Let $V_4$ be the vector space of all polynomials of degree less than or equal to 4 with the inner product $$\langle p(x),q(X) \rangle = \int_{-1}^{1} p(X)q(X)dX$$ calculate the angles between $(1,X),(X,X^2),(X^2,X^3),(X^3,X^4)$.

My question is are all these polynomials orthogonal to each other? I get $\langle 1,X \rangle =\int_{-1}^{1} XdX=0 , \langle X,X^2 \rangle =\int_{-1}^{1} X^3dX=0, \langle X^2,X^3 \rangle =\int_{-1}^{1} X^5dX=0, \langle X^3,X^4 \rangle = \int_{-1}^{1} X^7dX=0$

but I don't think that they are orthogonal to each other. Can someone please confirm whether they are orthogonal or what I'm doing wrong.

  • 1
    $\begingroup$ For sake of curiousity, check $\langle X, X^3 \rangle$. This is simple observation: if you integrate odd function over symmetric domain you'll always get zero value of integral. $\endgroup$ – Evgeny Jun 10 '14 at 14:00

You only showed that

• $1$ is orthogonal to $X$, and that

• $X$ is orthogonal to $X^2$ and that

• $X^2$ is orthogonal to $X^3$ and that

• $X^3$ is orthogonal to $X^4$.

This because the scalar product of these polynomials gives $0$. But there is a difference between saying that $(1,X,X^2,X^3,X^4)$ are orthogonal to each other and saying that $1 \bot X, X\bot X^2, X^2 \bot X^3, X^3 \bot X^4$.

Indeed if you have a basis $\phi_i = (\phi_1,\cdots,\phi_n)$ of a vector space you can say that the component of the basis are orthogonal to each other $:\iff \langle \phi_i, \phi_j\rangle= \delta_{ij} = \begin{cases}1 & if \quad i = j \\ 0 & if \quad i \neq j\end{cases}$ But in your cases this is not true, indeed:

$$\langle1,X^2\rangle = \int_{-1}^1X^2dX = \frac23\neq0$$


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You will note that $PQ\perp AB$ and $AB\perp RS$, yet the three lines are not mutually orthogonal.


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