# Basis of W

Discussion in 'Linear and Abstract Algebra' started by Displayer18, May 5, 2020.

1. ### Displayer18

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2. For p(x), q(x) in P2, the vector space of polynomials of the form ax2 + bx + c, define <(px),q(x)>× = p(-1)q(-1) + p(0)q(0)+ p(1)q(1). Assume that this is an inner product. Let W be the subspace spanned by x + 1.

a) Describe the elements of W.

b) Give a basis for Wó (orthogonal complement of W). (You do not need to prove that your set is a basis. Wó is defined on page 336.)

Displayer18, May 5, 2020
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2. ### Displayer18

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This is what I have now.
$$\begin{array}{l} \mathrm{W}^{\perp}=\{p \in P 2 |\langle p, x+1\rangle=0\} \\ \langle p x+1\rangle=p(-1) x+1(-1)+p(0) x+1(0)+p(1) x+1(1)=p(-1) \\ (-1+1)+p(0)(0+1)+p(1)(1+1)=p(0)+2 p(1)=2 \mathrm{p}(1) \end{array}$$
since we are looking for polynomials such that $\mathrm{p}(0)=2 \mathrm{p}(1),$ and with the definition of $\mathrm{P}^1$ all
polynomials $a x^2+b x+c$ such that $c=2(a+b+c),$ so the numbers a,b,c with 2a+2b+c=0. In terms of linear algebra and the null space of $A=[2,2,1]$ which is dimension 2 and generates the vectors
$\begin{bmatrix}1\\0\\-2\end{bmatrix}$ and $\begin{bmatrix}0\\1\\-2\end{bmatrix}$

Which converts back into polynomials to get
$W^{\perp}=\left\{\mathrm{x}^2-2, \mathrm{x}-2\right\}$
Did I solve this question correctly?

Displayer18, May 5, 2020

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