I have seen a lot of $CD - DC = I$ questions, which is fairly easy to solve using trace, in here. But is it possible for $CD + DC = I$? Given that both C and D are square matrices, I have only managed to show that $Tr(CD) = n/2$. But I feel that this isn't good enough to conclusively prove that it is possible. Any advice on how to continue?
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$\begingroup$Assume that $C\in M_n(\mathbb{C})$ is a generic matrix and let $spectrum(C)=(\lambda_i)$. If you prefer, randomly choose $C=[c_{i,j}]$ where the $c_{i,j}$ are independent and follow the same normal law. Then, with probability $1$, the $\lambda_i+\lambda_j,i,j$ are non-zero complex numbers.
Consider the function: $f:X\in M_n\rightarrow CX+XC\in M_n$. Note that $f$ is linear and that $spectrum(f)=\{\lambda_i+\lambda_j|i,j\}$; thus $f$ is bijective and there is a unique $D$ s.t. $CD+DC=I$.
In other words, to each generic matrix $C$ you can associate a unique matrix $D$ s.t. $CD+DC=I$. Also, the complex algebraic set $\{(C,D)|CD+DC=I\}$ has dimension $n^2$.
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