Question

Question:Let \(V\) be the space \(C\left( { - 2,2} \right)\) with the inner product of Example 7. Find an orthogonal basis for the subspace spanned by the polynomials 1, \(t\), and \({t^2}\).

Short answer

The orthogonal basis is \(\left\{ {1,\,t,\,3{t^2} - 4} \right\}\).

Step-by-step solution

Use the given information

If the pair of vectors\(\left\langle {f,g} \right\rangle \)belong to vector space \(V\) defined in \(C\left( { - 2,2} \right)\), then their inner product is given by \(\left\langle {f,g} \right\rangle = \int_{ - 2}^2 {f\left( t \right)g\left( t \right)dt} \).

It is given that \(1,\,t,\,{t^2}\) are orthogonal, then \(\left\langle {1,t} \right\rangle \) must be 0, that is\(\int_{ - 2}^2 {tdt} = 0\).

Use Gram Schmidt process

According to Gram Schmidt process, the third element in the orthogonal basis span\(\left\{ {1,t,{t^2}} \right\}\)can be defined as\({t^2} - \frac{{\left\langle {{t^2},1} \right\rangle }}{{\left\langle {1,1} \right\rangle }}1 - \frac{{\left\langle {{t^2},t} \right\rangle }}{{\left\langle {t,t} \right\rangle }}t\).

Find\(\left\langle {{t^2},1} \right\rangle ,\,\left\langle {1,1} \right\rangle ,\,\left\langle {t,t} \right\rangle \), as follows:

\(\begin{aligned}{}\left\langle {t,t} \right\rangle &= \int_{ - 2}^2 {{t^2}dt} \\ &= \left( {\frac{{{t^3}}}{3}} \right)_{ - 2}^2\\ &= \left( {\frac{8}{3} - \left( { - \frac{8}{3}} \right)} \right)\\ &= \frac{{16}}{3}\end{aligned}\)

\(\begin{aligned}{}\left\langle {1,1} \right\rangle &= \int_{ - 2}^2 {1dt} \\ &= \left( t \right)_{ - 2}^2\\ &= \left( {2 - \left( { - 2} \right)} \right)\\& = 4\end{aligned}\)

\(\begin{aligned}{}\left\langle {{t^2},t} \right\rangle &= \int_{ - 2}^2 {{t^3}dt} \\ &= \left( {\frac{{{t^4}}}{4}} \right)_{ - 2}^2\\ &= \left( {\frac{{16}}{4} - \frac{{16}}{4}} \right)\\ &= 0\end{aligned}\)

Simplify for Gram Schmidt inequality

Plug the above obtained values in \({t^2} - \frac{{\left\langle {{t^2},1} \right\rangle }}{{\left\langle {1,1} \right\rangle }}1 - \frac{{\left\langle {{t^2},t} \right\rangle }}{{\left\langle {t,t} \right\rangle }}t\) and simplify as follows:

\(\begin{aligned}{}{t^2} - \frac{{\left\langle {{t^2},1} \right\rangle }}{{\left\langle {1,1} \right\rangle }}1 - \frac{{\left\langle {{t^2},t} \right\rangle }}{{\left\langle {t,t} \right\rangle }}t &= {t^2} - \frac{{16/3}}{4}1 - \left( 0 \right)t\\ &= {t^2} - \frac{4}{3}\end{aligned}\)

The third element in the orthogonal basis for subspace spanned by the polynomials \(1,\,t,\,{t^2}\)can be obtained by scaling\({t^2} - \frac{4}{3}\)by 3, to get it as\(3{t^2} - 4\).

Thus, the orthogonal basis is \(\left\{ {1,\,t,\,3{t^2} - 4} \right\}\).