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Romanian Mathematical Olympiad

Algebra Difficulty 6.5 National Olympiad Prove it Romania

Let $m \ge n \ge 2$ be integers, and let $A \in \mathcal{M}_{m,n}(\mathbb{C})$ and $B \in \mathcal{M}_{n,m}(\mathbb{C})$ , be two matrices such that there exist $k \in \mathbb{N}^*$ , and $a_0, a_1, \dots, a_k \in \mathbb{C}$ , such that $a_k(AB)^k + a_{k-1}(AB)^{k-1} + \dots + a_1(AB) + a_0I_m = O_m$ , and $a_k(BA)^k + a_{k-1}(BA)^{k-1} + \dots + a_1(BA) + a_0I_n \ne O_n$ . Prove that $a_0 = 0$ .

Solution

Suppose that $a_0 \ne 0$ . Rewriting the given equality as
$AB \left( -\frac{a_k}{a_0}(AB)^{k-1} - \frac{a_{k-1}}{a_0}(AB)^{k-2} - \dots - \frac{a_1}{a_0}I_m \right) = I_m,$
we find that $AB$ is invertible, hence $\mathrm{rank}(AB) = m \ge n$ .
On the other hand, we have $m = \mathrm{rank}(AB) \le \min\{\mathrm{rank}A, \mathrm{rank}B\} \le n$ , therefore $m = n$ and, furthermore, $A, B$ and $BA$ are all invertible.
Now, consider the equality $a_k(AB)^k + a_{k-1}(AB)^{k-1} + \dots + a_1(AB) + a_0I_n = O_n$ and left multiply by $B$ and right multiply by $A$ . This yields $a_k(BA)^{k+1} + a_{k-1}(BA)^k + \dots + a_0(BA) = O_n$ . Multiplying by $(BA)^{-1}$ gives $a_k(BA)^k + a_{k-1}(BA)^{k-1} + \dots + a_0I_n = O_n$ , a contradiction.

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