Maybe I could post this as a linear algebra problem but I'll give some context.
I know that if $(U, x_1, \ldots, x_n)$ is a local chart of a smooth manifold $M$ I can write locally a Riemannian metric on $M$ as: $$g=\sum_{i, j=1}^m g_{ij}dx_i\otimes dx_j.$$ Here we have a matrix $(g_{ij})$, $i, j=1, \ldots, m$. I read this matrix is positive definite but I can't see why. I know I must check that if $v=(v_1, \ldots, v_m)\neq 0$ then $$\sum_{i, j=1}^m g_{ij} v_iv_j>0.$$ But for me it is not clear why the above sum will be positive. The problem is the terms $g_{ij}v_iv_j$ when $i\neq j$.
Can anyone help me?
I just encounter the same question, and I think the reason is $g$ by definition is a inner product. So let $(\frac{\partial}{\partial x_i})$ by basis of $T_pM$, and $g_{ij}=g(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}),v=(v_1, \ldots, v_m)^t$ then
$$v^t(g_{ij})v=\sum_{i, j=1}^m g_{ij} v_iv_j=g(v,v)\ge0$$
The first equality holds for matrix multiplication and the second holds since $g$ by definition is inner product hence bilinear, and the inequality becomes equality iff $v=0$. Hence $( g_{ij})$ is positive definite.