Question

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

14. $\mathbf{\left[}\begin{array}{c}\mathbf{1}\\ \mathbf{7}\\ \mathbf{1}\\ \mathbf{7}\end{array}\mathbf{\right]}\mathbf{,}\mathbf{\left[}\begin{array}{c}\mathbf{0}\\ \mathbf{7}\\ \mathbf{2}\\ \mathbf{7}\end{array}\mathbf{\right]}\mathbf{,}\mathbf{\left[}\begin{array}{c}\mathbf{1}\\ \mathbf{8}\\ \mathbf{1}\\ \mathbf{6}\end{array}\mathbf{\right]}$

Short answer

The orthonormal vectors of the sequence$\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right],\left[\begin{array}{c}0\\ 7\\ 2\\ 7\end{array}\right],\left[\begin{array}{c}1\\ 8\\ 1\\ 6\end{array}\right]$is $1/10\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right],1/\sqrt{2}\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right],1/\sqrt{2}\left[\begin{array}{c}0\\ 1\\ 0\\ -1\end{array}\right]$.

Step-by-step solution

Determine the Gram-Schmidt process.

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

Then

${\overrightarrow{\mathit{u}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\left|\left|{\overrightarrow{v}}_1\right|\right|}{\overrightarrow{\mathit{v}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathit{u}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\left|\left|{{\overrightarrow{v}}_2}^{\perp }\right|\right|}{{\overrightarrow{\mathit{v}}}_{\mathbf{2}}}^{\mathbf{\perp }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\overrightarrow{\mathit{u}}}_{\mathit{j}}\mathbf{=}\frac{\mathbf{1}}{\left|\left|{{\overrightarrow{v}}_j}^{\perp }\right|\right|}{{\overrightarrow{\mathit{v}}}_{\mathit{j}}}^{\mathbf{\perp }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\overrightarrow{\mathit{u}}}_{\mathit{m}}\mathbf{=}\frac{\mathbf{1}}{\left|\left|{{\overrightarrow{v}}_m}^{\perp }\right|\right|}{{\overrightarrow{\mathit{v}}}_{\mathit{m}}}^{\mathbf{\perp }}$

Apply the Gram-Schmidt process

Obtain the value of ${\overrightarrow{u}}_1$,${\overrightarrow{u}}_2$ and${\overrightarrow{u}}_3$ according to Gram-Schmidt process.

$$\begin{gathered} {\overrightarrow{u}}_1=\frac{{\overrightarrow{v}}_1}{\left|\left|\overrightarrow{v}\right|\right|}....\left(1\right) \\ {\overrightarrow{u}}_2=\frac{{\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{u}}_2\right){\overrightarrow{u}}_1}{\left|\left|{\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{u}}_2\right){\overrightarrow{u}}_1\right|\right|}....\left(2\right) \\ {\overrightarrow{u}}_3=\frac{{\overrightarrow{v}}_3-\left({\overrightarrow{u}}_1.{\overrightarrow{u}}_3\right){\overrightarrow{u}}_1-\left({\overrightarrow{u}}_2.{\overrightarrow{u}}_3\right){\overrightarrow{u}}_2}{\left|\left|{\overrightarrow{v}}_3-\left({\overrightarrow{u}}_1.{\overrightarrow{u}}_3\right){\overrightarrow{u}}_1-\left({\overrightarrow{u}}_2.{\overrightarrow{u}}_3\right){\overrightarrow{u}}_2\right|\right|}....\left(3\right) \end{gathered}$$

Find ${\overrightarrow{u}}_1$.

$$\begin{gathered} {\overrightarrow{u}}_1=\frac{1}{\sqrt{100}}\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right] \\ =\frac{1}{\sqrt{10}}\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right] \end{gathered}$$

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

$$\begin{gathered} {\overrightarrow{u}}_1.{\overrightarrow{v}}_2=10 \\ \left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right){\overrightarrow{u}}_1=\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right] \\ {\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right){\overrightarrow{u}}_1=\left[\begin{array}{c}0\\ 7\\ 2\\ 7\end{array}\right]-\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right]=\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right] \\ \left|\left|{\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right){\overrightarrow{u}}_1\right|\right|=\sqrt{2} \end{gathered}$$

