So, I read something in a proof that I didn't quite understand. It seems like the author uses an argument that, writing it as a statement, looks the following way:
Assume $V$ is a real vector space with a finite dimension and $\langle - , - \rangle$ a symmetric bilinear form on $V$. Furthermore, the gramian matrix of $\langle - , - \rangle$ is a diagonal matrix that only posesses positive elements. Then, the bilinear form is positive definite.
The conclusion simply means that $\langle v , v \rangle \gt 0$ for every $v \in V, v \neq 0$. But why is that? The question itself seems rather simple, but I don't see the connection between the elements of the gramian matrix being positive and the image of every $v$ being positive.
Let $\{v_{1},\ldots,v_{n}\}$ be a basis of $V$. Since the $(i,j)$-th entry of the Gramian matrix is $\langle v_{i},v_{j}\rangle$, we know that $$\langle v_{i},v_{j}\rangle=0\ \text{for}\ i\neq j$$ and $$\langle v_{i},v_{i}\rangle> 0\ \text{for all}\ i.$$
Let $v=a_{1}v_{1}+\cdots+a_{n}v_{n}$ be a nonzero vector. Using bilinearity of $\langle\ ,\ \rangle$ and the observations just made: $$\langle v, v\rangle= \langle a_{1}v_{1}+\cdots+a_{n}v_{n},a_{1}v_{1}+\cdots+a_{n}v_{n}\rangle=a_{1}^{2}\langle v_{1},v_{1}\rangle +\cdots + a_{n}^{2}\langle v_{n},v_{n}\rangle>0.$$