Prove that there exist constants $u,v$ such that $uA+vB$ is positive definite. $A, B$ are $n$ by $n$ symmetric real matrices where $x^TAx=x^TBx=0$ implies $x=0$.
Prove that there are real numbers $u, v$ making $uA+vB$ postive-definite.
I am not sure whether it is true or not. I tried to disprove it but found it hard to find an $x$ on unit sphere killing both forms. Any help is appreciated. Thx.

It isn't true. Counterexample:
$$
A=\pmatrix{1&0\\ 0&-1},\ B=\pmatrix{0&1\\ 1&0}.
$$
Clearly, $x^TAx=0$ if and only if $x=(t,\pm t)^T$, and $x^TBx=0$ if and only if $x=(t,0)^T$ or $(0,t)^T$. So, the only solution to $x^TAx=x^TBx=0$ is the zero vector. However, $uA+vB$ is never positive definite because it has a zero trace.
