Question

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

12.$\left[\begin{array}{c}2\\ 3\\ 0\\ 6\end{array}\left|\begin{array}{c}4\\ 4\\ 2\\ 13\end{array}\right.\right]$

Short answer

The orthonormal vectors of the sequence is $\left|\begin{array}{c}222\\ 3\\ 0\\ [6\end{array}\right|\left.\begin{array}{c}4\\ 4\\ 2\\ 13\end{array}\right]\mathrm{is}$$\left[\begin{array}{c}2/7\\ 3/7\\ 0\\ 6/7\end{array}\right],\frac{1}{3}\left[\begin{array}{c}0\\ -2\\ 2\\ 1\end{array}\right]$.

Step-by-step solution

Determine the Gram-Schmidt process.

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

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

Apply the Gram-Schmidt process

The given vectors are ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{1}}=\left[\begin{array}{c}2\\ 3\\ 0\\ 6\end{array}\right],{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_2=\left[\begin{array}{c}4\\ 4\\ 2\\ 13\end{array}\right]$.

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

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

$$\begin{gathered} \mathrm{Find}. \\ {\stackrel{\rightharpoonup }{\mathrm{u}}}_1=\frac{1}{\sqrt{2^2+3^2+0^2+6^2}}\left[\begin{array}{c}2\\ 3\\ 0\\ 6\end{array}\right] \\ =\frac{1}{7}\left[\begin{array}{c}2\\ 3\\ 0\\ 6\end{array}\right] \end{gathered}$$

Find u⇀2

Now, here is need to find out the values ${\stackrel{\rightharpoonup }{v}}_2-\left({\stackrel{\rightharpoonup }{u}}_1.u_2\right)\mathrm{and}||{\stackrel{\rightharpoonup }{\mathrm{v}}}_2-\left({\stackrel{\rightharpoonup }{\mathrm{u}}}_1.{\stackrel{\rightharpoonup }{\mathrm{u}}}_2\right){\stackrel{\rightharpoonup }{\mathrm{u}}}_1||$of and to obtain the value of ${\stackrel{\rightharpoonup }{\mathrm{u}}}_2$.

$\begin{array}{l}{\stackrel{\rightharpoonup }{u}}_{1-}{\stackrel{\rightharpoonup }{u}}_2=14\\ \left({\stackrel{\rightharpoonup }{u}}_{1-}{\stackrel{\rightharpoonup }{u}}_2\right){\stackrel{\rightharpoonup }{u}}_2=\left[\begin{array}{c}4\\ 6\\ 0\\ 12\end{array}\right]\end{array}$

And,

$$\begin{gathered} {\stackrel{\rightharpoonup }{v}}_2-\left({\stackrel{\rightharpoonup }{u}}_1.{\stackrel{\rightharpoonup }{u}}_2\right)=\left[\begin{array}{c}4\\ 4\\ 2\\ 3\end{array}\right]-\left[\begin{array}{c}4\\ 6\\ 0\\ 12\end{array}\right] \\ \left[\begin{array}{c}0\\ -2\\ 2\\ 1\end{array}\right] \end{gathered}$$

Then,

$$\begin{gathered} ||{\stackrel{\rightharpoonup }{v}}_2-\left({\stackrel{\rightharpoonup }{u}}_1-{\stackrel{\rightharpoonup }{u}}_2\right){\stackrel{\rightharpoonup }{u}}_1||=\sqrt{0+4+4+1} \\ =\sqrt{9} \\ =3 \\ \mathrm{Now},\mathrm{find}{\stackrel{\rightharpoonup }{\mathrm{u}}}_2. \\ {\stackrel{\rightharpoonup }{\mathrm{u}}}_2=\frac{1}{3}\left[\begin{array}{c}0\\ -2\\ 2\\ 1\end{array}\right] \\ \mathrm{Thus},\mathrm{the}\mathrm{values}\mathrm{of}{\stackrel{\rightharpoonup }{\mathrm{u}}}_1,\mathrm{and}{\stackrel{\rightharpoonup }{\mathrm{u}}}_2\mathrm{are}\left[\begin{array}{c}2/7\\ 3/7\\ 0\\ 6/7\end{array}\right],\frac{1}{3}\left[\begin{array}{c}0\\ -2\\ 2\\ 1\end{array}\right] \\ \mathrm{Hence},\mathrm{the}\mathrm{orthonormal}\mathrm{vectors}\mathrm{are}\left[\begin{array}{c}2/7\\ 3/7\\ 0\\ 6/7\end{array}\right],\frac{1}{3}\left[\begin{array}{c}0\\ -2\\ 2\\ 1\end{array}\right]. \end{gathered}$$