91Ó°ÊÓ

Let us find out subspace and its basis then we will change that basis to the orthonormal basis.

Consider the equations below.

Thus, a basis of the subspace will be$\left\{v_1\left(t\right)=1,v_2\left(t\right)=t^2\right\}$.

Now, apply the Gram-Schmidt process to get an orthonormal basis. Observe that if it takes $u_1\left(t\right)=v_1\left(t\right)$,

$$\begin{gathered} ||u_1\left(t\right)||=\frac{1}{2}{∫}_{-1}^{1}g{\left(t\right)}^{2}dt \\ ={∫}_{-1}^{1}1dt \\ =\frac{1}{2}×2 \\ =1 \end{gathered}$$

Consider a basis of a subspace Vof${\mathit{R}}^{\mathbf{n}}$ for j= 2,..., m we resolve the vector${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}}$into its components parallel and perpendicular to the span of the preceding vectors ${\stackrel{\mathit{→}}{\mathit{v}}}_{\mathit{1}}\mathit{,}\mathit{.}\mathit{.}\mathit{.}\mathit{,}{\stackrel{\mathit{→}}{\mathit{v}}}_{\mathit{j}\mathit{-}\mathit{1}}$,

Then

${\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{||{\stackrel{→}{v}}_1||}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{→}{v}}_2}^{âŠ\yen }||}{{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{2}}}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{j}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{→}{v}}_m}^{âŠ\yen }||}{{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}}}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{→}{v}}_j}^{âŠ\yen }||}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{m}}$

Now,Gram-Schmidt process allows us to find the second vector.

Further solve the above expression,

$$\begin{gathered} u_2\left(t\right)=\frac{t^2-{\displaystyle \frac{1}{3}}}{||t^2-{\displaystyle \frac{1}{3}}||} \\ =\frac{\sqrt{5}}{2}\left(3t^2-1\right) \end{gathered}$$

Hence, the orthonormal basis for subspace is$\left\{1,\frac{\sqrt{5}}{2}\left(3t^2-1\right)\right\}$ .