Showing that a bilinear form is positive-definite

Question

Let $V$ be a finite-dimensional $\mathbb{R}$-vector space and $\beta$ a symmetrical Bilinear form on $V$. $\beta(v,v) \neq 0$ for $v \neq 0$. Also there is a certain $v_0 \in V$ , such that $\beta(v_0,v_0)=2022$. Now I have to show that this Bilinear form is positive-definite.

I know that positive-definite means that $\langle v,v \rangle > 0$. Now this is true for this certain $v_0$ because $\langle v_0,v_0 \rangle=2022 > 0$ , but how would I show that this is true for all $v \in V , v \neq 0$. Is there a property that I'm missing that leads to the answer?

Answer 1

Since $V$ is finite-dimensional, you can use the Gram-Schmidt algorithm to construct an orthonormal basis $\{v_1,\ldots,v_n\}$ with respect to $V$. Then, assuming that any $v\in V$ satisfies $\beta(v,v)<0$, there must be some $v_i$ for which $\beta(v_i,v_i)<0$. See if you can work it out from there.

Answer 2

If a quadric form vanishes only for $v=0$, hence it is positive (or negative) definite. Since it is positive in $v_0$, it is positive everywhere.

Indeed, by contradiction, assume it exists $w\in V$ such that $-\alpha:=\beta(w,w)<0$. Remark that $v_0$ and $w$ are linearly indipendent, since $\beta(\lambda v_0,\lambda v_0)=\lambda^2\beta(v_0,v_0)\ge0$, $\forall\lambda\in\mathbb R$. It follows that $x_t=tw+v_0\ne0$, $\forall t\in\mathbb R$.

$$\beta(x_t,x_t)=-\alpha t^2+2t\,\beta(w,v_0)+2022.$$ Since the discriminant $\Delta=b^{2}-4ac=b^{2}+2022\alpha>0$, it exists $t$ such that $\beta(x_t,x_t)=0$, which contradicts the hypohesis $\beta(v,v)=0 \iff v=0$.

Answer 3

Take any $v$ in the space, different from any multiple of $v_0. $ The function $$f(t)= \beta((1-t)v_0+tv, (1-t)v_0+tv), \quad 0\le t\le 1$$ is continuous and $f(0)>0.$ By assumption $f(t)$ does not vanish, because $(1-t)v_0+tv\neq 0.$ Hence $f(t)$ constantly positive. In particular $\beta(v,v)=f(1)>0.$