comparison of eigenvalue of solution of Riccati equation

Question

I'm trying to understand the lecture notes comparison theorems in Riemannian geometry of Eschenburg and have a problem about the proof of Rauch I. Let $E$ be an euclidean vector space and $S(E)$ the space of its self adjoint endomorphisms with a partial ordering by putting $ A \leq B $ if $ \langle Ax,x\rangle \leq \langle Bx,x\rangle , \forall x \in E$.

If $A_{i}: [0, t_{i}) ,\ i = 1, 2 $ are solutions of the Riccati equations $ A_{i} ' + A^2 _{i} + R_{i} = 0 $ with $ \lambda_{-} (R_{1} )\geq \lambda_{+} (R_{2} ) $, $ \lambda_{-} $ the lowest eigenvalue and $ \lambda_{+} $ the highest eigenvalue of the self adjoint operator $ R_{i}$, such that $ U := A_{2} -A_{1} $ has a continuous extension to $ 0$ with $U(0) \geq 0 $. Why does it hold that $ \lambda_{+} (A_{1} (t)) \leq \lambda_{-} (A_{2}(t)) $ on $ (0, t_{0} ) $?

Thanks a lot.

Answer 1

A classical proof can be found here:

H. L. Royden, MR 948079 Comparison theorems for the matrix Riccati equation, Comm. Pure Appl. Math. 41 (1988), no. 5, 739--746.

Actually, more general version of this statement holds for more general version of the matrix Riccati equation. A self-contained and perhaps shorter proof, including your case, can be found in Appendix A (Thms. 38 and 40) of

D. Barilari, L. Rizzi, MR3491778 Comparison theorems for conjugate points in sub-Riemannian geometry, ESAIM Control Optim. Calc. Var. 22 (2016), no. 2, 439--472.