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In Exercises \(27-30\), find all of the eigenvalues of the matrix A over the complex numbers \(\mathrm{C}\). Give bases for each of the corresponding eigenspaces. $$A=\left[\begin{array}{cc} 4 & 1-2 i \\ 1+2 i & 0 \end{array}\right]$$

Short Answer

Expert verified
The eigenvalues are \( \lambda_1 = 5 \) and \( \lambda_2 = -1 \). The corresponding eigenspaces have bases \( \{\begin{bmatrix} 2+2i \\ 1 \end{bmatrix}\} \) and \( \{\begin{bmatrix} -1+2i \\ 5 \end{bmatrix}\} \), respectively.

Step by step solution

01

Define the Matrix and Characteristic Equation

Consider the matrix \( A \), given by \[A = \begin{bmatrix} 4 & 1-2i \ 1+2i & 0 \end{bmatrix}.\]To find the eigenvalues, we need to calculate the determinant of \( A - \lambda I \), where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix of the same order as \( A \). The characteristic equation is given by\[\det(A - \lambda I) = 0.\]
02

Calculate A - λI

Subtract \( \lambda \) times the identity matrix from \( A \):\[A - \lambda I = \begin{bmatrix} 4 & 1-2i \ 1+2i & 0 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \ 0 & \lambda \end{bmatrix} = \begin{bmatrix} 4-\lambda & 1-2i \ 1+2i & -\lambda \end{bmatrix}.\]
03

Determine the Determinant

Find the determinant of \( A - \lambda I \):\[\det(A - \lambda I) = \det \begin{bmatrix} 4-\lambda & 1-2i \ 1+2i & -\lambda \end{bmatrix} = (4-\lambda)(-\lambda) - (1-2i)(1+2i).\]
04

Simplify the Determinant Expression

Calculate \((1-2i)(1+2i)\):\[(1-2i)(1+2i) = 1^2 - (2i)^2 = 1 - (-4) = 5.\]Substitute back:\[\det(A - \lambda I) = -\lambda(4-\lambda) - 5 = \lambda^2 - 4\lambda - 5.\]
05

Solve Characteristic Equation

Solve the characteristic equation:\[\lambda^2 - 4\lambda - 5 = 0.\]Factor the quadratic equation:\[(\lambda - 5)(\lambda + 1) = 0.\]Thus, the solutions are \( \lambda_1 = 5 \) and \( \lambda_2 = -1 \).
06

Find Eigenspaces

For each eigenvalue, solve \( (A - \lambda I)\mathbf{x} = 0 \) to find the corresponding eigenspace. **For \( \lambda_1 = 5 \):**\[A - 5I = \begin{bmatrix} -1 & 1-2i \ 1+2i & -5 \end{bmatrix}.\]Solve:\[\begin{bmatrix} -1 & 1-2i \ 1+2i & -5 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}.\]One basis vector is \( \begin{bmatrix} 2+2i \ 1 \end{bmatrix}. \)**For \( \lambda_2 = -1 \):**\[A + I = \begin{bmatrix} 5 & 1-2i \ 1+2i & 1 \end{bmatrix}.\]Solve:\[\begin{bmatrix} 5 & 1-2i \ 1+2i & 1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}.\]One basis vector is \( \begin{bmatrix} -(1-2i) \ 5 \end{bmatrix}. \) or simplified to \( \begin{bmatrix} -1+2i \ 5 \end{bmatrix}. \)

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Key Concepts

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

Complex Numbers
Complex numbers are an extension of the real numbers and are used to solve equations that have no real solutions. A complex number has the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit. The imaginary unit has the property \( i^2 = -1 \). Complex numbers can be represented as points in a two-dimensional plane, known as the complex plane.

In this exercise, complex numbers come into play during calculations involving the elements \(1 - 2i\) and \(1 + 2i\) of the matrix \(A\). When finding eigenvalues and eigenspaces, complex numbers allow us to handle the imaginary components that arise. For instance, multiplying two complex numbers like \((1 - 2i)(1 + 2i)\) simplifies to a real number \(5\) by using the property that \(i^2 = -1\), resulting in \(1 - (-4)\).

