Question

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

13.$\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]\mathbf{,}\left[\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}\right]\mathbf{,}\left[\begin{array}{c}0\\ 2\\ 1\\ -1\end{array}\right]$

Short answer

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

Step-by-step solution

Determine the Gram-Schmidt process

Consider a basis of a subspace Vof${\mathit{R}}^{\mathbf{n}}$ for$\mathit{j}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{\dots }\mathbf{,}\mathit{m}$ 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

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

Apply the Gram-Schmidt process

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

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

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

$$\begin{gathered} {\overrightarrow{u}}_1=\frac{1}{\sqrt{1^2+1^2+1^2+1^2}}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \\ =\frac{1}{2}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \end{gathered}$$

Find u→2

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 ${\overrightarrow{u}}_2$.

$\begin{array}{l}{\overrightarrow{u}}_1.{\overrightarrow{v}}_2=1\\ \left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right){\overrightarrow{u}}_1=1/2\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]\\ {\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right){\overrightarrow{u}}_1=\left[\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}\right]-\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right]\\ =1/2\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right]\end{array}$

Then,

$\left|\left|{\overrightarrow{v}}_2·\left({\overrightarrow{u}}_2.{\overrightarrow{v}}_2\right){\overrightarrow{u}}_1\right|\right|=1$

Thus,

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

Now, consider the equation below to obtain the value of${\overrightarrow{u}}_3$in 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 find out the value of ${\overrightarrow{u}}_3$.

${\overrightarrow{u}}_1·{\overrightarrow{v}}_3=1$

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

Next, 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 $\left|\left|{\overrightarrow{v}}_3-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_3\right){\overrightarrow{u}}_1-\left({\overrightarrow{u}}_2.{\overrightarrow{v}}_3\right){\overrightarrow{u}}_2\right|\right|$.

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

Then,

$\left|\left|{\overrightarrow{v}}_3-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_3\right){\overrightarrow{u}}_1-\left({\overrightarrow{u}}_2.{\overrightarrow{v}}_3\right){\overrightarrow{u}}_2\right|\right|=1$

Find ${\overrightarrow{u}}_3$.

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

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

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