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Let the vector space $P_2$ have the inner product:

$\langle p,q\rangle=\int\limits_{-1}^{1}p(x)q(x)dx$

Apply the Gram-Schmidt process to transform the standard $S=\{1,x,x^2\}$ into an orthonormal basis.

The book does not provide solution for this problem and I do not know how to solve it.

1-How can I transform the S basis into an orthonormal basis if I need at least three vectors since the dimension of the vector space in cause is 3?

2-Can someone provide me a solution?

Thanks in advance!

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    $\begingroup$ There are three vectors, $x \mapsto 1, x \mapsto x, x \mapsto x^2$. $\endgroup$
    – copper.hat
    Jun 14, 2017 at 14:46

1 Answer 1

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First, normalize first vector of the basis $\;v_1=1\;$:

$$\langle v_1,v_1\rangle=\langle 1,1\rangle:=\int_{-1}^11\cdot dx=2\implies \color{red}{u_1=\frac{v_1}{\left\|v_1\right\|}=\frac1{\sqrt2}}$$

Next, orthogonalize second vector wrt the first one:

$$w_2:=v_2-\langle v_2,u_1\rangle u_1=x-\left\langle x,\frac1{\sqrt2}\right\rangle \frac1{\sqrt2}=x-\frac12\int_{-1}^1x\,dx=x-\left.\frac14x^2\right|_{-1}^1=x$$

Now, orthonormalize that last vector:

$$\langle x,x\rangle=\int_{-1}^1x^2dx=\left.\frac13x^3\right|_{-1}^1=\frac23\implies\color{red}{u_2=\frac{w_2}{\left\|w_2\right\|}}=\sqrt\frac32\,x$$

Last step, and this you will do: orthogonalize third vector wrt the first two:

$$w_3:=x^2-\langle x^2,u_1\rangle u_1-\langle x^2,u_2\rangle u_2$$

and then take

$$\color{red}{u_3=\frac{w_3}{\left\|w_3\right\|}}$$

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  • $\begingroup$ Sorry for disturbing. But could you provide me the last step solution. I am getting $x^2-\frac{7}{6}$ after the orthogonolization. However $||w_3||=(-\frac{69}{105})^{\frac{1}{2}}$ which would have an imaginary root. What might I have done wrong? $\endgroup$ Jun 15, 2017 at 16:20
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    $\begingroup$ $$w_3=x^2-\frac12\int_{-1}^1x^2dx-\overbrace{\frac32\int_{-1}^1x^3dx}^{=0}=x^2-\frac26=x^2-\frac13\implies$$$$\left\|w_3\right\|^2=\int_{-1}^1\left(x^2-\frac13\right)^2dx=\frac25+\frac29=\frac{28}{45}\implies u_3=\frac{w_3}{\left\|w_3\right\|}=\frac{3\sqrt5}{2\sqrt7}\left(x^2-\frac13\right)$$ $\endgroup$
    – DonAntonio
    Jun 15, 2017 at 19:53
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    $\begingroup$ I think a (-4/3) is missing in the resolution of the integral, or i am wrong? $\endgroup$ Feb 13, 2018 at 22:23
  • $\begingroup$ @ChanchoMuena What integral? $\endgroup$
    – DonAntonio
    Feb 14, 2018 at 7:28

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