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Vector of polynomial and basis.

Displayer1243

New member
May 5, 2020
1
If this question is in the wrong forum please let me know where to go.

For p
$(x), q(x) \in P_{2}$
, 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
$x+1$
.
a) Describe the elements of
$w$

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:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,736
Hello, and welcome to MHB! :)

I'm afraid I cannot help you personally, but there are many here who can, and it would be helpful to them to see what work you've done already. This way they can better guide you by perhaps pointing out any mistakes, or providing hints for you to continue, etc.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Frankly your notation is mystifying! You have a strange mixture of fonts. But my biggest problem is your definition of the inner product, "p(x), q(x)=p(-1) 1(-1)+p(0), q(0)+p(1) q(1)".

I can guess that "p(x), q(x)" was intended to be something like <p(x), q(x)> but what does "p(-1)1(-1)" mean? Was that supposed to be "p(-1)q(-1)"? That is, did you mean
"<p(x), q(x)>= p(-1)q(-1)+ p(0)q(0)+ p(1)q(1)"?

Now, W is the subspace spanned by the single vector x+ 1. I would think that you should know that vectors in a subspace spanned by a single vector are just multiples of that vector. All vectors in W are of the form a(x+ 1)= ax+ a where a can be any number.

Yes, the "orthogonal complement" of W is the set of all polynomials, [tex]p(x)= ax^2+ bx+ c[/tex] such that <p(x), x+ 1>= p(-1)(-1+1)+ p(0)(0+1)+ p(1)(1+ 1)= p(0)+ 2p(1)= 0
(you have "p(0)+ 2p(1)= 2p(1)". I have no idea how you got that!)
Since [tex]p(x)= .ax^2+ bx+ c[/tex] that is c+ 2(a+ b+ c)= 2a+ 2b+ 3c= 0. Taking a= 1, b= 0, that is 2+ 3c= 0 so c= -2/3. One vector in that set is [tex]x^2- 2/3[/tex]. Taking a= 0, b= 1, c= -2/3 so another vector is [tex]x- 2/3[/tex]. Those are not multiples so they are independent. The given vector space is clearly three dimensional and W, the subspace of multiples of x+ 1, is 1 dimensional so the orthogonal subspace has dimension 2. Those two independent vectors form a basis. Since this is given as a space or quadratic polynomials I see no reason to write them as "column vectors. [tex]\{x^2- 2/3, x- 2/3\}[/tex] is sufficient.