Let $u_1,...,u_k\in\mathbb{R}^n$ such that there is a non-zero quadratic form $Q$ satisfying $Q(u_i)=1$ for all $i=1,...,k$. Is there a positive definite quadratic form satisfying the same equations?
I know that the set of positive definite forms is open (and convex?) in the space of quadratic forms (considered as $\mathbb{R}^{\frac{1}{2}n(n+1)}$). Also, I tried considering the above equations as $k$ equations in $\frac{1}{2}n(n+1)$ variables, but not really sure how this helps, since these aren't homogenous equations.
Any help would be appreciated.
It seems the following.
A negative answer is suggested because a quadratic form $Q$ is positively definite iff a set $$\{u\in\Bbb R^n:Q(u)=1\}$$ is bounded, which can be seen considering a canonical form $$Q(x_1,\dots,x_n)=\lambda_1x_1^2+\dots +\lambda_nx_n^2.$$ Also we can present an explicit example. Let $n=2$ and $Q(x,y)=x^2-2y^2$. Then $$Q(1,0)=Q(3,2)=Q(17,12)=1.$$ Let $P(x,y)=ax^2+bxy+cy^2$ be a quadratic form such that $$P(1,0)=P(3,2)=P(17,12)=1.$$ Then $a=1$ and
$\cases{ 9+6b+4c=1\\ 289+204b+144c=1}$
$\cases{3b+2c=-4\\ 17b+12c=-24}$
$\cases{ 18b+12c=-24\\ 17b+12c=-24}$
so $b=0$ and $c=-2$. Hence $P=Q$.