I have a matrix $A\in R^{n\times n}$ which is not symmetric. I am considering a quadratic form $$ \mathbf{x}^{T} A \mathbf{x}$$ where $\mathbf{x} \in V \subset R^{n}$ ($V$ is a linear subspace of $R^n$). I can prove that the quadratic form is negative definite when it is restricted to $V$ (this is not true more generally in $R^n$). Is this enough to claim that $V$ is a subspace of the span of the eigenvectors of $A$ associated with negative-real-part eigenvalues?
More information: When considered on the entire space $R^n$ the matrix $A$ is only negative semi-definite, however when $\mathbf{x}$ is restricted to be in $V$ the quadratic form is negative definite. This is because the null space of $A$ is a complementary space to the range of $A$ in $R^n$ - this is something I have proved but it is a bit cumbersome. I want to figure out if just showing the fact that the quadratic form is negative definite on $V$ (this is something that is pretty easy for the problem I'm considering) is enough to conclude that the range of $A$ is associated with negative-real-part eigenvalues.