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I would like to know whether I have made mistakes in this Gram-Schmidt process as I kept getting mixed up with the vectors.

Using the vector space $P_1$ defined by the inner product

$<p, q> = ∫_{-1}^1 p(x)q(x)dx$ Find an orthonormal basis {$e_1, e_2$} using the Gram-Schmidt process for the set of vectors $u_1 = 1+ x$ $u_2 = 1 + 3x$

Using the process

$v_1 = u_1 = 1 + x$

$v_2 = u_2 - \frac{<v_1, u_2>}{||v_1||^2}v_1$ $= \frac{<1 + x, 1 + 3x>}{<1 + x, 1 + x>}(1 + x)$

$ = \frac{4}{8/3}(1 + x)$ $= \frac{3}{2}(1 + x)$ $= \frac{3}{2} + \frac{3}{2}x$

Now,

$ 1 + 3x - \frac{3}{2} + \frac{3}{2}x$ results in $\frac{-1}{2} + \frac{9}{2}x$

Thus, the orthogonal basis is $(1 + x, \frac{-1}{2} + \frac{9}{2}x)$

The orthonormal basis would then be $e_1 = \frac{v_1}{||v_1||} e_2 = \frac{v_2}{||v_2||}$

$e_1 = \frac{1 + x}{\sqrt8/3}$

$e_2 = \frac{-1/2 + 9x/2}{\sqrt14}$ which appears to be rather messy. So, what did I do wrong?

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1 Answer 1

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You have made a mistake while subtracting $(1+3x)$ and $\frac32+\frac32x$. The result is $v_2=-\frac12+\frac32x$ which is orthogonal to $v_1$.

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  • $\begingroup$ Ah yes I see that. So the orthonormal process would still be correct? $\endgroup$
    – DuncanK3
    Sep 27, 2020 at 6:03
  • $\begingroup$ The process till the subtraction is correct. Your error is trivial but it led to the wrong answer. $\endgroup$ Sep 27, 2020 at 13:54

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