IMC 2020 · Day 1 · P2

Let $X = AB - BA$. The first important observation is that $$\operatorname{trace}(X^2) = \operatorname{trace}(ABAB - ABBA - BAAB + BABA) = 2 \operatorname{trace}(ABAB) - 2 \operatorname{trace}(A^2 B^2)$$ using that the trace is cyclic. So we need to prove that $\operatorname{trace}(X^2) = n(n-1)$.

By assumption, $X + I$ has rank one, so we can write $X + I = v^t w$ for two vectors $v, w$. So $$X^2 = (v^t w - I)^2 = I - 2 v^t w + v^t w v^t w = I + (w v^t - 2) v^t w.$$ Now by definition of $X$ we have $\operatorname{trace}(X) = 0$ and hence $w v^t = \operatorname{trace}(w v^t) = \operatorname{trace}(v^t w) = n$ so that indeed $$\operatorname{trace}(X^2) = n + (n - 2) n = n (n - 1).$$ An alternative way to use the rank one condition is via eigenvalues: Since $X + I$ has rank one, it has eigenvalue 0 with multiplicity $n - 1$. So $X$ has eigenvalue $-1$ with multiplicity $n - 1$. Since $\operatorname{trace}(X) = 0$ the remaining eigenvalue of $X$ must be $n - 1$. Hence $$\operatorname{trace}(X^2) = (n-1)^2 + (n-1) \cdot 1^2 = n (n-1).$$