.5 AIME Prove it Romania

Let $X \in \mathcal{M}_2(\mathbb{C})$ be a matrix such that $X^{2023} = X^{2022}$ . Prove that $X^3 = X^2$ .

Solution

Let us denote $d = \det(X)$ and $\text{t = (X)}$ . From the assumption $X^{2023} = X^{2022}$ , we obtain $d^{2023} = d^{2022}$ . Therefore, $d \in \{0, 1\}$ .

If $d = 1$ , then $X$ is invertible. Then $X^{2022}$ is also invertible. We find $X = I_2$ . So, the relation $X^3 = X^2$ is verified.

Assume now $d = 0$ . The Cayley-Hamilton theorem implies $X^2 = tX$ . So we obtain $X^{n+1} = t^n X, \forall n \in \mathbb{N}^*$ . Hence $t^{2022}X = t^{2021}X$ . We get $t = 0$ or $t = 1$ or $X = O_2$ .

If $t = 0$ , then $X^2 = O_2$ , so $X^3 = X^2 = O_2$ .

If $t = 1$ , then $X^2 = X$ , so $X^3 = X^2$ .

If $X = O_2$ , then clearly $X^3 = X^2 = O_2$ .

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