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Expert-verified Found in: Page 224 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Using paper and pencil, perform the Gram-Schmidt process on the sequences of vectors given in Exercises 1 through 14.7. $\left[\begin{array}{c}2\\ 0\\ 1\end{array}\right]{\mathbf{,}}\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]{\mathbf{,}}\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]$

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

See the step by step solution

## Step 1: Determine the Gram-Schmidt process.

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

Then

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

## Step 2: Apply the Gram-Schmidt process

Obtain the value of $u{⃗}_{1},u{⃗}_{2}andu{⃗}_{2}$ according to Gram-Schmidt process.

$u{⃗}_{1}=\frac{v{⃗}_{1}}{||v⃗||}.........\left(1\right)\phantom{\rule{0ex}{0ex}}u{⃗}_{2}=\frac{v{⃗}_{2}-\left(u{⃗}_{1}.v{⃗}_{2}\right)u{⃗}_{1}}{||v{⃗}_{2}-\left(u{⃗}_{1}.v{⃗}_{2}\right)u{⃗}_{1}||}..........\left(2\right)\phantom{\rule{0ex}{0ex}}u{⃗}_{3}=\frac{v{⃗}_{3}-\left(u{⃗}_{1}.v{⃗}_{3}\right)u{⃗}_{1}-\left(u{⃗}_{2}.v{⃗}_{3}\right)u{⃗}_{2}}{||v{⃗}_{3}-\left(u{⃗}_{1}.v{⃗}_{3}\right)u{⃗}_{1}-\left(u{⃗}_{2}.v{⃗}_{3}\right)u{⃗}_{2}||}....\left(3\right)$

It can be observed that, $v{⃗}_{1}\cdot v{⃗}_{2}=0⇒v{⃗}_{1}\perp v{⃗}_{2}.$

Now, find $u{⃗}_{1}.$

role="math" localid="1659438380748" ${\stackrel{⇀}{u}}_{1}=\frac{1}{{2}^{2}+{2}^{2}+1}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right]$

Similarly,

role="math" localid="1659439075349" ${\stackrel{⇀}{u}}_{1}=\frac{1}{\sqrt{{\left(2\right)}^{2}+{1}^{2}+{2}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right]\phantom{\rule{0ex}{0ex}}$

## Step 3: 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 $v{⃗}_{3}-\left(u{⃗}_{1}\cdot v{⃗}_{3}\right)u{⃗}_{1}-\left(u{⃗}_{2}\cdot v{⃗}_{3}\right)u{⃗}_{2}$and $‖v{⃗}_{3}-\left(u{⃗}_{1}\cdot v{⃗}_{3}\right)u{⃗}_{1}-\left(u{⃗}_{2}\cdot v{⃗}_{3}\right)u{⃗}_{2}‖$to obtain the value of $u{⃗}_{3}$.

• role="math" localid="1659438669755" $u{⃗}_{1}\cdot v{⃗}_{3}=12$
• $u{⃗}_{2}\cdot v{⃗}_{3}=-12$

$\left(u{⃗}_{1}\cdot v{⃗}_{3}\right)u{⃗}_{1}=\left[\begin{array}{c}8\\ 8\\ 4\end{array}\right]\phantom{\rule{0ex}{0ex}}\left(u{⃗}_{2}\cdot v{⃗}_{3}\right)u{⃗}_{2}=\left[\begin{array}{c}8\\ -4\\ -8\end{array}\right]\phantom{\rule{0ex}{0ex}}v{⃗}_{3}-\left(u{⃗}_{1}\cdot v{⃗}_{3}\right)u{⃗}_{1}-\left(u{⃗}_{2}\cdot v{⃗}_{3}\right)u{⃗}_{2}=\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]-\left[\begin{array}{c}8\\ 8\\ 4\end{array}\right]-\left[\begin{array}{c}8\\ -4\\ -8\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}2\\ -4\\ 4\end{array}\right]\phantom{\rule{0ex}{0ex}}=2\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right]$

Then,

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

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 $\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right],\left[\begin{array}{c}-2/3\\ 1/3\\ 2/3\end{array}\right],\frac{1}{3}\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right]$. ### Want to see more solutions like these? 