# 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

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

3. ### HallsofIvy

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Nov 6, 2021
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The subspace spanned by a single vector is the space of all scalar multiples of that vector. Here that is the space of all vectors a(x+ 1)= ax+ a for a any number.

The "orthogonal complement" of a vector space is the space of vector such that their inner product with any member of that subspace is 0.

Any vector can be written bx^2+ cx+ d, for some numbers x, y, and z while any vector in W is ax+ a.
Taking p(x)= bx^2+ cx+ d, p(-1)= b- c+ d, p(0)= d, and p(1)= b+ c+ d.
Taking q(x)= ax+ a, q(-1)= -a+ a= 0, q(0)= a, and q(1)= a+ a= 2a.

Their inner product, using the definition given here, p(-1)q(-1) + p(0)q(0)+ p(1)q(1) is
(b- c+ d)(0)+ da+ (b+ c+ d)(2a)= (2b+ 2c+ 3d)a= 0. Here, a must be any number- it is not in general 0 so we must have 2b+ 2c+ 3d= 0. We can solve that for d: d= (-2/3)(b+ c).

Of course W={a(x+ 1)} depends on the single number, a, so is 1 dimensional. That means, since the vector space of quadratic polynomials, ax^2+ bx+ c, is 3 dimensional, the orthogonal complement of W is 2 dimensional. In fact, any vector in the orthogonal complement of W can be written bx^2+ cw+ d= bx^2+ cw+ (-2/3)(b+ c)= b(x^2- 2/3)+ c(x- 2/3).

So a basis is {x^2- 2/3, x- 2/3}.

That is NOT the same as (x^2- 2, x- 2).

HallsofIvy, Nov 23, 2021