I have been working on this problem, we are given the below system of vectors

$f_{1} = x, f_{2} = \cos(x), f_{3}= \sin(x)$ from the inner product of $C_{\mathbb{R}}[-1,1]$

and we have to orthogonalise the system

I know that the Gram-Schmidt algorithm works like below

$w_{1} = v_{1}$

$w_{2} = v_{2} - \operatorname{proj}_{w_{1}}(v_{2})$

and so on, and I am just struggling how to use all of this to orthgonalise the vectors. Do I have to integrate over the inner product space? Or am I just simply substituting into the Gram- Schmidt Algorithm?

Any help would be much appreciated

  • $\begingroup$ You need to use the inner product. Both for projection and for normalization $\endgroup$ – Andrei Aug 5 '18 at 2:40
  • $\begingroup$ How do I do that? $\endgroup$ – Tom Heeley Aug 5 '18 at 3:00
  • $\begingroup$ The projection in an inner product space could be written explicitly using inner products. References could be linear algebra textbooks. $\endgroup$ – xbh Aug 5 '18 at 3:24
  • $\begingroup$ Start by normalizing $f_1$: $w_1=\frac{f_1}{||f_1||}$, where $||f||=\int_{-1}^{1}f^2(x)dx$ $\endgroup$ – Andrei Aug 5 '18 at 3:25
  • $\begingroup$ Which inner product of $C_{\Bbb{R}}[-1,1]$ are you talking about? There are several! The simplest is $\langle f,g\rangle=\int_{-1}^1 f(x)g(x)\,dx$, but you can also have a positive definite weight function $w(x)$ as a third factor in the integrand. This observation actually leads up to an answer of sorts. The outcome depends on the choice of the inner product so, yes, you absolutely must calculate those inner products. No other way. $\endgroup$ – Jyrki Lahtonen Aug 5 '18 at 6:21

Do I have to integrate over the inner product space?

The vector space of real functions whose domain is an closed interval [a,b] with inner product $\langle f_1, f_2 \rangle $ is $\int_{a}^{b} f_1 f_2\,dx$. So yes, you need to integrate over the inner product space.

You need to start by normalizing $f_1$. Let's say $w_1$ is $f_1$ normalized.

$$ w_1 = \frac{f_1}{||f_1||}\, ||f_1||^2=\int_{a}^{b}f_1^2(x)dx \qquad(1) $$

After normalizing $f_1$, you need to find a vector orthogonal to $w_1$. We can use $f_2$ to find this vector $W_2$ as

$$ W_2 = f_2-\bigl(\;\langle\, w_1,f_2\,\rangle\;\bigr)w_1\; $$ where $\langle\, w_1,f_2\,\rangle$ is the inner product of $w_1$ and $f_2$ $$ W_2 = \sin x - \Bigl(\int_a^bw_1\sin x\,dx\Bigr)w_1 \qquad(2) $$

$W_2$ is not normalized and it can be normalized into $w_2$ in the same way as in eq(1).

Next $W_3$ and $w_3$ is found in the same way as in eq(2), which will give you 3 orthonormal vectors

  • $\begingroup$ See the edits I made in the source for a few tips about using TeX-spacings as well as inserting regular text inside a piece of displayed math. Mind you, I'm not sure it is a good idea to have that sentence where $\langle w_1,f_2\rangle$ ... on the displayed line as opposed to as regular text on the next line, but as you wanted to do it that way, I wanted to show how to do it right. $\endgroup$ – Jyrki Lahtonen Aug 5 '18 at 6:16
  • $\begingroup$ Thank you for the edit. I am new to the TeX so I will need this guidance from time to time. And yeah I think you are right about the "where... " sentence, I will edit that to a new line $\endgroup$ – artha Aug 5 '18 at 6:27
  • $\begingroup$ And regarding the inner product space, $C_{\Bbb{R}}[-1,1]$, Its difficult to integrate sinx and cosx over this range. If the OP's question is from a text, I doubt the author would have given this interval to solve the question. May be OP has got the range wrong. I don't have enough rep to comment on OP's, so instead of solving for this range, I just assumed [a,b]. $\endgroup$ – artha Aug 5 '18 at 6:37

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