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\(\begin{array}{lll}\text { Exercise } & \mathbf{3 . 3 . 5} & \text { Show that the eigenvalues of }\end{array}\) \(\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]\) are \(e^{i \theta}\) and \(e^{-i \theta} .\) (See Appendix A)

Short Answer

Expert verified
The eigenvalues are \(e^{i\theta}\) and \(e^{-i\theta}\).

Step by step solution

01

Understanding the Matrix and Eigenvalues

The given matrix is a rotation matrix:\[A = \begin{bmatrix}\cos \theta & -\sin \theta \ \sin \theta & \cos \theta\end{bmatrix}\]We need to find its eigenvalues, which are scalars \( \lambda \) satisfying the equation \( \det(A - \lambda I) = 0 \).
02

Writing Down the Characteristic Equation

The characteristic equation is\[\det(\begin{bmatrix}\cos \theta & -\sin \theta \ \sin \theta & \cos \theta\end{bmatrix} - \lambda \begin{bmatrix}1 & 0 \ 0 & 1 \end{bmatrix}) = 0\]This expands to\[\det(\begin{bmatrix}\cos \theta - \lambda & -\sin \theta \ \sin \theta & \cos \theta - \lambda \end{bmatrix}) = 0\]
03

Computing the Determinant

We compute the determinant:\[(\cos \theta - \lambda)(\cos \theta - \lambda) - (-\sin \theta)(\sin \theta) = 0\]Simplifying, we get:\[(\cos \theta - \lambda)^2 + \sin^2 \theta = 0\]
04

Simplifying the Equation

Combine and simplify:\[\lambda^2 - 2\lambda \cos\theta + \cos^2\theta + \sin^2\theta = 0\]Using \( \cos^2 \theta + \sin^2 \theta = 1 \), we get:\[\lambda^2 - 2\lambda \cos\theta + 1 = 0\]
05

Solving the Quadratic Equation

Solve the quadratic equation \( \lambda^2 - 2\lambda \cos\theta + 1 = 0 \) using the quadratic formula:\[\lambda = \frac{2\cos\theta \pm \sqrt{(2\cos\theta)^2 - 4}}{2}\]Simplifying gives:\[\lambda = \cos \theta \pm \sqrt{\cos^2\theta - 1}\]
06

Applying Complex Roots Simplification

Recognize that the term \( \sqrt{\cos^2\theta - 1} \) simplifies to \( i\sin\theta \), since \( \cos^2\theta - 1 = -\sin^2\theta \):\[\lambda = \cos \theta \pm i\sin\theta\]
07

Verifying the Complex Exponential Form

Recall Euler’s formula \( e^{i \theta} = \cos \theta + i\sin \theta \), thus:\[\lambda_1 = e^{i\theta}, \quad \lambda_2 = e^{-i\theta}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotation Matrices
Rotation matrices are fundamental in linear algebra used to rotate vectors in a plane. The general form of a 2D rotation matrix is:
  • \[A = \begin{bmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{bmatrix}\]
Here, the angle \( \theta \) represents the counterclockwise rotation of the vector. When applied to a vector, this matrix alters the direction but not the magnitude, hence the term "rotation."
It's important to understand that these matrices preserve lengths and angles, making them orthogonal, with a determinant of 1. This feature ensures that the rotation itself is rigid, i.e., it doesn't stretch or compress the original vector.
Characteristic Equation
The characteristic equation is central to finding the eigenvalues of a matrix. This involves the determinant of the matrix subtracted by the product of a variable, \( \lambda \), and an identity matrix. Formally, for a matrix \( A \), the characteristic equation is given by:
  • \[\det(A - \lambda I) = 0\]
This equation is derived by setting the determinant of the original matrix minus \( \lambda \) times the identity matrix to zero, yielding a polynomial equation in \( \lambda \). The roots of this polynomial are the eigenvalues.
In the context of rotation matrices, solving this quadratic gives insight into the behavior of the transformation applied by the matrix.
Complex Numbers
Complex numbers are crucial when solving problems involving rotations, especially when the solutions involve imaginary components. A complex number is expressed as:
  • \[a + bi\]
where \( a \) is the real part and \( b \) is the imaginary part, involving \( i = \sqrt{-1} \).
For rotation matrices, the appearance of complex numbers is linked to the trigonometric identities and Pythagorean identities that govern the functions sine and cosine. When solving for the eigenvalues of a rotation matrix, the real and imaginary parts reflect the cosine and sine terms in the corresponding expressions, ultimately vital for expressing the eigenvalues in exponential form via Euler's formula.
Euler's Formula
Euler's formula elegantly bridges complex numbers and trigonometry, showing the close relationship between exponential and trigonometric functions. It is expressed as:
  • \[e^{i\theta} = \cos \theta + i\sin \theta\]
This formula is particularly useful to showcase how rotation transformations can be expressed with complex exponentials.
Euler’s formula facilitates converting the eigenvalues derived from the characteristic equation into a simpler form that uses the exponential function. This conversion helps solve problems involving wave functions, oscillations, and quantum mechanics. In the case of the rotation matrix, the eigenvalues \( e^{i\theta} \) and \( e^{-i\theta} \) align with Euler's formula, highlighting the deep connection between rotations and exponential growth.

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Most popular questions from this chapter

If \(u\) is a number and \(A\) is an \(n \times n\) matrix, prove that \(\operatorname{det}(u A)=u^{n}\) det \(A\) by induction on \(n,\) using only the definition of \(\operatorname{det} A\).

a. Show that the only diagonalizable matrix \(A\) that has only one eigenvalue \(\lambda\) is the scalar matrix \(A=\lambda I\) b. Is \(\left[\begin{array}{cc}3 & -2 \\ 2 & -1\end{array}\right]\) diagonalizable?

Let \(A\) and \(B\) denote invertible \(n \times n\) matrices. Show that: a. \(\operatorname{adj}(\operatorname{adj} A)=(\operatorname{det} A)^{n-2} A\) (here \(\left.n \geq 2\right)\) [Hint: See Example 3.2.8.] b. \(\operatorname{adj}\left(A^{-1}\right)=(\operatorname{adj} A)^{-1}\) c. \(\operatorname{adj}\left(A^{T}\right)=(\operatorname{adj} A)^{T}\) d. \(\operatorname{adj}(A B)=(\operatorname{adj} B)(\operatorname{adj} A)\) [Hint: Show that \(A B \operatorname{adj}(A B)=A B \operatorname{adj} B \operatorname{adj} A .]\)

Let \(A^{2}=I,\) and assume that \(A \neq I\) and \(A \neq-I\) a. Show that the only eigenvalues of \(A\) are \(\lambda=1\) and \(\lambda=-1\) b. Show that \(A\) is diagonalizable. [Hint: Verify that \(A(A+I)=A+I\) and \(A(A-I)=-(A-I)\), and then look at nonzero columns of \(A+I\) and of \(A-I .]\) c. If \(Q_{m}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) is reflection in the line \(y=m x\) where \(m \neq 0\), use (b) to show that the matrix of \(Q_{m}\) is diagonalizable for each \(m\). d. Now prove (c) geometrically using Theorem 3.3.3.

If \(A\) is diagonalizable, show that each of the following is also diagonalizable. a. \(A^{n}, n \geq 1\) b. \(k A, k\) any scalar. c. \(p(A), p(x)\) any polynomial (Theorem 3.3.1) d. \(U^{-1} A U\) for any invertible matrix \(U\). e. \(k I+A\) for any scalar \(k\).

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