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For certain polynomials $p(t), q(t)$ and $r(t)$, say we are given the following inner products.

$\begin{array} {|c|c|c|c|} \hline \langle, \rangle & p & q & r \\ \hline p & 4 & 0 & 8 \\ \hline q & 0 & 1 & 0 \\ \hline r & 8 & 0 & 50 \\ \hline \end{array}$

For example, $\langle q, r \rangle = \langle r, q \rangle = 0$. Let $E$ be the span of $p$ and $q$. Find the orthogonal projection $\text{Proj}_E r$ (express it as linear combination of $p$ and $q$) and an orthonormal basis of the span of that three polynomials in linear combination form.

Usually, I encounter a problem that the polynomials already given and then it can be easier to determine the orthonormal basis in that case. However, I have read my textbook and about orthogonal projection I don't have much idea to show. How that is and what about the orthonormal basis? How can we determine it by just looking at the table of inner products given above?

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  • $\begingroup$ For the projection onto a one-dimensional subspace like $\text{span}\{p\}$, the formula is $\text{Proj}_{\text{span}\{p\}}(r) = \frac{\langle r, p \rangle}{\langle p, p\rangle} p$. The answer to your question will be similar: it will be a linear combination of $p$ and $q$, with coefficients obtained by numbers from the table of inner products. $\endgroup$
    – angryavian
    Mar 13, 2018 at 2:49

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You use the definition: since $p,q$ are mutually orthogonal,

$$ \text{Proj}_Er=\frac{\langle r,p\rangle} {\langle p,p\rangle}\,p +\frac{\langle r,q\rangle} {\langle q,q\rangle}\,q. $$

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  • $\begingroup$ How about the orthonormal basis, Sir? $\endgroup$
    – Shane Dizzy Sukardy
    Mar 13, 2018 at 2:57
  • $\begingroup$ @ShaneDizzySukardy You have an orthogonal basis. That’s good enough. $\endgroup$
    – amd
    Mar 13, 2018 at 4:00

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