Displayer1234
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- May 5, 2020
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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 [MATH]W^{perp}[/MATH]. (You do not need to prove that your set is a basis.)
To be straight forward I read the guidelines, I am clueless on absolutely everything but this is my shot at it. Any help will be greatly appreciated.
[MATH]W^{perp}[/MATH]={p [MATH]in[/MATH] [MATH]P^2[/MATH]| (p,x+1)=0}
<p,x+1> = 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)+2p(1)
p(0)=2p(1)
Form of [MATH]ax^2[/MATH]+bx+c
c=(2([MATH]ax^2[/MATH]+bx+c))
c=2(a(1)+b(1)+c)
c=2a+2b+2c
0=2a+2b+c
The Elements of W are (2,2,1)
I know I am not showing B but that is only because I don't know if a is correct or if what I have written contains both a and b.
a) Describe the elements of W.
b) Give a basis for [MATH]W^{perp}[/MATH]. (You do not need to prove that your set is a basis.)
To be straight forward I read the guidelines, I am clueless on absolutely everything but this is my shot at it. Any help will be greatly appreciated.
[MATH]W^{perp}[/MATH]={p [MATH]in[/MATH] [MATH]P^2[/MATH]| (p,x+1)=0}
<p,x+1> = 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)+2p(1)
p(0)=2p(1)
Form of [MATH]ax^2[/MATH]+bx+c
c=(2([MATH]ax^2[/MATH]+bx+c))
c=2(a(1)+b(1)+c)
c=2a+2b+2c
0=2a+2b+c
The Elements of W are (2,2,1)
I know I am not showing B but that is only because I don't know if a is correct or if what I have written contains both a and b.