Q15E In Exercises 13-16, use a rectan... [FREE SOLUTION]

91影视

Linear Algebra and its Applications

David C. Lay, Steven R. Lay and Judi J. McDonald

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 483 pages

ISBN: 978-03219822384

Chapter 1: Q15E (page 1)

In Exercises 13-16, use a rectangular coordinator system to plot $u = \left[ {\begin{array}{{20}{c}}5\\2\end{array}} \right]$, $v = \left[ {\begin{array}{{20}{c}}{ - 2}\\4\end{array}} \right]$ and their images under the given transformation $T$. (Make a separate and reasonably large sketch for each exercise.) Describe geometrically what $T$ does to each vector $x$ in ${\mathbb{R}^2}$.
$T\left( x \right) = \left[ {\begin{array}{{20}{c}}0&0\\0&1\end{array}} \right]\left[ {\begin{array}{{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]$

Short Answer

Expert verified

The transformation $T\left( x \right)$ represents the projection on the ${x_2}$ axis.

Step by step solution

Finding the rectangular coordinate

For the rectangular coordinate$u = \left[ {\begin{array}{*{20}{c}}5\\2\end{array}} \right]$, find the coordinate after thetransformation $T\left( x \right) = \left[ {\begin{array}{*{20}{c}}0&0\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]$.

$\begin{aligned} T\left( x \right) &= \left[ {\begin{array}{*{20}{c}}0&0\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5\\2\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{0 \times 5 + 0 \times 2}\\{0 \times 5 + 1 \times 2}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}0\\2\end{array}} \right]\end{aligned}$

Finding the rectangular coordinate

For therectangular coordinate $v = \left[ {\begin{array}{*{20}{c}}{ - 2}\\4\end{array}} \right]$, find the coordinate after the transformation $T\left( x \right) = \left[ {\begin{array}{*{20}{c}}0&0\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]$.

$\begin{aligned} T\left( x \right) &= \left[ {\begin{array}{*{20}{c}}0&0\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 2}\\4\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{0 \times \left( { - 2} \right) + 0 \times 4}\\{0 \times \left( { - 2} \right) + 1 \times 4}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}0\\4\end{array}} \right]\end{aligned}$

Finding the rectangular coordinate

The transformed coordinates $\left[ {\begin{array}{*{20}{c}}0\\2\end{array}} \right]$ and $\left[ {\begin{array}{*{20}{c}}0\\4\end{array}} \right]$ can be plotted as follows: