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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?

Larrrrrrrrrry
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  • Such matrices exist, the $4 \times 4$ matrices given by the gamma matrices that generate a Clifford algebra are defined under the anticommutation relation $$ {\gamma^\mu, \gamma^\nu } = \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu} I_4 $$ where ${ , }$ is the anticommutator and $\eta^{\mu \nu}$ is the Minkowski metric with signatuure $(+ − − −)$, $I_4$ is the $4 × 4$ identity matrix. –  Sep 29 '16 at 14:14
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    $C=D=I/sqrt(2)$ – Empy2 Sep 29 '16 at 14:16

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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 $n^2$.