I saw a solution of an olympiad problem that I didn't understand and I'm confused if it's actually correct. The problem is:
Let $n\in\mathbb{Z}_{\geq2}$ and $A,B\in M_n(\mathbb{C})$ so that there exists an idempotent matrix $C\in M_n(\mathbb{C})$ such that C*$=AB-BA$. Prove that $AB-BA=O_n$. (C* is the adjoint/reciprocal of the matrix, not the conjugate transpose).
The solution begins by considering the Jordan form of C, $C=S^{-1}JS$, where J is a diagonal matrix with $0$'s or $1$'s on the diagonal(because C is idempotent). If there is only $1$'s on the diagonal, $C=I_n$ and there is a contradiction taking the trace in the statement.
Here comes the part I don't understand. It says that if J has one $0$ on the diagonal, its adjoint has one $1$ on the diagonal and the rest are $0$'s, resulting in it having trace $1$ in contradiction with C*$=AB-BA$, and that if it has more than one zero on the diagonal, its adjoint is $O_n$, which supposedly gives the conclusion. I don't get what the connection between J* and C* is that makes us able to conclude what the solution says.
Using $\operatorname{adj}(XY)=\operatorname{adj}(Y)\operatorname{adj}(X)$, we get \begin{aligned} \operatorname{adj}(S^{-1}JS) &=\operatorname{adj}(S)\operatorname{adj}(J)\operatorname{adj}(S^{-1})\\ &=\left(\det(S)\frac{\operatorname{adj}(S)}{\det(S)}\right)\operatorname{adj}(J)\left(\det(S^{-1})\frac{\operatorname{adj}(S^{-1})}{\det(S^{-1}}\right)\\ &=\det(S)\det(S^{-1})\left(S^{-1}\operatorname{adj}(J)S\right)\\ &=S^{-1}\operatorname{adj}(J)S. \end{aligned} Alternatively, recall that if $p$ is the characteristic polynomial of a linear operator $A$ defined on an $n$-dimensional vector space, then $\operatorname{adj}(A)=q(A)$ where $q$ is the polynomial defined by $q(x)=(-1)^{n-1}\frac{p(x)-p(0)}{x-0}$. Now let $p$ be the common characteristic polynomial of $J$ and $S^{-1}JS$. Then $$ \operatorname{adj}(S^{-1}JS) =q(S^{-1}JS) =S^{-1}q(J)S =S^{-1}\operatorname{adj}(J)S. $$