Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 14E from Chapter 6.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 14E
Chapter:
Problem:
Write y as the sum of a vector in Span {u} and a vector orthogonal to u...
Step-by-Step Solution
Step 1
Given Vectors:\[\begin{array}{l}{\bf{y}} = \left[ {\begin{array}{*{20}{c}}2\\6\end{array}} \right]\\{\bf{u}} = \left[ {\begin{array}{*{20}{c}}7\\1\end{array}} \right]\end{array}\]We have to write the vector $y$ as the sum of vector $u$ and a vector orthogonal to $u$.
Step 2: The projection of $y$ onto $u$
\[\begin{array}{l}{\bf{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over y} }} = \dfrac{{{\bf{\vec y}}{\bf{.\vec u}}}}{{{\bf{\vec u}}{\bf{.\vec u}}}}{\bf{\vec u}}\\{\bf{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over y} }} = \dfrac{{\left[ {\begin{array}{*{20}{c}}2&6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}7\\1\end{array}} \right]}}{{\left[ {\begin{array}{*{20}{c}}7&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}7\\1\end{array}} \right]}}{\bf{\vec u}}\\ = \dfrac{{14 + 6}}{{{7^2} + {1^1}}}{\bf{\vec u}}\\ = \dfrac{{20}}{{50}}{\bf{\vec u}}\\ = \dfrac{2}{5}\left[ {\begin{array}{*{20}{c}}7\\1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{\dfrac{{14}}{5}}\\{\dfrac{2}{5}}\end{array}} \right]\end{array}\]
Step 3: The component of $y$ orthogonal to $u$
\[\begin{array}{l}{\bf{y}} - {\bf{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over y} }} = \left[ {\begin{array}{*{20}{c}}2\\6\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{\dfrac{{14}}{5}}\\{\dfrac{2}{5}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - \dfrac{4}{5}}\\{\dfrac{{28}}{5}}\end{array}} \right]\end{array}\]
ANSWERS
The required vector as sum of two vectors \[{\bf{y}} = \left[ {\begin{array}{*{20}{c}}{\dfrac{{14}}{5}}\\{\dfrac{2}{5}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - \dfrac{4}{5}}\\{\dfrac{{28}}{5}}\end{array}} \right]\]
Given Vectors:\[\begin{array}{l}{\bf{y}} = \left[ {\begin{array}{*{20}{c}}2\\6\end{array}} \right]\\{\bf{u}} = \left[ {\begin{array}{*{20}{c}}7\\1\end{array}} \right]\end{array}\]We have to write the vector $y$ as the sum of vector $u$ and a vector orthogonal to $u$.
Step 2: The projection of $y$ onto $u$
\[\begin{array}{l}{\bf{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over y} }} = \dfrac{{{\bf{\vec y}}{\bf{.\vec u}}}}{{{\bf{\vec u}}{\bf{.\vec u}}}}{\bf{\vec u}}\\{\bf{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over y} }} = \dfrac{{\left[ {\begin{array}{*{20}{c}}2&6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}7\\1\end{array}} \right]}}{{\left[ {\begin{array}{*{20}{c}}7&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}7\\1\end{array}} \right]}}{\bf{\vec u}}\\ = \dfrac{{14 + 6}}{{{7^2} + {1^1}}}{\bf{\vec u}}\\ = \dfrac{{20}}{{50}}{\bf{\vec u}}\\ = \dfrac{2}{5}\left[ {\begin{array}{*{20}{c}}7\\1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{\dfrac{{14}}{5}}\\{\dfrac{2}{5}}\end{array}} \right]\end{array}\]
Step 3: The component of $y$ orthogonal to $u$
\[\begin{array}{l}{\bf{y}} - {\bf{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over y} }} = \left[ {\begin{array}{*{20}{c}}2\\6\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{\dfrac{{14}}{5}}\\{\dfrac{2}{5}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - \dfrac{4}{5}}\\{\dfrac{{28}}{5}}\end{array}} \right]\end{array}\]
ANSWERS
The required vector as sum of two vectors \[{\bf{y}} = \left[ {\begin{array}{*{20}{c}}{\dfrac{{14}}{5}}\\{\dfrac{2}{5}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - \dfrac{4}{5}}\\{\dfrac{{28}}{5}}\end{array}} \right]\]