Linear Algebra question using Evaluation Inner Product Points

  • #1
Question from my last LA.II assignment. I have no idea what to with it. It looked simple but now I think I don't even understand the question.



Homework Statement



Consider the Inner Product Space P2 with the evaluation inner product at points x0=-1, x1=0, and x2=2, and consider the subspace W=span(p1,p2) where p1=1+x+x2 and p2=2-3x2. Express the polynomial p=2+x+x2 in the form w=w1+w2 where w1 is an element of W and w2 is an element of W orthogonal.



Homework Equations



Evaluation inner product points are defined such that <p,q>=p(x0)q(x0)+p(x1)q(x1)+...p(xn)q(xn)




The Attempt at a Solution



First I have to find an orthogonal basis for W. But I can't use three basis vectors because the question demands a combination of two vectors. — I have no idea where to go. I used Gram-Smidt to come up with two orthogonal bases v1 and v2 . . . but I can't use just these two vectors to solve c1v1+c2v2=w.

And now I'm stuck. I assume evaluation inner products are used to find the solution, but seeing as we never used them in class, and all the textbook gives is their definition, I have no idea as to how they may be useful.
 
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Answers and Replies

  • #2
lanedance
Homework Helper
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so you have p1 & p2, first check if the are orthogonal, if not use gram schmidt. You should now have a orthogonal basis, and it should be a simple matter to make it orthonormal, say q1 & q2.

now find the projection of p onto W, using dot products with each of the basis vectors

so you can write the component of P in w as
p_w = w1 = a q1 + b q2, for scalars a,b

then have a think about how to find the perpendicular component w2..
 
  • #3
Can't all of this be done without the use of evaluation inner product points though? — I still don't understand why she brought them up. I'm assuming she wants the question to be answered using the x0=-1, x1=0, x2=2 somehow, but I don't see why they are of any use in answering this question.
 
  • #4
lanedance
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don't they specify the inner product to be used in the calcs?
 
  • #5
My textbook defines an Evaluation Inner Product at x0, x1, x2...xn

to be:

<p,q>=p(x0)q(x0)+p(x1)q(x1)+...p(xn)q(xn).

Other than that, I copied the question as-is, and we've never used the concept in class before. She usually likes to include a question or two that require a bit of research, but with this one I really don't understand how evaluation inner products are at all related to the question. It seems perfectly answerable without their use.

I tried using the above definition when taking the inner product of two vectors, but then I end up with vectors that are non orthogonal (and really messy) . . . :confused:
 
  • #6
lanedance
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It seems perfectly answerable without their use.
but you would be making some assumption about the inner product that hasn't been specified... namely that the inner product is the "usual" one.

Now I have to admit I haven't done much with evaluation inner products, but assumed they retain the properties of a normal inner product, which a quick google confirms.
http://books.google.com.au/books?id...=onepage&q="Evaluation Inner Product"&f=false

ok so with
[tex] x_0=-1, \ x_1=0,\ x_2=2 [/tex]

so start with the basis vectors
[tex] <p1,p1> =<1+x+x^2,1+x+x^2> = 1.1 + 1.1 + 7.7 = 51 [/tex]
[tex] <p2,p2> =<3-x^2,3-x^2> = 2.2 + 3.3 + (-1).(-1) = 14 [/tex]

and the dot product between them
[tex] <p1,p1> =<1+x+x^2,3-x^2> = 1.2 + 1.3 + 7.(-1) = -2 [/tex]

using those vaues you should be able to find a orthonormal basis for W using ordinary gram schmidt
 
  • #7
lanedance
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a good insight about polynomials witha n evaluation inner product space is that say you have a nonzero polynomial u then
[tex] <u,u> \neq 0 [/tex]
as u can have at most n roots, and there are n+1 evaluation points
 
  • #8
That's one of the things I tried originally, but the resulting vectors I get after Gram-Scmidting using that inner product are not orthogonal.

p2=2-3x2 (correction) so <p1, p2>= -69.

So
v1 = (1,1,1)
v2 = (1,0,-3)-(-69/51)(1,1,1) = (40/17, 23/17, -28/17)

And these are not orthogonal.

If they're supposed to come out orthogonal, I don't know what I'm doing wrong.

Bleh, I guess I'll just answer the question using the usual inner product and hope for partial marks.

Thanks for the help!

(p.s: that book you linked to is actually the book we're using in class :smile:)
 
  • #9
lanedance
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sorry for the mistake before

i think you've hit it though, the poynomials given and the evaluation inner product are equivalent to working with vectors in R3 and the normal dot product, so consider the vectors

so using eval points
[tex] x_0=-1, \ x_1=0,\ x_2=2 [/tex]

the vectors are
[tex] p_1 = 1 + x +x ^2 \to (1, 1, 7) [/tex]
[tex] p_2 = 2-3x ^2 \to (-1, 2,-10) [/tex]

and the dot products are
[tex] <p_1, p_2> = 1^2 + 1^2 + 7^2 [/tex]
[tex] <p_2,p_2>=(-1)^2 +2^2+(-10)^2 [/tex]
[tex] <p_1,p_2>=1(-1) +1.2+7(-10) [/tex]

using these it should be easy to do normal gram schmidt and find a vector p3 in the form
[tex] p_3 = a.p_1 + b.p_2 [/tex]

with
[tex] <p_3,p_1> = 0 [/tex]

to convert p3 back to polynomial form, just substitute in the polynomial forms for p1 & p2, and similarly for the rest of the question
 
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  • #10
Oh I got it, I see where I went wrong. Thanks man!
 

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