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Q43E

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Found in: Page 217

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# In Exercises 40 through 46, consider vectors$\stackrel{\mathbf{\to }}{{\mathbf{V}}_{\mathbf{1}}}{\mathbf{,}}\stackrel{\mathbf{\to }}{{\mathbf{V}}_{\mathbf{2}}}{\mathbf{,}}\stackrel{\mathbf{\to }}{{\mathbf{V}}_{\mathbf{3}}}$ in${{\mathbf{R}}}^{{\mathbf{4}}}$ ; we are told that$\stackrel{\mathbf{\to }}{{\mathbf{V}}_{\mathbf{i}}}{\mathbf{,}}\stackrel{\mathbf{\to }}{{\mathbf{V}}_{\mathbf{j}}}$ is the entry ${{\mathbf{a}}}_{{\mathbf{ij}}}$of matrix A.${\mathbf{A}}{\mathbf{=}}\left[\begin{array}{ccc}3& 5& 11\\ 5& 9& 20\\ 11& 20& 49\end{array}\right]$Find ${{\mathbf{proj}}}_{\stackrel{\mathbf{\to }}{\mathbf{v}}}\left({\stackrel{\to }{v}}_{1}\right)$, expressed as a scalar multiple of${\stackrel{\mathbf{\to }}{\mathbf{v}}}_{{\mathbf{2}}}$ .

The required projection is $pro{j}_{{\stackrel{\to }{v}}_{2}}\left({\stackrel{\to }{v}}_{1}\right)=\frac{5}{9}{\stackrel{\to }{v}}_{2}$

See the step by step solution

## Step 1: Formula for the orthogonal projection.

If V is a subspace of ${{\mathbf{R}}}^{{\mathbf{n}}}$with an orthonormal basis ${\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{1}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{,}}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{m}}}$thenlocalid="1659451887578" ${{\mathbf{proj}}}_{{\mathbf{v}}}\stackrel{\mathbf{\to }}{\mathbf{x}}{\mathbf{=}}{\stackrel{\mathbf{\to }}{\mathbf{x}}}^{{\mathbf{ll}}}{\mathbf{=}}\mathbf{\left(}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{\mathbf{1}}\mathbf{.}\stackrel{\mathbf{\to }}{\mathbf{x}}\mathbf{\right)}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{1}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}\mathbf{\left(}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{\mathbf{1}}\mathbf{.}\stackrel{\mathbf{\to }}{\mathbf{x}}\mathbf{\right)}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{m}}}$

For all $\stackrel{\to }{x}$in localid="1659436140715" ${R}^{n}$.

Let us write the matrix in the $\stackrel{\to }{{V}_{i}}.\stackrel{\to }{{V}_{i}}$notation.

Consider the terms below.

localid="1659451940364" $A=\left[\begin{array}{ccc}3& 5& 11\\ 5& 9& 20\\ 11& 20& 49\end{array}\right]=\left[\begin{array}{ccc}{\stackrel{\to }{v}}_{1}.{\stackrel{\to }{v}}_{1}& {\stackrel{\to }{v}}_{1}.{\stackrel{\to }{v}}_{2}& {\stackrel{\to }{v}}_{1}.{\stackrel{\to }{v}}_{3}\\ {\stackrel{\to }{v}}_{2}.{\stackrel{\to }{v}}_{1}& {\stackrel{\to }{v}}_{2}.{\stackrel{\to }{v}}_{2}& {\stackrel{\to }{v}}_{2}.{\stackrel{\to }{v}}_{3}\\ {\stackrel{\to }{v}}_{3}.{\stackrel{\to }{v}}_{1}& {\stackrel{\to }{v}}_{3}.{\stackrel{\to }{v}}_{2}& {\stackrel{\to }{v}}_{3}.{\stackrel{\to }{v}}_{3}\end{array}\right]=\left[\begin{array}{ccc}{||{\stackrel{\to }{v}}_{1}||}^{2}& {\stackrel{\to }{v}}_{1}.{\stackrel{\to }{v}}_{2}& {\stackrel{\to }{v}}_{1}.{\stackrel{\to }{v}}_{3}\\ {\stackrel{\to }{v}}_{2}.{\stackrel{\to }{v}}_{1}& {||{\stackrel{\to }{v}}_{2}||}^{2}& {\stackrel{\to }{v}}_{2}.{\stackrel{\to }{v}}_{3}\\ {\stackrel{\to }{v}}_{3}.{\stackrel{\to }{v}}_{1}& {\stackrel{\to }{v}}_{3}.{\stackrel{\to }{v}}_{2}& {||{\stackrel{\to }{v}}_{3}||}^{2}\end{array}\right]$

## Step 2: obtain the orthonormal basis for span(v→2)

To obtain the span. Since it is a singleton vector. So by normalizing it. Therefore, an orthonormal basis will be.

$\frac{{\stackrel{\to }{v}}_{2}}{||{\stackrel{\to }{v}}_{2}||}=\frac{{\stackrel{\to }{v}}_{2}}{3}=\stackrel{\to }{u}$.

So, according to the above theorem

$pro{j}_{{\stackrel{\to }{v}}_{2}}\left({\stackrel{\to }{v}}_{1}\right)=\left(\stackrel{\to }{u}.{\stackrel{\to }{v}}_{1}\right)\stackrel{\to }{u}\phantom{\rule{0ex}{0ex}}=\frac{{\stackrel{\to }{v}}_{2}}{3}.{\stackrel{\to }{v}}_{1}\frac{{\stackrel{\to }{v}}_{2}}{3}\phantom{\rule{0ex}{0ex}}=\frac{{\stackrel{\to }{v}}_{2}.{\stackrel{\to }{v}}_{1}}{3}\frac{{\stackrel{\to }{v}}_{2}}{3}\phantom{\rule{0ex}{0ex}}=\frac{5}{9}{\stackrel{\to }{v}}_{2}$

Hence, the role="math" localid="1659438206334" $pro{j}_{{\stackrel{\to }{v}}_{2}}\left({\stackrel{\to }{v}}_{1}\right)=\frac{5}{9}{\stackrel{\to }{v}}_{2}$.