I'm trying to prove or disprove the following claim:
Let $A\in \mathbb R ^{n\times n}$ be some symmetric real matrix. Suppose that there exist two linearly independent vectors $u$ and $v$ such that $u^TAu<0$ and $v^TAv<0$.
Can we deduce that $A$ has at least two negative eigenvalues? I did show that there is at least one negative eigenvalue, by using the spectral decomposition theorem (we have $0>u^TAu=\sum \lambda_i (A)\tilde u^2$ for a vector $\tilde u$, which implies that at least one of the eigenvalues is negative). Now I'm trying to use the linear independence property of the two vectors, but with no result.
I've encountered this while trying to classify the Hessian matrix of a function in one of my models. Any help will be much appreciated.
No: consider $$ A = \begin{pmatrix} \epsilon & 0 \\ 0 & -1 \end{pmatrix} , $$ $0<\epsilon<1$. Then $u = (1,-1)$ has $u^TAu = \epsilon -1$, $v=(1,1)$ has the same value, and clearly $u$ and $v$ are linearly independent (orthogonal, even) but $A$ has only one negative eigenvalue.