Trying to prove that the Eigenvalues here are positive

Question

I'm giving a gram matrix $G$ where the first row of G is defined like this $ <v_1,v_1>,<v_1,v_2>...,<v_1,v_n>$and the second row is $<v_1,v_2>,<v_2,v_2>...<v_2,v_n> $ and so on ,with an inner product basis $B=\{v_1...v_n\}$ and also $<v,u>=[v]_B^TG[u]_B$ .I'm supposed to use this given to show that the Egienvalue of G must be positive ,so i thought ill do this $<x,x>=[x]_B^TG[x]_B=[x]_B^T\lambda x=\lambda||x||^2 \implies \lambda>0$,when $x$ is an Egienvector. But i'm not sure of this solution because of the change of coordinates with respect to the basis B.

Answer 1

You know that for any vector $v$, we have $$ \langle v,v \rangle = [v]_B^T G [v]_B $$ Now, suppose that $x$ is an eigenvector of $G$ with $\|x\| = 1$. There exists a $v$ such that $x = [v]_B$ (in particular, take $v = \sum_i x_i v_i$). We note that $$ \lambda = \lambda \|x\|^2 = x^T(\lambda x) = [v]_B^T G[v]_B = \langle v,v \rangle > 0 $$ So, all eigenvalues must be positive.