I've been working with Riemannian manifolds $(M,g)$ of dimension $n$, with Riemannian connection $\nabla$, and I've encountered a problem that I can't seem to solve. I would appreciate any help you could provide.
Here's the scenario:
Given are two vector fields $X$ and $Y$ on the manifold, and a constant, invertible $n \times n$ matrix $A$. The Levi-Civita connection (i.e., the Riemannian connection) $\nabla$ provides a way to differentiate vector fields along other vector fields. So I understand that if we have two vector fields $X$ and $Y$, the Riemannian connection $\nabla_XY$ will give us a new vector field.
My question pertains to how this connection interacts with the multiplication of a vector field by a constant matrix.
Specifically, suppose we have a vector field $Y$ and we multiply it by a constant matrix $A$ to get $AY$. If we then apply the Riemannian connection $\nabla_X$ to this product, will the result be the same as if we first applied the connection to $Y$ and then multiplied the result by $A$?
Formally, my question is: Can we say that $\nabla_X(AY)=A\nabla_XY$ for all vector fields $X$ and $Y$ and for all constant matrices $A$?
I understand that the Riemannian connection is a local derivative operator, so intuitively, I would expect that it can be distributed over constant matrix multiplication like this, but I haven't been able to prove it, nor find a counterexample. I would be grateful if you could confirm whether this equation holds in general, and if so, could you please provide a detailed explanation or proof? If it doesn't, a counterexample would be highly appreciated.
Thank you for your help.