Question

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

7.$\left[\begin{array}{c}2\\ 0\\ 1\end{array}\right]\mathbf{,}\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]\mathbf{,}\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]$

Short answer

The orthonormal vectors of the sequence $\left[\begin{array}{c}2\\ 0\\ 1\end{array}\right],\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right],\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]is\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right],\left[\begin{array}{c}-2/3\\ 1/3\\ 2/3\end{array}\right],\frac{1}{3}\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right].$

Step-by-step solution

Determine the Gram-Schmidt process.

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

Then

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

Apply the Gram-Schmidt process

Obtain the value of $u{⃗}_1,u{⃗}_{2}andu{⃗}_2$according toGram-Schmidt process.

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

It can be observed that, $v{⃗}_1\cdot v{⃗}_2=0\Rightarrow v{⃗}_1\perp v{⃗}_2.$

Now, find $u{⃗}_1.$

role="math" localid="1659438380748" $$\begin{gathered} {\stackrel{\rightharpoonup }{u}}_1=\frac{1}{2^2+2^2+1}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right] \\ =\frac{1}{3}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right] \\ =\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right] \end{gathered}$$

Similarly,

role="math" localid="1659439075349" $$\begin{gathered} {\stackrel{\rightharpoonup }{u}}_1=\frac{1}{\sqrt{{\left(2\right)}^2+1^2+2^2}} \\ =\frac{1}{3}\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right] \\ =\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right] \end{gathered}$$

Find u⃗3      

Consider the equation below to obtain the value of $u{⃗}_3$in equation (3).

Here is need to find out the values of $v{⃗}_3-(u{⃗}_1\cdot v{⃗}_3)u{⃗}_1-(u{⃗}_2\cdot v{⃗}_3)u{⃗}_2$and $\Vert v{⃗}_3-(u{⃗}_1\cdot v{⃗}_3)u{⃗}_1-(u{⃗}_2\cdot v{⃗}_3)u{⃗}_2\Vert$to obtain the value of $u{⃗}_3$.

• role="math" localid="1659438669755" $u{⃗}_1\cdot v{⃗}_3=12$
  
• $u{⃗}_2\cdot v{⃗}_3=-12$
  

$$\begin{gathered} \left(u{⃗}_1\cdot v{⃗}_3\right)u{⃗}_1=\left[\begin{array}{c}8\\ 8\\ 4\end{array}\right] \\ \left(u{⃗}_2\cdot v{⃗}_3\right)u{⃗}_2=\left[\begin{array}{c}8\\ -4\\ -8\end{array}\right] \\ v{⃗}_3-\left(u{⃗}_1\cdot v{⃗}_3\right)u{⃗}_1-\left(u{⃗}_2\cdot v{⃗}_3\right)u{⃗}_2=\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]-\left[\begin{array}{c}8\\ 8\\ 4\end{array}\right]-\left[\begin{array}{c}8\\ -4\\ -8\end{array}\right] \\ =\left[\begin{array}{c}2\\ -4\\ 4\end{array}\right] \\ =2\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right] \end{gathered}$$

Then,

role="math" localid="1659439148633" $\stackrel{\rightharpoonup }{u_3}=\frac{1}{3}\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right]$

Thus, the values of ,$u{⃗}_1,u{⃗}_{2}andu{⃗}_3$ are role="math" localid="1659439162227" $\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right],\left[\begin{array}{c}-2/3\\ 1/3\\ 2/3\end{array}\right],\frac{1}{3}\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right]$.

Hence, the orthonormal vectors are$\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right],\left[\begin{array}{c}-2/3\\ 1/3\\ 2/3\end{array}\right],\frac{1}{3}\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right]$.