I am trying to solve the following question which I came across when studying root system in euclidean spaces, with positive definite symmetric bilinear form.
Statement: Given a basis $\{v_1,\cdots, v_n\}$ of $\mathbb{R}^n$, $\exists$ $v\in\mathbb{R}^n$ such that $(v,v_i)>0$ for all $i$
My answer: Let $v=a_1v_1 + \cdots + a_nv_n$. Then $$(v,v_i)=a_1(v_1,v_i) + a_2(v_2,v_i) + \cdots + a_n(v_n,v_i), \hskip5mm i=1,2,\cdots,n.$$ Writing this in matrix form we get $$ \begin{bmatrix} (v,v_1) \\ \vdots \\ (v,v_n)\end{bmatrix} = \begin{bmatrix} (v_1,v_1) & \cdots & (v_n,v_1)\\ \vdots & & \vdots \\ (v_1,v_n) & \cdots & (v_n,v_n) \end{bmatrix} \begin{bmatrix} a_1 \\ \vdots \\ a_n\end{bmatrix}. $$ We want to find $a_1,\cdots,a_n$ such that the left column vector contains positive entries only. Let $A$ be the square matrix appearing above; it is matrix of positive definite (symmetric) inner product w.r.t given basis, so it is invertible over $\mathbb{R}$. Take $$ \begin{bmatrix} a_1 \\ \vdots \\ a_n\end{bmatrix} := A^{-1}\begin{bmatrix} 1 \\ \vdots \\ 1\end{bmatrix}.$$ This is non-zero vector, and for this choice of $a_i$'s we get the vector $v=a_1v_1+\cdots + a_nv_n$ with desired properties.
Q.1 Is the above statement and proof correct?
Q.2 Is there geometric way to prove the statement?
Your idea looks fine, but as a comment has pointed out, there are loose ends (such as why the inner product matrix is invertible) to tie up.
For a geometric proof, just do something akin to Gram-Schmidt orthogonalisation. Take $v=v_1$, so that $(v,v_1)>0$. Let $u_2=v_2-\frac{(v_1,v_2)}{(v_1,v_1)}v_1$ be the component of $v_2$ that is orthogonal to $v_1$. Then $u_2\ne0$, or else $v_1$ and $v_2$ are linearly dependent. Add $c_2u_2$ to $v$ for a sufficiently large $c_2>0$. Then $(v,v_i)>0$ for $i=1,2$. Continue in this manner, you get a desired $v$.