# Proving that g1,g2,g3 are linearly independent

1. Feb 7, 2009

### vikkisut88

1. The problem statement, all variables and given/known data
Let V = {differentiable f:R -> R}, a vector space over R. Take g1,g2,g3 in V where g1(x) = e$$^{}x$$, g2(x) = e$$^{}2x$$ and g3(x) = e$$^{}3x$$.
Show that g1, g2 and g3 are distinct.

2. Relevant equations
If g1-g3 are linearly independent, it means that for any constant, k in F (field) then they all = 0 when g1k1 + g2k2 + g3k3 = 0.

3. The attempt at a solution
I have the idea to choose a value of x in R so that g1(x), g2(x), g3(x) are distinct, but I'm not exactly sure where to go from there because apart from choosing x=0 then for the rest of the time then they must be distinct as they all have different values. I'm just not sure how I prove this.

2. Feb 7, 2009

### CompuChip

If they are linearly dependent, there should be choices of {k1, k2, k3} such that k1 g1(x) + k2 g2(x) + k3 g3(x) = 0 for all x (by definition of the vector (k1 g1 + k2 g2 + k3 g3).

So probably the easiest way is to take x = 0 and x = ln(2), for example, and write down a system of equations for {k1, k2, k3} and show that only the trivial solution k1 = k2 = k3 = 0 is possible.

3. Feb 7, 2009

### HallsofIvy

Another way to do that is to use the Wronskian. If f(x), g(x), and h(x) are dependent functions, then there exist a, b, c, not all 0, so that af(x)+ bg(x)+ ch(x)= 0 for all x. One way to see if that is true is to choose 3 values for x, as CompuChip suggests, so you have three numeric equations you can solve and see if a=b=c= 0 is the only solution (independent) or if there are other solutions (dependent).

But with functions, you can also get three equations by differentiating. If af(x)+ bg(x)+ ch(x)= 0 for all x, then it is a constant and so its derivative is also 0: af'(x)+ bg'(x)+ ch'(x)= 0. Since that is also identically 0, it is a constant and so its derivative is 0" af"(x)+ bg"(x)+ ch"(x)= 0. Taking x to be any convenient value, say 0 here, we have the three equations af(0)+ bg(0)+ ch(0)= 0, af'(0)+ bg'(0)+ ch'(0)= 0, and af"(0)+ bg"(0)+ ch"(0)= 0. That system of equations will have a unique solution, and the functions will be independent, if and only if the determinant of the array of coefficients,
$$\left|\begin{array}{ccc}f(0) & g(0) & h(0) \\ f'(0) & g'(0) & h'(0) \\ f"(0) & g"(0) & h"(0)\end{array}\right|$$
the "Wronskian" of the functions at x= 0, is non-zero and the functions will be dependent if that determinant is 0.