Basis of W

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

  1. Displayer18

    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

    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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