Does $tr{\mathscr{L}_X}A=0$ and $tr(A{\mathscr{L}_X}A)=0$ imply $tr A=0$?

Question

Let $A$ be a $(1,1)$-tensor on Riemannian manifold $(M,g)$ and $X$ a Killing vector field. Does $tr{\mathscr{L}_X}A=0$ and $tr(A{\mathscr{L}_X}A)=0$ imply $tr A=0$ where $\mathscr{L}_X$ denote Lie derivation along vector field $X$.

Answer 1

Certainly not. For example, if $A$ is the identity endomorphism of the tangent bundle, then $\mathscr L_X A\equiv 0$ for any vector field $X$, but $\operatorname{tr} A$ is equal to the dimension of the manifold.