Answer to Question #278687 in Linear Algebra for Ted

Question

Check that {1,(x+1),(x+1)^2} is a basis of the vector space of polynomial over R of degree at most 2. Find the coordinate of 3+x+2x^2 with respect to the basis.

Answer 1

Wronskian of the functions $\{1, (x+1), (x+1)^2\}$ is

$W = \begin{vmatrix} 1 & (x+1) & (x+1)^2 \\ 0 & 1 & 2 (x+1) \\ 0 & 0 & 2 \end{vmatrix} = 1*1*2 = 2 \ne 0$ ,

so the functions are independent and hence form the basis.

To find a coordinate of 3+x+2x^2 find such values of A, B, and C such that

$3 + x + 2 x^2 = A\cdot1 + B\cdot (x+1) + C \cdot (x+1)^2$

$A\cdot1 + B\cdot (x+1) + C \cdot (x+1)^2 = A + B \cdot (x+1) + C \cdot (x^2 + 2 x + 1) = (A+B+C) + (B + 2 C)\cdot x + C \cdot x^2$

therefore C = 2;

(B + 2 C) = 1and B = 1- 2C = -3;

(A + B + C) = 3 and A = 3 - B - C = 4.

So the coordinates are {4, -3, 2}