Linear combination of two vectors that zeroes out the quadratic form of matrix. Let $A$ be a given $N\times N$ hermitian matrix. Let $v$ be a vector such that $v^HAv&gt;0$ and $w$ be a vector such that $w^HAw&lt;0$ ($w^H$ denotes conjugate transpose of $w$). Assume both $v$ and $w$ are unit-norm. Does there exist any non-zero vector $x$ which is a linear combination of $v$ and $w$, such that $x^HAx=0$?
A non-constructive proof:  

Let $Q(\mathbf u)=\mathbf u^HA\mathbf u$. Since $Q(k\mathbf u)=\left|k\right|^2Q(\mathbf u)$, we know that $\mathbf w$ is not a scalar multiple of $\mathbf v$. Consider a parametrization $\mathbf c(t)$ ($t\in\mathbb R$) of the unit circle in the plane spanned by $\mathbf v$ and $\mathbf w$ with $\mathbf c(a)=\mathbf v$ and $\mathbf c(b)=\mathbf w$. $Q(\mathbf c(t))$ is a continuous function of $t$, so there must be some $t_0\in(a,b)$ for which $Q(\mathbf c(t_0))=0$.
A constructive proof:

$\exists \lambda = -\frac{v^HAv}{w^HAw} \in \mathbb{R}^*_+$ such as $v^HAv+\lambda w^HAw = 0$.

if $w^H Av =0$ then we can construct $u=v+\lambda^{1/2}w$ which verify:

$u^HAu=v^HAv+\lambda w^HAw + \lambda^{1/2}(w^H Av+v^H Aw)=0$

else let $\tilde w = i(w^HAv) w$ then $ \tilde w^H A \tilde w  = (-i)(i)|w^H Av|^2 w^H Aw &lt; 0$

so there exists $v$ and $\tilde w$ and $\tilde \lambda= -\frac{v^HAv}{\tilde w^HA\tilde w}  \in \mathbb{R}^*_+$ such as $v^HAv+\lambda \tilde w^HA\tilde w =0$ in one hand.

In other hand  $\tilde w^HAv = (-i) (\overline{w^HAv}) (w^HAv) =- \overline{\tilde w^HAv}$.

So $(v+{\lambda}^{1/2}\tilde w)^HA(v+{\lambda}^{1/2}\tilde w) = v^HAv+\tilde \lambda \tilde w^HA\tilde w + \tilde \lambda^{1/2}(\tilde w^HAv + \overline{\tilde w^HAv}) = 0$