I have a function space $\mathcal {F}([-1,1],\mathbb R)$ and the subspace $\mathcal{P_2}:=$ $(x\mapsto a_o+a_1x+a_2x^2| a_0,a_1,a_2 \in \mathbb R )$ for all polynomials with degree $\le2$. In this subspace the dot product and norm are defined as follows:



I have four polynomials $p_1,...,p_4$





I have to show that $p_1,p_2,p_3$ are linearly independant and I have to create an Orthonormal basis $q_1,q_2,q_3$

1) The first step is "trivial". They are linearly independant.

2) I want to use Gram-Schmidt to create an orthonormal basis.


$q_2'=p_2-<q_1,p_2> \cdot q_1 \Rightarrow q_2=\frac{q_2'}{||q_2'||}$

$q_3'=p_3-<q_1,p_3>\cdot q_1-<q_2,p_3>\cdot q_2 \Rightarrow q_3=\frac{q_3'}{||q_3'||}$

$\vec{p_1}=\begin{pmatrix}0\\4\\3\end{pmatrix}; ||\vec{p_1}||=\sqrt{0^2+4^2+3^2}=5 \Rightarrow \color{red}{\vec{q_1}=\frac{1}{5}\begin{pmatrix}0\\4\\3\end{pmatrix}}$


As you can see, the numbers are getting extremely ugly. The next step is even worse. Am I doing something wrong?

I would be grateful for any help you can give me.


  • $\begingroup$ Just that the numbers "get ugly" is no reason to believe it is wrong. Why not check if what you have so far is indeed orthonormal? $\endgroup$
    – quid
    Dec 13, 2014 at 13:36
  • $\begingroup$ @quid If I am correct then the dot product of two orthogonal vectors should be 0. I am getting $\frac{11}{15}$ though. $\endgroup$
    – qmd
    Dec 13, 2014 at 13:41
  • $\begingroup$ Yes two vectors are called orthogonal if their dot-product is 0. $\endgroup$
    – quid
    Dec 13, 2014 at 13:47
  • $\begingroup$ Yeah. Since I am getting $\frac{11}{5}$ they are not orthogonal. I can't really see where I have made a mistake though. $\endgroup$
    – qmd
    Dec 13, 2014 at 13:52
  • 1
    $\begingroup$ I just need to find a basis for $p_1, p_2, p_3$ Edit: I get it now. Just like you and Lee said, I used the wrong definition of the dot product. Thank you very much for your help. $\endgroup$
    – qmd
    Dec 13, 2014 at 14:03

1 Answer 1


Because there is a mistake on $\vec{p_{1}}$.

$\vec{p_{1}}=\frac{\vec{q_{1}}}{||\vec{q_{1}}||} = \frac{\vec{q_{1}}}{\sqrt{<\vec{q_{1}},\vec{q_{1}}>}} $

$<\vec{q_{1}},\vec{q_{1}}>=∫^{1}_{-1}(4x+3)(4x+3)dx $

  • $\begingroup$ Ah, of course!!!! Thats what quid was saying too but somehow I didn't pick up on that. Thanks! $\endgroup$
    – qmd
    Dec 13, 2014 at 14:05

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