Show that the set ${f_1,\dots, f_n}$ is linearly dependent on $[a,b]$ if $\text{det}\left(\int_a^bf_i(x)\space f_j(x)\right)=0$ Let $f_1, f_2 \dots f_n$ be continuous real valued functions on $[a,b]$. Show that the set ${f_1,\dots, f_n}$ is linearly dependent on $[a,b]$ if 
$$\text{det}\left(\int_a^bf_i(x)\space f_j(x)\right)=0$$
If the determinant of this matrix (call it $M_{ij})$ is 0 then exists nonzero $v$ such that $Mv=0$. Then what? I tried forming the expression $0 = v^{T}Mv$ but not sure what that gets me....
 A: Your instinct of $Mv=0$ is a good one. The next leap to make is realizing that $(f, g) \mapsto \int^a_b fg$ is an inner product on the $\mathbb{R}$-space $C([a, b], \mathbb{R})$ (otherwise known as the set of continuous functions from $[a, b]$ to $\mathbb{R}$). We shall henceforth sensibly denote $\int^a_b fg$ as $\langle f, g \rangle$.
Our proof is going to rely on synchronizing two equations together.
The first one is an elaboration on what this "$Mv=0$" looks like. Remember that $M$ is of course our "Gram-Matrix"
$$\begin{bmatrix}
\langle f_1, f_1 \rangle & \dots &  \langle f_1, f_n \rangle \\
\vdots & \ddots & \vdots \\
\langle f_n, f_1 \rangle & \dots &  \langle f_n, f_n \rangle
\end{bmatrix},$$
and $v$ is some nonzer0 real-valued $n$-dimensional row vector $(v_1, \dots, v_n)$. As you observed, our hypothesis gives us
$$\begin{bmatrix}
\langle f_1, v_1f_1+ \dots+v_nf_n \rangle \\
\vdots \\
\langle f_n, v_1f_1+ \dots+v_nf_n \rangle
\end{bmatrix}=\begin{bmatrix}
v_1\langle f_1, f_1\rangle+ \dots+v_n\langle f_1, f_n \rangle \\
\vdots \\
v_1\langle f_n, f_1\rangle+ \dots+v_n\langle f_n, f_n \rangle
\end{bmatrix}=Mv=\begin{bmatrix}
0 \\
\vdots \\
0
\end{bmatrix},$$
for some $v$.
Our next equation is based on a backwards step on what we want to prove. Remember in this inner-product mumbo-jumbo that all we want is some $a_1, \dots, a_n$ such that $a_p \neq 0$ for some $p \in [n]$ such that $a_1f_1+\dots+a_nf_n=0$. This condition holds for $a_1, \dots a_n$ if and only if
$$\sum^n_{i=1}a_i\langle f_i, a_1f_1+\dots+a_nf_n \rangle=||a_1f_1+\dots+a_nf_n||^2=0.$$
Now once you stomach the notation, then the proof will simply be choosing your $v \neq 0$ such that $Mv=0$ and synchronizing the two equations together.
Hope this helps.
