Question 6.7.8: Let V be the space C[0, 1] with the inner product of Example...
Let V be the space C[0, 1] with the inner product of Example 7, and let W be the subspace spanned by the polynomials p_{1}(t)=1, p_{2}(t)=2 t-1, and p_{3}(t)=12 t^{2}. Use the Gram–Schmidt process to find an orthogonal basis for W .
\left\langle p_{2}, q_{1}\right\rangle=\int_{0}^{1}(2 t-1)(1) d t=\left.\left(t^{2}-t\right)\right|_{0} ^{1}=0.
\text { So } p_{2} \text { is already orthogonal to } q_{1} \text {, and we can take } q_{2}=p_{2} \text {. For the projection of } p_{3}\text { onto } W_{2}=\operatorname{Span}\left\{q_{1}, q_{2}\right\}, \text { compute }
\left\langle p_{3}, q_{1}\right\rangle=\int_{0}^{1} 12 t^{2} \cdot 1 d t=\left.4 t^{3}\right|_{0} ^{1}=4.
\left\langle q_{1}, q_{1}\right\rangle=\int_{0}^{1} 1 \cdot 1 d t=\left.t\right|_{0} ^{1}=1.
\left\langle p_{3}, q_{2}\right\rangle=\int_{0}^{1} 12 t^{2}(2 t-1) d t=\int_{0}^{1}\left(24 t^{3}-12 t^{2}\right) d t=2.
\left\langle q_{2}, q_{2}\right\rangle=\int_{0}^{1}(2 t-1)^{2} d t=\left.\frac{1}{6}(2 t-1)^{3}\right|_{0} ^{1}=\frac{1}{3}.
Then
\operatorname{proj}_{W_{2}} p_{3}=\frac{\left\langle p_{3}, q_{1}\right\rangle}{\left\langle q_{1}, q_{1}\right\rangle} q_{1}+\frac{\left\langle p_{3}, q_{2}\right\rangle}{\left\langle q_{2}, q_{2}\right\rangle} q_{2}=\frac{4}{1} q_{1}+\frac{2}{1 / 3} q_{2}=4 q_{1}+6 q_{2}.
and
q_{3}=p_{3}-\operatorname{proj}_{W_{2}} p_{3}=p_{3}-4 q_{1}-6 q_{2}.
\text { As a function, } q_{3}(t)=12 t^{2}-4-6(2 t-1)=12 t^{2}-12 t+2 \text {. The orthogonal basis }\text { for the subspace } W \text { is }\left\{q_{1}, q_{2}, q_{3}\right\} .