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Romanian Mathematical Olympiad

Algebra Difficulty 5.5 AIME, harder Prove it Romania

Suppose $A, B \in \mathcal{M}_n(\mathbb{C})$ are such that $A + B = AB + BA$ . Prove that:
a) if $n$ is odd, then $\det(AB - BA) = 0$ ;
b) if $\det(A) \ne \det(B)$ , then $\det(AB - BA) = 0$ .

Solution

a) Define $C = 2A - I_n$ and $D = 2B - I_n$ . We have $CD - I_n = -(DC - I_n) = 2(AB - BA)$ . As $n$ is odd, $\det(CD - I_n) = -\det(DC - I_n)$ .
As for any two square matrices $X, Y$ , we have $\det(XY - I_n) = \det(YX - I_n)$ , in our case we get $\det(CD - I_n) = -\det(CD - I_n)$ , that is $\det(CD - I_n) = 0$ . Consequently $\det(AB - BA) = 0$ .

b) Define $E = AB - BA$ . Suppose $\det(E) \neq 0$ . We have $AE + EA = A^2B - BA^2 = A(A + B - BA) - (A + B - AB)A = AB - BA = E$ . As $E$ is invertible, $E^{-1}AE + A = I_n$ . Using the property $\operatorname{tr}(E^{-1}AE) = \operatorname{tr}(A)$ , we get $\operatorname{tr}(A) = \frac{n}{2}$ . By symmetry, $\operatorname{tr}(B) = \frac{n}{2}$ , implying $\operatorname{tr}(A) = \operatorname{tr}(B)$ , in contradiction with the condition b). That is $\det(AB - BA) = 0$ .

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