Answer to Question #102809 in Linear Algebra for Abhishek misra

Question

Check that B={1,x+1,(x+1)(x+1)} is a basis for p2.
Find the co-ordinates of 2+x+3x2 with respect to this basis.

Answer 1

Consider vectors

$b_1=1, b_2=x+1, b_3=(x+1)^2$

Let's form a linear combination

$\alpha_1b_1+\alpha_2 b_2+\alpha_3 b_3=0$

and show that equality is possible if

$\alpha_1=\alpha_2=\alpha_3=0$

$\alpha_11+\alpha_2(x+1)+\alpha_3(x^2+2x+1)=0\\ 1\alpha_1+x\alpha_2x+\alpha_2+x^2\alpha_3+2x\alpha_3+\alpha_3=0\\ x^2: \alpha_3=0\\ x: \alpha_2+2\alpha_3=0\implies\alpha_2=0\\ x^0: \alpha_1+\alpha_2+\alpha_3=0\implies\alpha_1=0$

So, basic vectors are

$b_1=1, b_2=x+1, b_3=(x+1)^2\\ f(x)= 3x^2+x+2$

decompose it in the basis $b_1, b_2, b_3$

according to the scheme of Horner:

$\begin{matrix} & 3&1&2 \\ -1 & 3&-2&4\\ -1&3&-5\\ -1&3\\ \end{matrix}$

$f(x)=3(x+1)^2-5(x+1)+4=\\ =4b_1-5b_2+3b_3=\\ =(4;-5;3)$