What is the intuition behind Gramian method for linear independence? and Is there $simple$ proof of it? I'm trying to figure out the intuition behind Gramian method to determine the linear independence of functions.
I searched the web for such simple intuitive explanation and found nothing.
I tried also to find $"simple"$ proof for it and found nothing.
I tried to prove it in a simple way and here what I did:
If $f(x)$ and $g(x)$ are lineary dependent on interval $(a,b)$, then $f(x)=cg(x):c\ne0$.
$$<f,f>=\int_a^bf.fdx=c^2\int_a^bg^2dx$$
$$<f,g>=\int_a^bf.gdx=c\int_a^bg^2dx$$
$$<g,f>=\int_a^bg.fdx=c\int_a^bg^2dx$$
$$<g,g>=\int_a^bg.gdx=\int_a^bg^2dx$$
Gramian determinant is then:
$$G_{(f,g)}=det\left(\begin{matrix} <f,f> & <f,g>\cr <g,f> & <g,g> \end{matrix}\right)=det\left(\begin{matrix} c^2\int_a^bg^2dx & c\int_a^bg^2dx\cr c\int_a^bg^2dx & \int_a^bg^2dx \end{matrix}\right)$$
$$G_{(f,g)}=\left(\left(c^2\int_a^bg^2dx\right) \int_a^bg^2dx\right)-\left(c\int_a^bg^2dx\right)^2=0$$
So, the question is: What is the intuition behind Gramian method? and is my work valid as a $"proof"$?
 A: Consider integrable functions of the form $f_{i}:[a,b]\to\mathbb{R}$ with $i:1,\,2\,\cdots,\,n$. Now assume that the set containing the functions $f_i$ is linearly dependent, then there exist real coefficients $c_i$ (with not all zero)  such that
$$
\sum_{i=1}^{n}c_{i}f_i(x)=0.
$$
Define a vector containing the coefficient as $\gamma=[c_1,\,c_2,\cdots,\,c_n]$, and a vector containing the functions as $F(x)=[f_1(x),\,f_2(x),\cdots,\,f_n(x)]$. Take the squared $\mathcal{L}_2$-norm of the previous expression, which yields
$$
\int_{a}^{b}\Big(\sum_{i=1}^{n}c_{i}f_i(x)\Big)^2\text{d}x=
\gamma\int_{a}^{b}F^\top(x)F(x)\text{d}x\,\gamma^\top=\gamma\, G\,\gamma^\top=0.
$$
This means that the (gramian) matrix $G$ is singular. For the contrary, suppose that the set is linearly independent, then such matrix has to be non-singular, otherwise there exist constants $c_i$ which make the quadratic form zero. That is why the non-singularity of the gramian is a sufficient and necessary condition for linear independence. 
A: you are keeping the functions a bit too long. 
$$G_{(f,g)}=\det\left(\begin{matrix} c^2\int_a^bg^2dx & c\int_a^bg^2dx\cr c\int_a^bg^2dx & \int_a^bg^2dx \end{matrix}\right)$$
You have a two by two matrix, a scalar multiplier, in your case $\int_a^bg^2dx,$ multiplies the determinant by its square. That is
$$G_{(f,g)}=\det\left(\begin{matrix} c^2\int_a^bg^2dx & c\int_a^bg^2dx\cr c\int_a^bg^2dx & \int_a^bg^2dx \end{matrix}\right) = \left(\int_a^bg^2dx \right)^2 \det\left(\begin{matrix} c^2 & c\cr c & 1  \end{matrix}\right) = 0$$
The integrals and functions are a bit of an illusion. You are dealing with an (unspecified) finite dimensional, call it $n,$ vector space of continuous functions. It has a basis, and this basis can be transformed to an orthonormal basis by the Gram-Schmidt process.  Then your $n$ by $n$ matrix of integrals is simply expressed in terms of matrices, as is the Gramian. If the matrix $P$ has $n$ columns, all real numbers, and each entry $p_{ij} = \langle e_i, v_j \rangle$ is where $e_i$ make an orthonormal basis, then $P$ has nonzero determinant if and only if the $v_j$ are independent, and the same holds for the Gramian $P^TP.$ However, in this case, the entries of the Gramian are  $p_{ij} = \langle v_i, v_j \rangle.$
All square matrices of numbers, the functions and integrals go away early.
