Question

Using paper and pencil, perform the Gram-Schmidt process on the sequences of vectors given in Exercises 1 through 14.

11.$\mathbf{\left[}\begin{array}{c}\mathbf{2}\\ \mathbf{3}\\ \mathbf{0}\\ \mathbf{6}\end{array}\mathbf{\right]}\mathbf{,}\mathbf{\left[}\begin{array}{c}\mathbf{4}\\ \mathbf{4}\\ \mathbf{2}\\ \mathbf{13}\end{array}\mathbf{\right]}$

Short answer

The orthonormal vectors of the sequence is $\left[\begin{array}{c}4\\ 0\\ 0\\ 3\end{array}\right],\left[\begin{array}{c}5\\ 2\\ 14\\ 10\end{array}\right]\mathrm{is}{\stackrel{\rightharpoonup }{\mathrm{u}}}_1=\frac{1}{5}\left[\begin{array}{c}4\\ 0\\ 0\\ 3\end{array}\right],{\stackrel{\rightharpoonup }{\mathrm{u}}}_2=\frac{1}{5}\left[\begin{array}{c}-3\\ 2\\ 14\\ 4\end{array}\right],$.

Step-by-step solution

Determine the Gram-Schmidt process.

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

localid="1659441812648" $$\begin{gathered} \mathbf{Then} \\ {\stackrel{\mathbf{\rightharpoonup }}{\mathbf{u}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{||{\stackrel{\rightharpoonup }{v}}_1||}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{u}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{\rightharpoonup }{v}}_1}^{\perp }||}{{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_2}^{\mathbf{\perp }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{u}}}_{\mathbf{j}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{\rightharpoonup }{v}}_j}^{\perp }||}{{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_2}^{\mathbf{\perp }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{u}}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{||{{\stackrel{\rightharpoonup }{v}}_j}^{\perp }||}{{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{1}}}^{\mathbf{\perp }} \end{gathered}$$

Apply the Gram-Schmidt process

The given vectors are, ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{1}}=\left[\begin{array}{c}4\\ 0\\ 0\\ 3\end{array}\right],{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{1}}=\left[\begin{array}{c}5\\ 2\\ 14\\ 10\end{array}\right].$.

Obtain the values of ${\stackrel{\rightharpoonup }{u}}_1,{\stackrel{\rightharpoonup }{u}}_2$ according to the Gram-Schmidt process.

$$\begin{gathered} {\stackrel{\rightharpoonup }{u}}_1=\frac{{\stackrel{\rightharpoonup }{\mathrm{v}}}_1}{||\stackrel{\rightharpoonup }{\mathrm{v}}||}....\left(1\right) \\ {\stackrel{\rightharpoonup }{u}}_2=\frac{{\stackrel{\rightharpoonup }{\mathrm{v}}}_2-\left({\stackrel{\rightharpoonup }{u}}_1-{\stackrel{\rightharpoonup }{\mathrm{v}}}_2\right){\stackrel{\rightharpoonup }{u}}_1}{||{\stackrel{\rightharpoonup }{\mathrm{v}}}_2-\left({\stackrel{\rightharpoonup }{u}}_1-{\stackrel{\rightharpoonup }{\mathrm{v}}}_2\right){\stackrel{\rightharpoonup }{u}}_1||}.....\left(2\right) \end{gathered}$$

localid="1659441845864" $$\begin{gathered} \mathrm{Find}. \\ {\stackrel{\rightharpoonup }{\mathrm{u}}}_1=\frac{1}{\sqrt{4^2+0^2+0^2+3^2}}\left[\begin{array}{c}4\\ 0\\ 0\\ 3\end{array}\right] \\ =\frac{1}{5}\left[\begin{array}{c}4\\ 0\\ 0\\ 3\end{array}\right] \\ \mathrm{Next},\mathrm{find}{\stackrel{\rightharpoonup }{\mathrm{u}}}_2. \\ {\stackrel{\rightharpoonup }{\mathrm{u}}}_2=\frac{1}{15}\left(\left[\begin{array}{c}5\\ 2\\ 14\\ 10\end{array}\right]-\left[\begin{array}{c}8\\ 0\\ 0\\ 6\end{array}\right]\right) \\ {\stackrel{\rightharpoonup }{\mathrm{u}}}_2=\frac{1}{15}\left[\begin{array}{c}-3\\ 2\\ 14\\ 4\end{array}\right] \\ \mathrm{Thus},\mathrm{the}\mathrm{values}\mathrm{of}{\stackrel{\rightharpoonup }{\mathrm{u}}}_1=\frac{1}{5}\left[\begin{array}{c}4\\ 0\\ 0\\ 3\end{array}\right],{\stackrel{\rightharpoonup }{\mathrm{u}}}_2=\frac{1}{15}\left[\begin{array}{c}-3\\ 2\\ 14\\ 4\end{array}\right] \\ \mathrm{Hence},\mathrm{the}\mathrm{orthonormal}\mathrm{vectors}\mathrm{are}{\stackrel{\rightharpoonup }{\mathrm{u}}}_1=\frac{1}{5}\left[\begin{array}{c}4\\ 0\\ 0\\ 3\end{array}\right],{\stackrel{\rightharpoonup }{\mathrm{u}}}_2=\frac{1}{15}\left[\begin{array}{c}-3\\ 2\\ 14\\ 4\end{array}\right] \end{gathered}$$