Thread: Linear Independence problem.

Hi, I have this question to solve.

Let f=exp(t), g=t, and h= 2+3t. Are f, g, and h linearly independent or linearly dependent?

I'm confused as to how I should figure this out. I understand that a set of vectors to be linearly independent, say v1=(1,2) and v2=(2,3), that the only solution for the system a+2b=0, 2a+3b=0 is for a=b=0 (i don't know if that is the case in the example i'm given). However, i'm confused as to how to go about the given problem. Thank you.

It's really the same idea. Suppose a, b and c are such that af + bg + ch =0. This is really a function equation. That is it means for all t, af(t) +bg(t) + ch(t) = 0. Now one way to prove independence is to find 3 different t values to get 3 equations in a, b and c whose only solution is 0. If you happen to know about Wronskians, there's another approach.

Thank you for the reply.
So, I set the equation a*exp(t)+b*t+c(2+3t)=0 and I have to figure if the only solution to it is a=b=c=0 or not?

Yes, the only solution for which the equation holds for all t is a=b=c=0. So if you can find 3 explicit t values which then force a=b=c=0, then the function equation is 0 for all t only if a=b=c=0.

Okay, thank you!

As a "function" equation, that has to be true for all t. So setting t equal to three different numbers will give you three different equations to solve for a, b, and c. For example, it t= 0, you get a(exp(0))+ b(0)+ c(2+ 3(0)= a+ 2c= 0.
Now, set t equal to, say, 1 and -1. What equations do you get?