Let $L\colon E\to E$ be an endomorphism of complex Hermitian vector bundle over a smooth manifold $M$. Suppose $L$ is self-adjoint and its eigenvalues are constant. How do I see that $E$ can be decomposed into a direct sum of eigenbundles.
For each fiber $E_x$ where $x\in M$, I can decompose $E_x$ into eigenspaces. However, how do I see that these eigenspaces form subbundles of $E$? What if the ranks of these eigenbundles are not locally constant and jump around?
My idea: If I can smoothly choose an eigensection $\sigma$ of $E$, i.e., $L(x)\sigma(x)=\lambda \sigma(x)$ for every $x\in M$. Then, I can consider the line bundle generated by $\sigma$ and its orthogonal complement $W$. Since $L$ necessarily maps $W$ into itself, I can choose another eigensection again. By induction, I am done. However, I don't know how to smoothly choose an eigensection at the beginning.
If I understand well, you assume that the eigenvalues $a_1,\dots,a_k$ of $L_x$ are independent of the point $x$. Then you can sort things out by using the the projection onto an eigenspace of a diagonalizable linear operator can be written as a polynomial in the operator. For each $i=1,\dots,k$, define $$ P_i:=(\prod_{j\neq i}(a_i-a_j)^{-1})\prod_{j\neq i}(L-a_j\cdot id_E), $$ where the second product is a composition of vector bundle homomorphisms $E\to E$. Clearly, this defines a vector bundle homomorphism $E\to E$. Using that $L_x$ is diagonalizable with eigenvalues $a_1,\dots,a_k$ you see that $(P_i)_x$ is the projection onto the $a_i$-eigenspace in $E_x$. In particular, in each point $x$, the ranks of the maps $(P_i)_x$ add up to the fiber dimension of $E$. Since locally, none of these ranks can drop, they have to be locally constant, so for each $i$, the image of $P_i$ is a smooth subbundle of $E$. (The same argument applies if the eigenvalues do depend on the point, as long as the stay pairwise different.)