Consider $P_3$ with the inner product $\langle p(x),q(x)\rangle = \int_{0}^1 p(x)q(x) dx$.

Use Gram Schmidt to find an orthonormal basis for the space $$ U= \operatorname{Span}(\{x+1, x^2-x,x^3\})$$

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    $\begingroup$ You already know a basis of U; it's given by the three vectors that span it, as they are linearly independent (by considerations of degree). So just apply Gram-Schmidt to that basis. $\endgroup$ Apr 7 '13 at 23:15


For example

$$||1+x||^2:=\langle 1+x\,,\,1+x\rangle:=\int\limits_0^1(1+x)^2\,dx=\left.\frac{1}{3}(1+x)^3=\right|_0^1=\frac{1}{3}(8-1)=\frac{7}{3}\implies$$

$$\implies u_1:=\frac{1+x}{||1+x||}=\sqrt\frac{3}{7}\,(1+x)$$


$$v_2:=x^2-x-\frac{\langle x^2-x\,,\,1+x\rangle}{||1+x||}(1+x)=(x^2-x)-u_1\int\limits_0^1(x^2-x)(1+x_\,dx=\ldots\implies$$

$$\implies u_2=\frac{v_2}{||v_2||}\ldots\;\text{etc.}$$

  • $\begingroup$ Thank you guys. It makes sense now $\endgroup$ Apr 7 '13 at 23:31
  • $\begingroup$ Hi DonAntonio, Ive noticed youve been answering numeros questions in math.stackexchange can you tell me what you do for a living and where you went to school to learn math. Thank you $\endgroup$ Apr 7 '13 at 23:49
  • $\begingroup$ Well, I'm a mathematician and I teach maths, make some research and I and my wife try to raise three kids (and myself). My higher studies were at the Hebrew University in Jerusalem, Israel. There, no more disclosure! :) $\endgroup$
    – DonAntonio
    Apr 8 '13 at 1:29

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