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 7E from Chapter 5.5 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 7E
Chapter:
Problem:
In Exercises 7–12, use Example 6 to list the eigenvalues of A. In each case, the transformation is the composition of a rotation and a scaling. Give the angle ' of the rotation, where and give the scale factor r.
Step-by-Step Solution
Given Information
We are given with a matrix A: \[ A = \left[ \begin{array} { c c } { \sqrt { 3 } } & { - 1 } \\ { 1 } & { \sqrt { 3 } } \end{array} \right] \] We have to find the eigenvalues of A and give the angle $\phi$ of the rotation and give the scale factor $r$
Step-1:
The matrix $A = \left[ \begin{array} { l l } { a } & { - b } \\ { b } & { a } \end{array} \right]$ represents a rotation through an angle $\phi$ and scaling by a factor $\lambda$, The rotation angles are given by: \[ \cos \varphi = \dfrac { a } { \sqrt { a ^ { 2 } + b ^ { 2 } } } , \sin \varphi = \dfrac { b } { \sqrt { a ^ { 2 } + b ^ { 2 } } } \] and the scale factor is: \[ | \lambda | = \sqrt { a ^ { 2 } + b ^ { 2 } } \]
Step-2: The rotation angle
On comparing the matrix A with standard matrix:
$a = \sqrt { 3 }$ and $b=1$.
The scaling factor is: \[\begin{array}{l} |\lambda | = \sqrt {{a^2} + {b^2}} \\ = \sqrt {3 + 1} \\ = 2 \end{array}\]
Step-3: The eigenvalues
The eigenvalues are: \[a \pm bi = \sqrt 3 \pm i\]
Step-4: The rotation angle
\[\begin{array}{l} \cos \phi = \dfrac{a}{{\sqrt {{a^2} + {b^2}} }}\\ \\ \cos \phi = \dfrac{{\sqrt 3 }}{{\sqrt {{{\left( {\sqrt 3 } \right)}^2} + {1^2}} }}\\ \\ \cos \phi = \dfrac{{\sqrt 3 }}{{\sqrt 4 }}\\ \\ \cos \phi = \dfrac{{\sqrt 3 }}{2}\\ \\ \phi = 30^\circ \end{array}\] Therefore,
The angle of rotation is $30^\circ$ and scaling factor is 2
We are given with a matrix A: \[ A = \left[ \begin{array} { c c } { \sqrt { 3 } } & { - 1 } \\ { 1 } & { \sqrt { 3 } } \end{array} \right] \] We have to find the eigenvalues of A and give the angle $\phi$ of the rotation and give the scale factor $r$
Step-1:
The matrix $A = \left[ \begin{array} { l l } { a } & { - b } \\ { b } & { a } \end{array} \right]$ represents a rotation through an angle $\phi$ and scaling by a factor $\lambda$, The rotation angles are given by: \[ \cos \varphi = \dfrac { a } { \sqrt { a ^ { 2 } + b ^ { 2 } } } , \sin \varphi = \dfrac { b } { \sqrt { a ^ { 2 } + b ^ { 2 } } } \] and the scale factor is: \[ | \lambda | = \sqrt { a ^ { 2 } + b ^ { 2 } } \]
Step-2: The rotation angle
On comparing the matrix A with standard matrix:
$a = \sqrt { 3 }$ and $b=1$.
The scaling factor is: \[\begin{array}{l} |\lambda | = \sqrt {{a^2} + {b^2}} \\ = \sqrt {3 + 1} \\ = 2 \end{array}\]
Step-3: The eigenvalues
The eigenvalues are: \[a \pm bi = \sqrt 3 \pm i\]
Step-4: The rotation angle
\[\begin{array}{l} \cos \phi = \dfrac{a}{{\sqrt {{a^2} + {b^2}} }}\\ \\ \cos \phi = \dfrac{{\sqrt 3 }}{{\sqrt {{{\left( {\sqrt 3 } \right)}^2} + {1^2}} }}\\ \\ \cos \phi = \dfrac{{\sqrt 3 }}{{\sqrt 4 }}\\ \\ \cos \phi = \dfrac{{\sqrt 3 }}{2}\\ \\ \phi = 30^\circ \end{array}\] Therefore,
The angle of rotation is $30^\circ$ and scaling factor is 2