Understanding operations with complex numbers, such as addition, subtraction, multiplication, and division, is crucial since they frequently appear in mathematical problems concerning eigenvalues and eigenspaces.
Characteristic Equation
The characteristic equation is a fundamental concept when determining eigenvalues of a matrix. This equation is derived from the characteristic polynomial, which is obtained by finding the determinant of the matrix \(A - \lambda I\), where \(\lambda\) represents the potential eigenvalues and \(I\) is the identity matrix of the same size as \(A\).

The matrix \(A = \begin{bmatrix} 4 & 1-2i \ 1+2i & 0 \end{bmatrix}\) leads to the characteristic equation through the equation \(\det(A - \lambda I) = 0\). It is essential because solving this equation gives the eigenvalues. In our steps, the determinant simplifies to \(\lambda^2 - 4\lambda - 5\), providing a quadratic characteristic equation. Solving \(\lambda^2 - 4\lambda - 5 = 0\) yields the eigenvalues \(\lambda_1 = 5\) and \(\lambda_2 = -1\).

The characteristic equation highlights the intrinsic relationship between matrices and polynomials, illustrating how algebraic solutions facilitate the understanding of matrix properties.
Matrix Determinants
Matrix determinants play a crucial role when finding eigenvalues and are especially important in calculating the characteristic equation. A determinant is a special number that can be calculated from a square matrix. It provides useful information about the matrix, such as whether it is invertible and how transformations based on the matrix alter areas or volumes.

For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant is \(ad - bc\). In the context of our exercise, when finding \(\det(A - \lambda I)\), calculations involve expressions like \((4-\lambda)(-\lambda)\) and complex number multiplication such as \((1-2i)(1+2i)\), which simplifies to 5. Therefore, the determinant reduces to \(-\lambda (4-\lambda) - 5\), leading to the simplified polynomial \(\lambda^2 - 4\lambda - 5\).

Understanding how to compute determinants is vital because they lead to the characteristic polynomial, and ultimately, to the eigenvalues.

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Most popular questions from this chapter

A scientist places two strains of bacteria, \(X\) and \(Y\), in petri dish. Initially, there are 400 of \(X\) and 500 of \(Y\) The two bacteria compete for food and space but do not feed on each other. If \(x=x(t)\) and \(y=y(t)\) are the numbers of the strains at time \(t\) days, the growth rates of the two populations are given by the system \\[ \begin{array}{l} x^{\prime}=1.2 x-0.2 y \\ y^{\prime}=-0.2 x+1.5 y \end{array} \\] (a) Determine what happens to these two populations by solving the system of differential equations. (b) Explore the effect of changing the initial populations by letting \(x(0)=a\) and \(y(0)=b .\) Describe what happens to the populations in terms of \(a\) and \(b\)

Prove that if \(A\) is diagonalizable, so is \(A^{T}\)

If a Leslie matrix has a unique positive eigenvalue \(\lambda_{1}\) what is the significance for the population if \(\lambda_{1}>1 ?\) \(\lambda_{1}<1 ? \lambda_{1}=1 ?\)

Consider the dynamical system \(\mathbf{x}_{k+1}=A \mathbf{x}_{k}\) (a) Compute and plot \(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\) for \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\\ 1\end{array}\right]\) (b) Compute and plot \(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\) for \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\\ 0\end{array}\right]\) (c) Using eigenvalues and eigenvectors, classify the origin as an attractor, repeller, saddle point, or none of these. (d) Sketch several typical trajectories of the system. $$A=\left[\begin{array}{ll} 2 & 1 \\ 0 & 3 \end{array}\right]$$

Solve the recurrence relation with the given initial conditions. $$a_{0}=4, a_{1}=1, a_{n}=a_{n-1}-a_{n-2} / 4 \text { for } n \geq 2$$

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