Verify that two or more functions are linearly independent on an interval If we want to verify that 1 and x are linearly independent on the interval [0,1]  , can we solve it as following ?
let
 $$c_1 +c_2  x=0$$
at x=0 
$$c_1=0$$
at x=1 
$$c_2=0$$
So 1 and x are linearly independent on the interval [0,1]
This is how my teacher solved it but I am not convinced because I think it is not sufficient to try two values of x only ! 
I know that we can prove that 1 and x are independent using Wronskian . 
My problem is : 
can we use the method of  $$c_1 y_1(x) + c_2 y_2(x) =0$$ to prove that two functions are independent on an interval by trying only 2 values of x and getting that all constants are zero ?! are two values of x sufficient ?! why ?! and must we try the values at beginning and end of the interval ?
 A: the statement $ay+by=0$ is an equality of functions which is a shortcut for saying $ay(x)+by(x)=0$ for every $x$. Thus by choosing $N$ different points $x$, you can create a system of $N$ simultaneous equations, which can then let you find your coefficients.
A: The demonstration for $1$ and $x$ is correct. To prove that two functions are linearly independent, you want to show that $c_1f(x) + c_2g(x)=0$ for all $x$ in some interval implies that $c_1 = c_2 = 0$. So clearly it suffices to find particular values of $x$ that force $c_1 = c_2 = 0$; this is what was done in the given example.
In the general case, if you want to show that two functions $y_1(x)$ and $y_2(x)$ are linearly independent, yes, it will always suffice to try two values of $x$, but they have to be the "right" values. For example, to show that $x$ and $x^2$ are linearly independent, you can't use $0$ and $1$; this gives $0=0$ and $c_1+c_2=0$. But if you choose $\frac{1}{2}$ and $1$, you will get the system
$$ \frac{1}{2}c_1 + \frac{1}{4}c_2 = 0,\quad c_1+c_2=0,$$
which as a pair of linear equations in $c_1$ and $c_2$ has the unique solution $c_1 =c_2 = 0$.
So in general, if you can find two values of $x$ such that the resulting system in $c_1$ and $c_2$ has nonzero determinant, so that it has only the trivial solution, you are done.
A: Two values are sufficient provided that we get two linearly independent equations. If we want to find $c_1$ and $c_2$ given that $c_1y_1(x)+c_2y_2(x)=0$, substituting two different values, say $0$ and $1$, gives
$$c_1y_1(0)+c_2y_2(0)=0$$
and
$$c_1y_1(1)+c_2y_2(1)=0$$
We can then solve these two equations simultaneously (provided they are not scalar multiples of each other) to find $c_1$ and $c_2$.
It is not necessary to try the two values at the beginning and end of the interval; any two values which give different equations will work.
In your original example: $c_1+c_2x=0$, you could instead choose $x=\frac{1}{3}$ and $x=\frac{1}{2}$, which yields the equations
$$c_1+\frac{1}{3}c_2=0 \text{ and } c_1 + \frac{1}{2}c_2=0.$$
Solving these two equations also gives $c_1=0, c_2=0$.
