# Linear Algebra Lagrange polynomials & Basis

1. Jan 21, 2008

http://img100.imageshack.us/img100/9016/linalggp1.jpg [Broken]
for (a): does that mean i must compute l0(t), l1(t) and l2(t), and i wasn't sure how to do this with the lagrange polynomial formula given, so i found one online and did it, i'm not sure if this is correct, but my l0(t) looks like this:
= (t-a1)(t-a2)/(ao-a1)(ao-a2)
= (t-2)(t-3)/(1-2)(1-3)
=1/2(t-2)(t-3)

for (b) how do i go about proving this?

and also (c) what does it mean to deduce, i have looked up the definition but i still don't understand.

Last edited by a moderator: May 3, 2017
2. Jan 24, 2008

### HallsofIvy

Staff Emeritus
Yes, your polynomial for (a) is correct. I'm not sure why you say you "weren't sure how to do it with the given polynomial", it looks like you have done exactly that.
With i= 0, the fraction is $(t- a_1)(t-a_2)/(a_0-a1)(a_0-a2)= (t- 2)(t-3)/(1-2)(1-3)= (t-2)(t-3)/(-1)(-2)= -1/2 (t- 2)(t- 3)$.
With i= 1, the fraction is $(t- a_0)(t-a_2)/(a_1-a_0)(a_1-a_2)= (t-1)(t-3)/(2-1)(2-3)= (t-1)(t-3)/(1)(-1)= -1(t-1)(t-3)$.
With i= 2, the fraction is $(t- a_0)(t-a_1)/(a_2-a_0)(a_2-a_1)= (t-1)(t-2)/(3-1)(3-2)= (t-1)(t-2)/(2)(1)= 1/2(t-1)(t-2)$
Multiply those out and add.

To prove that a set is a "basis", you just prove that any function can be written as a "linear combination" of those (the sum of a number times each of those, as in (c)) in only one way. An easy way to do that is to note that there exist a unique nth order polynomial through any given n+1 points. Do you see that "lj(t)" is an l-1 degree polynomial? Given any n degree polynomial, what do you get if you evaluate that function at a1, a2, etc.?

Once you have proved (b), (c) follows easily. Saying that the {li} forms a basis means that any such function can be written as a linear combination. In fact, you really need to show (c) in order to prove (b)!