# Spectral decomposition of the matrix ODE

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. $f(X,t) = AX + XA^{T} + XBX + C.$

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Perhaps if you have a specific form for $f$, but certainly not in general. This is because that if there were ODEs on $R$ and $D$ you will have that $R$ and $D$ must be continuously differentiable functions, but it is well-known that for an arbitrary smooth family of symmetric matrices, the collection of eigenfunctions (which is encoded in $R$) need not be even continuous. –  Willie Wong Mar 6 '12 at 9:20
@WillieWong, I think we can consider only linear-quadratic $f(X,t)$. –  Nimza Mar 6 '12 at 9:33
That's a time-varying Riccati equation and that's been studied quite extensively. –  user13838 Mar 6 '12 at 9:47
@percusse, so for this equation there exists such decomposition? –  Nimza Mar 6 '12 at 10:38