Given a smooth symmetric bilinear form over the trivial normal bundle of the oblique manifold $\mathcal{OB}(n,m)$ imbedded in $\mathbb{R}^{n \times m}$ is there an efficient way to determine whether it is a definite form?
The specific form I'm looking at is $\langle x, x A \rangle_F$, $A \in \mathbb{R}^{m \times m}$, $A^T = A$ and $\langle, \rangle_F$ is the Frobenius inner product.
I know how to do this for the form $\langle x, Ax \rangle$ on the normal bundle of the sphere $S^{n-1}$: since the normal bundle to the sphere contains every unit vector, the question reduces to determining whether A is positive definite over $\mathbb{R}^n$ (which can be determined from the eigenvalues of A).
However, I was finding it difficult to find a similar approach for the oblique manifold. An idea I had was finding a set of orthogonal vectors from the normal bundle, then performing a change of basis $P$ such that these orthogonal vectors are eigenvectors of $P^{-1} A P$. My thought was then definiteness would simply be determined by the eigenvalues of $P^{-1} A P$, but I wasn't sure if this would work.
Edit 1: I've been working on this some more, would Sylvester's law of inertia imply definiteness of the form is equivalent to the definiteness of A over $\mathbb{R}^{n \times m}$?
Edit 2: It turns out my thought from edit 1 is incorrect. I've found an interesting result, for the 2x2 case, if $A = \begin{bmatrix}a & b \\c & d \end{bmatrix}$, A being positive semidefinite over the normal bundle of the oblique manifold is equivalent to $a + d > |b+c|$. I'll see if there is a simple way to generalize this.