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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 14.8. $\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right]{\mathbf{,}}\left[\begin{array}{c}3\\ 6\\ 7\\ -2\end{array}\right]$

The orthonormal vectors of the sequence $\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right],\left[\begin{array}{c}3\\ 6\\ 7\\ -2\end{array}\right]is\left[\begin{array}{c}5/7\\ 4/7\\ 2/7\\ 2/7\end{array}\right],\left[\begin{array}{c}-2/7\\ 2/7\\ 5/7\\ -4/7\end{array}\right]$.

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 ${\mathbit{v}}{{\mathbf{⃗}}}_{{\mathbf{1}}}$ into its components parallel and perpendicular to the span of the preceding vectors ${\mathbit{v}}{{\mathbf{⃗}}}_{{\mathbf{1}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{,}\mathbf{v}}_{\mathbf{j}\mathbf{-}\mathbf{1}}{\mathbf{⃗}}$,

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

Let the given vectors are $v{⃗}_{1}=\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right],v{⃗}_{2}=\left[\begin{array}{c}3\\ 6\\ 7\\ -2\end{array}\right]$.

Obtain the values of , according to the Gram-Schmidt process.

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

Find $u{⃗}_{1}$.

$u{⃗}_{1}=\frac{1}{\sqrt{{5}^{2}+{4}^{2}+{2}^{2}+{2}^{2}}}\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{7}\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right]$

## Step 3: Find u⃗2

Now, here is need to find out the values of $v{⃗}_{2}-\left(u{⃗}_{1}.v{⃗}_{2}\right)u{⃗}_{1}and||v{⃗}_{2}-\left(u{⃗}_{1}.v{⃗}_{2}\right)u{⃗}_{1}||$ to obtain the value of $u{⃗}_{2}$.

Consider the equations below.

role="math" localid="1659441637408" $\stackrel{\to }{{u}_{1}}·\stackrel{\to }{{v}_{1}}=7\phantom{\rule{0ex}{0ex}}\left(\stackrel{\to }{{u}_{1}}·\stackrel{\to }{{v}_{1}}\right)\stackrel{\to }{{u}_{1}}=\left[\begin{array}{c}5\\ 4\\ 2\\ 3\end{array}\right]\phantom{\rule{0ex}{0ex}}\stackrel{\to }{{v}_{1}}-\left(\stackrel{\to }{{u}_{1}}·\stackrel{\to }{{v}_{1}}\right)\stackrel{\to }{{u}_{1}}=\left[\begin{array}{c}3\\ 6\\ 7\\ -2\end{array}\right]-\left[\begin{array}{c}5\\ 4\\ 2\\ 2\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}-2\\ 2\\ 5\\ -4\end{array}\right]\phantom{\rule{0ex}{0ex}}$

Then,

$||\stackrel{\to }{{v}_{1}}-\left(\stackrel{\to }{{u}_{1}}·\stackrel{\to }{{v}_{1}}\right)\stackrel{\to }{{u}_{1}}||=\sqrt{4+4+25+16}\phantom{\rule{0ex}{0ex}}=\sqrt{49}\phantom{\rule{0ex}{0ex}}=7$

Now find role="math" localid="1659441716297" $\stackrel{\to }{{u}_{2}}$.

$\stackrel{\to }{{u}_{2}}=\frac{1}{7}\left[\begin{array}{c}-2\\ 2\\ 5\\ -4\end{array}\right]$

Thus, the values of $u{⃗}_{1},u{⃗}_{2}$ are $\left[\begin{array}{c}5/7\\ 4/7\\ 2/7\\ 2/7\end{array}\right],\left[\begin{array}{c}-2/7\\ 2/7\\ 5/7\\ -4/7\end{array}\right]$.

Hence, the orthonormal vectors are $\left[\begin{array}{c}5/7\\ 4/7\\ 2/7\\ 2/7\end{array}\right],\left[\begin{array}{c}-2/7\\ 2/7\\ 5/7\\ -4/7\end{array}\right]$.