Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME, harder Prove it Romania

Let $A$ , $B \in \mathcal{M}_2(\mathbb{C})$ be two non-zero matrices with $AB + BA = O_2$ and $\det(A + B) = 0$ . Prove that $\text{tr}(A) = \text{tr}(B) = 0$ .

Solution

So, subtracting we get $\text{tr}(A)B = \text{tr}(B)A$ .
If $\text{tr}(A) = 0$ , then $\text{tr}(B) = 0$ for else $A = O_2$ , a contradiction.
Hence $A = \lambda B$ , so $AB + BA = O_2$ leads to $\lambda B^2 = O_2$ . Therefore $\lambda = 0$ , which yields $\text{tr}(A) = \text{tr}(B) = 0$ .

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