Next, solve for${\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1·{\overrightarrow{v}}_2\right){\overrightarrow{u}}_1$and $\left|\left|{\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1·{\overrightarrow{v}}_2\right){\overrightarrow{u}}_1\right|\right|$.

role="math" localid="1659952093244" $$\begin{gathered} {\overrightarrow{u}}_1.{\overrightarrow{v}}_2=10 \\ \left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right){\overrightarrow{u}}_1=\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right] \\ {\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right){\overrightarrow{u}}_1=\left[\begin{array}{c}0\\ 7\\ 2\\ 7\end{array}\right]-\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right]=\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right] \\ \left|\left|{\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right){\overrightarrow{u}}_1\right|\right|=\sqrt{2} \end{gathered}$$

Thus,

${\overrightarrow{u}}_2=\frac{1}{\sqrt{2}}\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right]$

Find u→3

Now, consider the equation below to obtain the value ofin equation (3).

Here is need to find out the values of${\overrightarrow{v}}_3-\left({\overrightarrow{u}}_1·{\overrightarrow{v}}_3\right){\overrightarrow{u}}_1-\left({\overrightarrow{u}}_2·{\overrightarrow{v}}_3\right){\overrightarrow{u}}_2$ and$\left|\left|{\overrightarrow{v}}_3-\left({\overrightarrow{u}}_1·{\overrightarrow{v}}_3\right){\overrightarrow{u}}_1-\left({\overrightarrow{u}}_2·{\overrightarrow{v}}_3\right){\overrightarrow{u}}_2\right|\right|$to obtain the value of ${\overrightarrow{u}}_3$.

$$\begin{gathered} {\overrightarrow{u}}_1·{\overrightarrow{v}}_3=10 \\ \left({\overrightarrow{u}}_1·{\overrightarrow{v}}_3\right){\overrightarrow{u}}_1=1/2\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right] \\ {\overrightarrow{v}}_3.{\overrightarrow{u}}_2=0 \\ \left({\overrightarrow{u}}_2·{\overrightarrow{v}}_3\right)=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right] \end{gathered}$$

Now find, ${\overrightarrow{v}}_3-\left({\overrightarrow{u}}_1·{\overrightarrow{v}}_3\right){\overrightarrow{u}}_1-\left({\overrightarrow{u}}_2·{\overrightarrow{v}}_3\right){\overrightarrow{u}}_2$and $\left|\left|{\overrightarrow{v}}_3-\left({\overrightarrow{u}}_1·{\overrightarrow{v}}_3\right){\overrightarrow{u}}_1-\left({\overrightarrow{u}}_2·{\overrightarrow{v}}_3\right){\overrightarrow{u}}_2\right|\right|$.

$$\begin{gathered} {\overrightarrow{v}}_3-\left({\overrightarrow{u}}_1·{\overrightarrow{v}}_3\right){\overrightarrow{u}}_1-\left({\overrightarrow{u}}_2·{\overrightarrow{v}}_3\right){\overrightarrow{u}}_2=\left[\begin{array}{c}1\\ 8\\ 1\\ 6\end{array}\right]-\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right]-\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right] \\ =\left[\begin{array}{c}0\\ 1\\ 0\\ -1\end{array}\right] \\ \Rightarrow \left|\left|{\overrightarrow{v}}_3-\left({\overrightarrow{u}}_1·{\overrightarrow{v}}_3\right){\overrightarrow{u}}_1-\left({\overrightarrow{u}}_2·{\overrightarrow{v}}_3\right){\overrightarrow{u}}_2\right|\right|=\sqrt{2} \end{gathered}$$

Now, find ${\overrightarrow{u}}_3$.

${\overrightarrow{u}}_3=\frac{1}{\sqrt{2}}\left[\begin{array}{c}0\\ 1\\ 0\\ -1\end{array}\right]$

Thus, the values of ${\overrightarrow{u}}_1$,${\overrightarrow{u}}_2$ and${\overrightarrow{u}}_3$ are $1/10\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right],1/\sqrt{2}\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right],1/\sqrt{2}\left[\begin{array}{c}0\\ 1\\ 0\\ -1\end{array}\right]$.

Hence, the orthonormal vectors are $1/10\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right],1/\sqrt{2}\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right],1/\sqrt{2}\left[\begin{array}{c}0\\ 1\\ 0\\ -1\end{array}\right]$.