Linear Independence of vectors. Dont understand this solution

Question

Show whether the following three vectors are linearly independent or not

 

V := C(R);  f₁(x) = e^x; f₂(x) = e^3x; f₃(x) = x(x−1)

 

Solution:

 

af₁ + bf₂ + cf₃ = 0

 

af₁(x) + bf₂(x) + cf₃(x) = 0

 

Therefore, for x = 0, x = 1, and x = 2, we obtain:

 

a + b =0

ae + be³ =0

ae² + be⁶ + 2c =0

 

Using the first two equations leads to:

 

b(−e + e³) =0

 

 

 

I dont understand where they got x=0, x=1 and x=2 and how to get that system of equations.

Answer 1

I see you have 3 functions:  f₁(x) = e^x; f₂(x) = e^3x; f₃(x) = x(x−1) and you want to show if they are linear independent or not. We say n elements f₁, f₂, ..., f_n  (not necessary vector or fucntions) are linear dependent if there exist  such constants a₁, a₂, ... a_n at least one of them not zero, such that a₁f₁+a₂f₂+..+a_nf_n=0.  (IND)

If (IND) is true only for a₁=a₂=...=a_n=0, then f₁.,...,f_n are linear independent.

So you need to show that 

af₁(x)+bf₂(x)+cf₃(x)=0 only for a=b=c=0 or  it true for some nonzero of a, b  or c.

Sso let' see

a e^x+be^3x+c x(x−1)=0.

You can test independence (dpendence) in different way. One of them, for example, is to calculate Wronsilan of these fucntions ar some point, say x=0.

As it is explained by Philip P. eventaully you have to obtain 3  equations for a, b and c. In case with Wronskian, you need second dderivatives of those 3 functions.

You try both ways. 

Now regarding the question why it was chosen those 3 points (0, 1, 2). I suppose it should be explained in class: due to continuity from independence (dependence) at some point follows independence (dependence) for the whole domain.

I hope I told something useful.

Answer 2

To show linear independence, you must have a = b = c = 0.  Since you have three unknowns (a, b, c) you need three equations. Since x is the independent variable, you are free to pick any three values for x to get the three equations. They chose x = 0, 1, and 2. You could have picked x = 3, 4, 5 or any other combination to get a system of three equations to solve. Some choices make simpler equations than others, but you get to pick the values of x.

 

From above:

 

b(-e+e³) = 0

b = 0

 

Plug b = 0 into the first equation:

 

a + b = 0

a + 0 = 0

a = 0

 

Plug a = b = 0 into the third equation:

 

ae² + be⁶ + 2c =0

2c = 0

c = 0

 

f₁, f₂, and f₃ are linearly independent.