 Solution

Write $D_t={\nabla }_{\dot{\gamma }}$ and let $P_t:T_{p}M\to T_{\gamma (t)}M$ denote parallel transport. Since $\gamma$ is a geodesic, $D_t\dot{\gamma }=0$, while the transported vectors satisfy $D_t{e}_i(t)=0$. The locally symmetric Riemannian manifold condition is $\nabla R=0$. Hence the Levi-Civita connection product rule yieldsBoth this vector field and ${\lambda }_i{e}_i(t)$ are parallel and agree at zero. Uniqueness of parallel transport therefore provesThe eigenvalues are independent of $t$. The original PDF has $K_{\dot{\gamma }(t)}$ here; the converted TeX omits the dot, which would incorrectly put a point rather than a tangent vector in the operator's subscript.