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Given $ P_{2}(\mathbb R)$ the vector space of real polynomials of degree less than or equal to 2 equipped with the inner product $$\langle f,g\rangle=f(-1)g(-1)+f(0)g(0)+f(1)g(1).$$ How would I use the Gram-Schmidt process to find an orthonormal basis ?

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closed as off-topic by Martin R, user91500, JonMark Perry, Daniel W. Farlow, Arnaldo May 21 '17 at 17:36

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  • $\begingroup$ The vector space of real polynomials has infinite dimension; Gram-Schmidt will not give you a result in a finite time. $\endgroup$ – Bib-lost May 21 '17 at 7:08
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    $\begingroup$ @Bib-lost Look again. The polynomial space in question is $P_2(\mathbb R)$. $\endgroup$ – amd May 21 '17 at 7:08
  • $\begingroup$ I would guess $P_2$ is the subspace of order maximum 2. $\endgroup$ – mathreadler May 21 '17 at 7:09
  • $\begingroup$ What exactly don’t you understand about how to apply G-S to this space? $\endgroup$ – amd May 21 '17 at 7:09
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    $\begingroup$ Projection which is used in Gram Schmidt algorithm is defined with respect to the inner product, so start with an arbitrary basis for $P_2(\mathbb R)$, express the projection in terms of inner products and then carry out the calculations. $\endgroup$ – mathreadler May 21 '17 at 7:10
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Take any basis, for instance the canonical basis $(1, X, X^2)$

First normalise $1$ :

We have $\langle 1,1\rangle = 1+1+1 = 3$ so the normalised vector is $e_1 = {1\over \sqrt{3}}$

Now find the part of $X$ orthogonal to $1\over \sqrt{3}$. That's easy since $\langle 1,X \rangle = 0$

And normalise $X$ : $$\langle X,X\rangle = 1+0+1 = 2$$ so the normalised vector is $$e_2={X\over \sqrt 2}$$

Finally find the part of $X^2$ orthogonal to $span(1,X)$, ie : $$\begin{align}X^2 - \langle X^2 ,{X\over \sqrt2}\rangle{X\over \sqrt 2} - \langle X^2, {1\over \sqrt 3}\rangle {1\over \sqrt 3} &= X^2 - {1\over 2}(-1+0+1){X} + {1\over 3}(1+0+1)\\ &= X^2 - {1\over 3}\end{align}$$

Now normalise this vector : $$\langle X^2 - {1\over 3}, X^2 - {1\over 3}\rangle = \langle X^2, X^2\rangle -2\langle X^2, {1\over 3}\rangle + \langle {1\over 3}, {1\over 3}\rangle = 2-{4\over 3} + {1\over 3} = 1$$

Thus $e_3 = X^2-{1\over 3}$

In conclusion Gram-Schmidt process applied to the canonical basis gives you the orthonormal basis : $$\left({1\over \sqrt 3}, {X\over \sqrt 2}, X^2 - {1\over 3}\right)$$

Note : I highly recommend spending some time on it to make sure you understand the idea behind it and are confortable enough about it to do it yourself. One good exercise you can now do is apply the process on the same basis but treating the vectors in a different order (say $X^2$, then $X$ and finally $1$). This will give you a different basis than the one we obtained here.

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  • $\begingroup$ How did you get <1,1> = 1+1+1 =3 ? Other than that very nice answer thank you. $\endgroup$ – april analysis May 21 '17 at 7:30
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    $\begingroup$ @aprilanalysis: see my hint for how to apply the scalar product. $\endgroup$ – mathreadler May 21 '17 at 7:31
  • $\begingroup$ @mathreadler got it thanks. $\endgroup$ – april analysis May 21 '17 at 7:36
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Hint I can help you with first part of the problem

( First we may want to convince ourselves that we are actually dealing with an inner product here, but let's assume the question creator is not devious enough to lie about it being one. )

Let us first define projection with the scalar product between two polynomials, $f$, $g$, having coefficients $f_2,f_1,f_0$ and $g_2,g_1,g_0$ respectively:

$$f(x) = f_2 x^2 + f_1 x + f_0 \\g(x) = g_2 x^2 + g_1 x + g_0 $$

$$\langle f,g \rangle = f(-1)g(-1) + f(0)g(0)+f(1)g(1)$$

$$\begin{align*}f(-1)g(-1) &= /\text{convince yourself}/ = (f_2-f_1+f_0)(g_2-g_1+g_0)\\ f(0)g(0) &= /\text{convince yourself}/ = (f_0)(g_0)\\ f(1)g(1) &= /\text{convince yourself}/ = (f_2+f_1+f_0)(g_2+g_1+g_0)\end{align*}$$

Now you can gather all terms up and simplify and you have the first piece of the puzzle done.

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  • $\begingroup$ still not sure actually. because this gives sqrt(11) when computed ? $\endgroup$ – april analysis May 21 '17 at 7:52
  • $\begingroup$ What? You don't even mention to me what you are trying to calculate then give me a random number. How would you expect me to be able to help you then? $\endgroup$ – mathreadler May 21 '17 at 7:55
  • $\begingroup$ sorry about that. For <1,1> = 1+1+9 = sqrt(11) with this method ? $\endgroup$ – april analysis May 21 '17 at 7:57
  • $\begingroup$ how do you insert the <1,1> in place of the <f(x),g(x)> $\endgroup$ – april analysis May 21 '17 at 7:58
  • $\begingroup$ what does <1,1> assign to the $g_2,g_1,g_0,f_2,f_1,f_0$ ? $\endgroup$ – mathreadler May 21 '17 at 7:58

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