I am reading a paper, Yano and Ishihara, “Submanifolds with Parallel Mean Curvature Vector” (MSN), where the authors have constructed a linear operator, say $A$, on vector fields. They claim that this operator having locally constant eigenvalues is enough to imply that the distribution generated by the eigenvectors of $A$ is integrable. I am having trouble understanding this.

In order to have integrability, I believe we must show that $[v,w]$ is an eigenvector for any eigenvectors $v,w$ of $A$. Say that $\lambda_1,\lambda_2$ are the eigenvalues associated to $v,w$, respectively. The constant eigenvalue condition seems to imply that $$\nabla(Av) = A(\nabla v) = \lambda_1\nabla v,$$ but I do not understand how this implies the distribution is closed under Lie brackets. In particular, it is clear from the above that $[Av,Aw] = \lambda_1\lambda_2[v,w]$, and it is also clear that the bracket is $A$-linear in both arguments separately. With that said, why is it that $[v,w]$ is also an eigenvector of $A$? I do not see it.

EDIT: The operator $A$ is (as I understand it) not a priori a homomorphism of Lie algebras. It is something related to the shape operator of an immersed submanifold. The precise statement of the problem above can be found in the final paragraph on pg 108 of Yano and Ishihara, “Submanifolds with Parallel Mean Curvature Vector”.