"Let $v_1, v_2, \ldots, v_m \in\mathbb{R}^n$ be linearly independent vectors. Define the matrix $A\in \mathbb{R}^{m\times m}$ by $a_{ i,j}=v_{i}^{T}v_{j}$, $i,j=1, 2, \ldots, m$. Show that $A$ is symmetric and positive definite."
I have a question about how I proceed in this activity, this matrix is symmetric due to the symmetric property of the Euclidean product $v_{i}^{T}v_{j}=v_{j}^{T}v_{i}$. But I don't know how I can show that it is definitely positive. Should I use the definition or some other result?
Following the definition, we have for $x\in \mathbb{R}^{m}-\{0\}$ that $x^TAx=(x_1v_{1}^T+x_{2}v_{2}^T+ \ldots+x_{m}v^T)(v_1x_1+v_2x_2+\ldots+v_mx_m)=\sum_{i=j=1}^{m}x_{i}^2\Vert v_i \Vert^2+\sum_ {i=1}^{m}\sum_{j=1,j\neq i}^{m}x_{i}v_{i}^Tv_{j}x_{j}$. The first term is positive, because the vectors $v_i$ are linearly independent, but couldn't the other term be negative?
We have $$(x_1v_{1}^T+x_{2}v_{2}^T+ \ldots+x_{m}v^T)(v_1x_1+v_2x_2+\ldots+v_mx_m) = (x_1v_{1}+x_{2}v_{2}+ \ldots+x_{m}v)^T(v_1x_1+v_2x_2+\ldots+v_mx_m)=l^Tl$$ where $l = x_1v_{1}+x_{2}v_{2}+ \ldots+x_{m}v.$ Now the result is easy: if $l=(y_1, ..., y_n)^T$, then $l^Tl=y_1^2+... +y_n^2$.