I have a matrix ODE $\dot X = f(X,t)$ on the symmetric matrix with the initial condition $X(0) = X_0$, where $X_0$ is the positive definite and symmetric. Then by the spectral decomposition $X_0 = R_0^T D_0 R_0$, where $D_0$ is the diagonal and $R_0$ is the ortogonal matrix. Is there a possibility to receive ODE's on R and D such that $X = R^TDR$? I need this to control the positive definiteness of $X$.
We can consider only linear-quadratic $f(X,t)$, i.e. $f(X,t) = AX + XA^{T} + XBX + C.$