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#### Displayer1243

##### New member

- May 5, 2020

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If this question is in the wrong forum please let me know where to go.

For p
, the vector space of polynomials to the form ax'2+bx+c. p(x), q(x)=p(-1) 1(-1)+p(0), q(0)+p(1) q(1), Assume that this is an inner product. Let W be the subspace spanned by
.

a) Describe the elements of

b) Give a basis for W (Orthogonal complement) ". ( You do not need to prove that your set is a basis)

I've been stuck on this question for quite a while and have made progress that I don't know is fully right if anyone could help so I can get clarification that would be awesome!! Thanks!

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 $\mathrm{P}^2$ 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?

For p

a) Describe the elements of

b) Give a basis for W (Orthogonal complement) ". ( You do not need to prove that your set is a basis)

I've been stuck on this question for quite a while and have made progress that I don't know is fully right if anyone could help so I can get clarification that would be awesome!! Thanks!

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 $\mathrm{P}^2$ 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?

Last edited: