Explanation of a scalar product calculation

Question

Let $f_1, f_2, \ldots, f_n$ be continuous real valued functions on $[a,b]$. Show that the set $\{f_1,\ldots,f_n\}$ is linearly dependent on $[a,b]$ if and only if $$\det\left(\int\limits_a^b f_i(x)f_j(x)dx\right)=0$$

I can't understand the solution to the sufficient condition of this problem. The solution says

Let $G$ be the matrix with entries $$G_{ij}=\int\limits_a^b f_i(x)f_j(x)dx.$$

If the determinant of $G$ vanishes, then $G$ is singular; let $a$ be a non-zero $n$-vector with $Ga=0$. Then $$0=a^TGa=\sum\limits_{i=1}^n\sum\limits_{i=j}^n\int\limits_a^b a_if_i(x)a_jf_j(x)dx=\int\limits_a^b\left(\sum\limits_{i=1}^n a_if_i(x)\right)^2dx,$$ so, since the $f_i$'s are continuous functions, the linear combination $\sum a_if_i$ must vanish identically.

I have troubles replicating the scalar product calculation above. Let $a_i$ be the $i$-th component of $a$, then the $i$-th component of $Ga$ is $\sum\limits_{j=1}^n\int\limits_a^b f_i(x)a_jf_j(x)dx$ and so $$a^TGa=\sum\limits_{i=1}^n\sum\limits_{j=1}^n\int\limits_a^b a_if_i(x)a_jf_j(x)dx,$$ i.e. my summation is $\sum\limits_{i=1}^n\sum\limits_{j=1}^n$ rather than $\sum\limits_{i=1}^n\sum\limits_{i=j}^n$. Can someone please spot my mistake and explain the calculation in more details?

Answer 1

I see nothing wrong in your computations. This is probably due to a typo in the original proof. But yours is fine.

Answer 2

One can also see the matrix as $G=[\langle f_i,f_j\rangle]=A^TA$ where $A$ consists of the coefficients of $f_i$ with respect to an orthonormal basis. Thus $\det G=\det(A)^2$ is zero iff $\det A=0$, i.e., there is a non-trivial solution $Av=0$, $\sum_iv_if_i=0$.