Invariance of the rank of the trace of Riemannian curvature under a change of frame

Question

Let $$ R=\begin{pmatrix} R_{11} & ... & R_{1n} \\ &...\\ R_{n1} &...& R_{nn} \end{pmatrix}, $$ where $R_{ikjl}$ is curvature tensor of a Riemannian manifold $(M,g)$ and $R_{ij}=g^{kl}R_{ikjl}$. How to show the rank of $R$ is invariant under changing frame? I just guessed the result, I really don't know how to start to prove it .

Answer 1

This is essentially a question of linear algebra. In fact, the rank of any bilinear form is independent of basis: Under a change of basis (with change-of-basis matrix $P$), the matrix representation $[R]$ of a bilinear form $R$ transforms as $$[R] \mapsto P^T [R] P.$$ Since $P$ is invertible, it has full rank (and hence so does $P^T$), so the matrix representations of $R$ w.r.t. the two bases satisfy $$\operatorname{rank} [R] = \operatorname{rank} (P^T [R] P)$$ as claimed.