Answer to Question #274149 in Linear Algebra for Nikhil Singh

Question #274149

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.

1
Expert's answer
2021-12-22T05:03:01-0500

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


W=1(x+1)(x+1)2012(x+1)002=112=20W = \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 AB, and C such that


3+x+2x2=A1+B(x+1)+C(x+1)23 + x + 2 x^2 = A\cdot1 + B\cdot (x+1) + C \cdot (x+1)^2


A1+B(x+1)+C(x+1)2=A+B(x+1)+C(x2+2x+1)=(A+B+C)+(B+2C)x+Cx2A\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}


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