Proving that g1,g2,g3 are linearly independent

In summary, the conversation discusses the linear independence of g1, g2, and g3 in the vector space V, using methods such as choosing specific values for x and using the Wronskian. The Wronskian can be used to determine the linear dependence of the functions at a specific value of x.
  • #1
vikkisut88
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Homework Statement


Let V = {differentiable f:R -> R}, a vector space over R. Take g1,g2,g3 in V where g1(x) = e[tex]^{}x[/tex], g2(x) = e[tex]^{}2x[/tex] and g3(x) = e[tex]^{}3x[/tex].
Show that g1, g2 and g3 are distinct.

Homework 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.

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.
 
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  • #2
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
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,
[tex]\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|[/tex]
the "Wronskian" of the functions at x= 0, is non-zero and the functions will be dependent if that determinant is 0.
 

FAQ: Proving that g1,g2,g3 are linearly independent

1. How do you prove that g1, g2, g3 are linearly independent?

The most common method for proving linear independence is by setting up a linear combination of the vectors g1, g2, g3 and solving for the coefficients. If the only solution is when all coefficients are equal to zero, then the vectors are linearly independent.

2. What is the definition of linear independence?

Linear independence refers to a set of vectors in which no vector can be written as a linear combination of the other vectors in the set.

3. Can you provide an example of proving linear independence?

Sure, let's say we have the vectors g1 = [1, 2, 3], g2 = [4, 5, 6], and g3 = [7, 8, 9]. We can set up the linear combination a*g1 + b*g2 + c*g3 and solve for the coefficients a, b, and c. If the only solution is a=0, b=0, and c=0, then the vectors are linearly independent.

4. Is there any other method for proving linear independence?

Another method is by using determinants. If the determinant of the matrix formed by the vectors g1, g2, and g3 is non-zero, then the vectors are linearly independent. However, this method is only applicable for a small number of vectors.

5. Why is it important to prove linear independence?

Proving linear independence is important in several areas of mathematics and science, particularly in linear algebra and differential equations. It allows us to determine if a set of vectors can form a basis for a vector space, which has many practical applications in fields such as physics, engineering, and computer science.

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